Lesson 44 – Keep waiting: The memoryless property of exponential distribution

Bob Waits for the Bus

As the building entrance door closes behind, Bob glances at his post-it note. It has the directions and address of the car dealer. Bob is finally ready to buy his first (used) car. He walks to the nearby bus stop jubilantly thinking he will seldom use the bus again. Bob is tired of the waiting. Throughout these years the one thing he could establish is that the average wait time for his inbound 105 at the Cross St @ Main St is 15 minutes.

Bob may not care, but we know that his wait time follows an exponential distribution that has a probability density function  f(t) = \lambda e^{-\lambda t} .

The random variable T, the wait time between buses is an exponential distribution with parameter  \lambda . He waits 15 minutes on average. Some days he boards the bus earlier than 15 minutes, and some days he waits much longer.

Looking at the function  f(t) = \lambda e^{-\lambda t} , and the typical information we have for exponential distribution, i.e., the average wait time, it will be useful to relate the parameter  \lambda to the average wait time.

The average wait time is the average of the distribution — the expected value E[.].

E[X] for a continuous distribution, as you know from lesson 24 is  E[X] = \int x f(x) dx.

Applying this using the limits of the exponential distribution, we can derive the following.

 E[T] = \int_{0}^{\infty} t f(t) dt

 E[T] = \int_{0}^{\infty} t \lambda e^{-\lambda t} dt

 E[T] = \lambda \int_{0}^{\infty} t e^{-\lambda t} dt

The definite integral is  \frac{1}{\lambda^{2}} .

So we have

E[T] = \frac{1}{\lambda}

The parameter  \lambda is a non-negative real number ( \lambda > 0), and represents the reciprocal of the expected value of T.

In Bob’s case, since the average wait time (E[T]) is 15 minutes, the parameter \lambda is 0.066.

Bob gets to the bus shelter, greets the person next to him and thinks to himself “Hope the wait will not exceed 10 minutes today.”

Please tell him the probability he waits more than 10 minutes is 0.5134.

 P(T > 10) = e^{-\lambda t} = e^{-10/15} = 0.5134

Bob is visibly anxious. He turns his hand and looks at his wristwatch. “10 minutes. The wait won’t be much longer.”

Please tell him about the memoryless property of the exponential distribution. The probability that he waits for another ten minutes, given he already waited 10 minutes is also 0.5134.

Let’s see how. We will assume t represents the first ten minutes and s represents the second ten minutes.

 P(T > t + s \mid T > t) = \frac{P(T > t \cap T > t+s)}{P(T > t)}

\hspace{10cm} = \frac{P(T > t+s)}{P(T > t)}

\hspace{10cm} = \frac{e^{-\lambda (t+s)}}{e^{-\lambda t}}

\hspace{10cm} = \frac{e^{-\lambda t} e^{-\lambda s}}{e^{-\lambda t}}

\hspace{10cm} = e^{-\lambda s}

 P(T > 10 + 10 \mid T > 10) = e^{-10\lambda} = 0.5134

The probability distribution of the remaining time until the event occurs is always the same regardless of the time that passed.

There is no memory in the process. The history is not relevant. The time to next arrival is not influenced by when the last event or arrival occurred.

This property is unique to the strictly decreasing functions: exponential and the geometric distributions.

The probability that Bob has to wait another s minutes (t + s) given that he already waited t minutes is the same as the probability that Bob waited the first s minutes. It is independent of the current wait time.

Bob Gets His First Chevy

Bob arrives at the dealers. He loves the look of the red 1997 Chevy. He looks over the window pane; “Ah, manual shift!” That made up his mind. He knows what he is getting. The price was reasonable. A good running engine is all he needed to drive it away.

The manager was young, a Harvard alum, as Bob identified from things in the room. “There is no guarantee these days with academic inflation, … the young lad is running a family business, … or his passion is to sell cars,” he thought to himself.

The manager tells him that the engine is in perfect running condition and the average breakdown time is four years. Bob does some estimates ($$$$) in his mind while checking out the car. He is happy with what he is getting and closes the deal.

Please tell Bob that there is a 22% likelihood that his Chevy manual shift will break down in the first year.

The number of years this car will run ~ exponential
distribution with a rate (\lambda) of 1/4.

Since the average breakdown time (expected value E[T]) is four years, the parameter \lambda = 1/4.

 P(T \le 1) = 1 - e^{-\lambda t} = 1 - e^{-(1/4)} = 0.22

Bob should also know that there is a 37% chance that his car will still be running fine after four years.

P(T > 4) = e^{-4\lambda} = e^{-4/4} = 0.37

Bob in Four Years

Bob used the car for four years now with regular servicing, standard oil changes, and tire rotations. The engine is great.

Since the average lifetime has passed, should he think about a new car? How long should we expect his car to continue without a breakdown? Another four years?

Since he used it for four years, what is the probability that there will be no breakdown until the next four years?

You guessed it, 37%.

 P(T > 8 \mid T > 4) = \frac{P(T > 8)}{P(T > 4)} = e^{-4\lambda} = 0.37

Now let’s have a visual interpretation of this memoryless property.

The probability distribution of the wait time (engine breakdown) for  \lambda = 1/4 looks like this.

Let us assume another random variable  T_{2} = T - 4, as the breakdown time after four years of usage. The lower bound for  T_{2} is 0 (since we measure from four years), and the upper bound is \infty.

For any values  T > 4 , the distribution is another exponential function — it is shifted by four years.

 f(t_{2}) = \lambda e^{-\lambda t_{2}} = \lambda e^{-\lambda (t-4)}

Watch this animation, you will understand it better.

The original distribution is represented using the black line. The conditional distribution  P(T > 4+s \mid T > 4) is shown as a red line using links.

The same red line with links (truncated at 4) is shown as the shifted exponential distribution (f(t_{2})=\lambda e^{-\lambda (t-4)}). So, the red line with links from t = 4 is the same as the original function from t = 0. It is just shifted.

The average value of T is four years. The average value of T_{2} = T - 4 is also four. They have the same distribution.

If Bob reads our lessons, he’d understand that his Chevy will serve him, on the average, another four years.

Just like the car dealer’s four-year liberal arts degree from Harvard is forgotten, Bob’s four-year car usage history is forgotten — Memoryless.

As the saying goes, some memories are best forgotten, but the lessons from our classroom are never forgotten.

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Lesson 43 – Wait time: The language of exponential distribution

Wednesday, no, the Waiting Day

November 29, 2017

6:00 AM

As the cool river breeze kisses my face, I hear the pleasant sound of the waves. “How delightful,” I think, as I drop into the abyss of eternal happiness. The sound of the waves continues to haunt me. I run away from the river; the waves run with me. I close my ears; the waves are still here.

It’s the time when your dream dims into reality. Ah, it’s the sound of the “waves” on my iPhone. Deeply disappointed, I hit the snooze and wait for my dream to come back.

6:54 AM

“Not again,” I screamed. I have 30 minutes to get ready and going. I-95 is already bustling. I can’t afford to wait long in the toll lane. Doctor’s check-in at 8 AM.

7:55 AM

“Come on, let’s go.” For the 48th time, waiting in the toll lane, I curse myself for not having gotten the EZ pass that week. “Let’s go, let’s go.” I maneuver my way while being rude to the nasty guy who tried to sneak in front of my car. Finally, I pay cash at the toll and drive off in a swift to my doctor’s.

8:15 AM

“Hi, I have an appointment this morning. Hope I am not late.” The pretty lady at the desk stared at me, gave me a folder and asked me to wait. Dr. D will be with you shortly. As I was waiting for my turn, I realized that the lady’s stare was for my stupid question. My appointment was at 8 AM after all.

8:50 AM

The doctor steps in; “Please come in” he said. A visibly displeased me walked-in instantly, all the way shaking my head for the delay. My boss will be waiting for me at the office. We are launching a new product today.

9:15 AM

“You are in perfect health. The HDLs and LDLs are normal. Continue the healthy eating and exercise practices you have. See you next time, but don’t wait too long for the next visit.”

9:25 AM

My wait continues, this time for the train. “The next downtown 1-train will arrive in 10 minutes,” said the man (pre-recorded).

10:00 AM

My boss expressed his displeasure at my delay in his usual sarcastic ways. “But, I told you I was going to be late today,” I said to myself. We get busy with work and the product launch.

1:00 PM

I am waiting in the teller line at the local bank. Essential bank formalities and some checks to deposit. There were already ten people before me; there is only one teller, and for some reason, she is taking her own sweet time to serve each customer.

The only other living being in the bank (bank employees of course) is the manager; she is busy helping a person with his mortgage. “Poor guy seems to be buying a house at the peak,” I thought as I start counting the time it is talking to serve each customer.

1:35 PM

“One extra-hot Cappuccino,” said Joe at Starbucks in his usual stern voice. The wait for my coffee was not as annoying. There’s something about coffee and me. Can wait forever ! or maybe it is Starbucks; I can’t say.

7:00 PM

After a long tiring, waiting day, I am still waiting for my train.

I waited 22 minutes. The train surely has to come in the next minute,” I said to myself.

The clock ticks, my energy drops, still no trail.

.

.

.

“The next uptown 1-train is now arriving. Please stand away from the platform edge.”

I step in and grab the one remaining seat. “Finally; no more waiting for the day,” I said to myself.

The wheels rattle, the brains muffle, and the eyes scuttle. Same beautiful abyss of happy, restful state from the morning.

8:00 PM

As I park my car and check my door mail, I realize that my day was filled with wait times. I said to myself, “Aren’t these the examples of exponential distribution that data analysis guy from college used to talk about? I finally understand it. You live and learn.”

9:00 PM

I start logging my Wednesday, no, the waiting day.

“Let me derive the necessary functions for the exponential distribution before I go to bed,” I said to myself.

The time between arrivals at service facilitates, time to failure of systems, flood occurrence, etc., can be modeled as exponential distributions.

Since I want to measure the time between events, I should think of time T as a continuous random variable,  t_{1}, t_{2}, t_{3} , etc., like this.

That means, this distribution is positive only, as  T \ge 0 (non-negative real numbers).

We can have a small wait time or a long wait time. It varies, and we are estimating the probability that T is less than or greater than a particular time, and between two times.

The distribution of the probability of these wait times is called the exponential distribution.

As I watch the events and wait times figure carefully, I can sense that there is a relation between the Poisson distribution and the Exponential distribution.

The Poisson distribution represents the number of events in an interval of time, and the exponential distribution represents the time between these events.

If N is the number of events during an interval ( a span of time) with an average rate of occurrence \lambda,

P(N = k) = \frac{e^{-\lambda t}(\lambda t)^{k}}{k!}

If T is measured as the time to next occurrence or arrival, then it should follow an exponential distribution.

The time to arrival exceeds some value t, only if N = 0 within t, i.e., if there are no events in an interval [0, t].

 P(T > t) = P(N = 0) = \frac{e^{-\lambda t}(\lambda t)^{0}}{0!} = e^{-\lambda t}

If  P(T > t) = e^{-\lambda t} , then  P(T \le t) = 1 - e^{-\lambda t} .

I know that P(T \le t) = F(t) is the cumulative density function. It is the integral of the probability density function. F(t) = \int_{0}^{t}f(t)dt.

The probability density function f(t) can then be obtained by taking the derivative of F(t).

f(t) = \frac{d}{dt}F(t) = \frac{d}{dt}(1-e^{-\lambda t}) = \lambda e^{-\lambda t}

The random variable T, the wait time between successive events is an exponential distribution with parameter \lambda.

Let me map this on to the experiences I had today.

If on average, 25 vehicles pass the toll per hour, \lambda=25 per hour. Then the wait time distribution for the next vehicle at the toll should look like this.

The probability that I will wait more than 5 minutes to pass the toll is  P(T > 5) = e^{-\lambda t} = e^{-25*(5/60)} = 0.125.

So, the probability that my wait time will be less than 5 minutes is 0.875. Not bad. I should have known this before I swore at the guy who got in my way.

It is clear that the distribution will be flatter if \lambda is smaller and steeper if \lambda is larger.

10:00 PM

I lay in my bed with a feeling of accomplishment. My waiting day was eventful; I checked off all boxes on my to-do list. I now have a clear understanding of exponential distribution.

10:05 PM

I am hoping that I get the same beautiful dream. My mind is still on exponential distribution with one question.

“I waited 22 minutes for the train in the evening, why did it not arrive in the next few minutes? Since I waited a long time, shouldn’t the train arrive immediately?”

A tired body always beats the mind.

It was time for the last thought to dissolve into the darkness. The SHIREBOURN river is “waiting” for me on the other side of the darkness.

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Lesson 42 – Bounded: The language of Beta distribution

Last week, in Lesson 41, we started toying with the idea of continuous probability distributions. When a random variable X is continuous (i.e., can be any Real number), we can compute the probability of X between any two values,  P(X \in (a,b)) = P(a \le X < b) using a continuous probability distribution function  f(x) .

The continuous probability density function (pdf) is the limiting shape of the frequency plot (histogram) of the data as the number of possible observations (n) goes to infinity. While the probability that the random variable X takes any specific value x is 0, the height of the smooth curve measures how dense the probability is at that point.

In the limit, as the number of observations approaches infinity (continuous), the proportion of observations that belong to an interval (a, b) is the probability that X is in this interval;  P(X \in (a,b)) = P(a \le X < b) .

This area under the curve is computed using the integral of the function over the range a to b.

 P(a < X < b) = \int_{a}^{b} f(x) dx

I left you with a few practice questions:

If X is a random variable with a probability distribution function defined as

 f(x) = 90x^{8}(1-x) for 0 < x < 1

  1. What is the probability that X is between 0.2 and 0.3?
  2. What is the probability that X will exceed 0.9?
  3. What is the median of X?

If you have solved these questions, more power to you. If you are waiting for the stars to align, this is that auspicious moment!

Let’s solve this problem step by step to understand the nuts and bolts of continuous distributions. At the end of the problem, I will lead you to our first type of continuous probability distribution functions, the Beta distribution and its special case, the uniform distribution. You will see why I selected this problem as the primer.

Our function is  f(x) = 90x^{8}(1-x) for 0 < x < 1.

The plot of this function reveals a bell-like shape. Notice that x is between 0 and 1 and the function is continuous.

Let’s take the first question: What is the probability that X is between 0.2 and 0.3?

 P(0.2 < X < 0.3) = \int_{0.2}^{0.3} f(x) dx

 = \int_{0.2}^{0.3} 90x^{8}(1-x) dx

 = \int_{0.2}^{0.3}90x^8 dx - \int_{0.2}^{0.3}90x^{9}dx

 = \frac{90}{9} x^{9}\Big|_{0.2}^{0.3} - \frac{90}{10} x^{10}\Big|_{0.2}^{0.3}

 = \frac{90}{9}(0.3^{9}-0.2^{9}) - \frac{90}{10}(0.3^{10}-0.2^{10})

 = 0.00014

Using the same procedure, we can solve for the probability that X will exceed 0.9?

 P(X > 0.9) = \int_{0.9}^{1} f(x) dx

 = \frac{90}{9} x^{9}\Big|_{0.9}^{1} - \frac{90}{10} x^{10}\Big|_{0.9}^{1}

 = \frac{90}{9}(1^{9}-0.9^{9}) - \frac{90}{10} (1^{10}-0.9^{10})

 = 0.264

We could have integrated the function from 0 to 0.9 and then subtracted this number from 1 because  P(X > x) = 1 - P(X \le x) and  P(X \le x) = \int_{0}^{x}f(x)dx in this case.

Also, remember that  P(X \le x) = F(x), the cumulative distribution function. We will be using the cumulative distribution function very often from now on.

Now, let us look at the third question: what is the median of X?

We know from order statistics that median is the 50th percentile, i.e., the value for which 50% of the values of X are below this number.

 P(X \le x_{median}) = F(x_{median}) = 0.5

\int_{0}^{x_{median}}f(x)dx = 0.5

 \int_{0}^{x_{median}}90x^{8}(1-x)dx = 0.5

This reduces to  10x_{median}^{9} - 9x_{median}^{10} = 0.5

We can use the Newton Raphson iterative method to find that the root of this equation is 0.84 when 0 < x < 1.

Hence,  x_{median} = 0.84

Beta Distribution

Now, look at the function I gave you carefully.
 f(x) = 90x^{8}(1-x) for 0 < x < 1

It is bounded between 0 and 1.

It has some exponents for x and (1-x); 8 and 1 in this case.

It has a constant, 90, acting as a multiplier.

The function we solved is a Beta distribution. The standard form, i.e., the probability density function of a Beta distribution is

 f(x) = cx^{a-1}(1-x)^{b-1} for 0 < x < 1.

As you can see, it is defined only in the 0 to 1 range. The beta distribution is a bounded distribution. The function is 0 everywhere else.

a and b are the parameters that control the shape of the distribution. They can take any positive real numbers; a > 0 and b > 0. In our example, a = 9 and b = 2. The distorted bell shape we have for the function is because of these two values.

c is called the normalizing constant. It ensures that the pdf integrates to 1. Take, for example, our function  f(x) = 90x^{8}(1-x) .

If we integrate the function  x^{8}(1-x) between 0 to 1 (over the range of x), we will get

 \int_{0}^{1} x^{8}(1-x)= \frac{1}{90}

For the pdf f(x) to integrate to unity, we need to multiply it with a constant 90. Hence, we had 90 as the multiplier for our function.

This normalizing constant c is called the beta function and is defined as the area under the graph of  x^{a-1}(1-x)^{b-1} between 0 and 1.

 c = \frac{1}{\int_{0}^{1} x^{a-1}(1-x)^{b-1} dx}

For integer values of a and b, this constant c is defined using the generalized factorial function.

 c = \frac{(a + b - 1)!}{(a-1)!(b-1)!}

In our example, a = 9 and b =2. Applying these numbers will give

 c = \frac{(9+2-1)!}{(9-1)!(2-1)!} = \frac{10!}{8!1!} = 90.

Did you observe that we just need the values of a and b to get the Beta distribution?

We call it the beta family as the curve will have different shapes depending on the values of a and b.

Substitute a = 1 and b = 1 in the standard function and see what you get.

 c = \frac{(1 + 1 - 1)!}{(1-1)!(1-1)!} = 1

 f(x) = 1x^{1-1}(1-x)^{1-1} = 1

A constant value 1 for all x. This flat function is called the uniform distribution. It is a special case of the beta distribution when a and b are 1. It looks like a rectangle or a flat line.

Now substitute a = 2 and b = 2 and see.

a = 0.5 and b = 0.5 will be a u-shape with asymptotic ends.

I want you to experiment with different values of a and b and visualize how the shape changes, like in the opening animation. Try it this week. Don’t wait till the R lesson. You have come this far, and you are already a good coder in R.

As you see here, the beta distribution is flexible to take on different shapes.

The uniform distribution is used for simulating data from different probability distributions. Again, meditate on this idea before we see it in an R lesson.

The beta distribution is also used as a probability distribution for the probability p of an outcome. The probability of the probability 😉
In other words, if we want to estimate the probability p of an outcome, we assume prior to having any data, that p follows a beta distribution (0 < p < 1). Once we have the data, we can update this knowledge using the Bayes rule.

The beta distribution is also typically used in project management when we want to estimate the probability of completing the project ahead of schedule. The duration of each job is a random variable that can be approximated using a beta distribution as it is bounded between the worst completion time (pessimistic) and best completion time (optimistic).

Knowing this about project management, I set out to complete several pending tasks during this Thanksgiving break. My initial estimated probability of completion was 0.91. After a somewhat lazy turkey day, I now realize that my lower bound (best completion time) should have been my upper bound (worst completion time). The fix is in. The probability of completing the pending tasks’ project in the Christmas break is 0.91.

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Lesson 41 – Struck by a smooth function

Review lesson 32.

If you assume X is a random variable that represents the number of successes in a Bernoulli sequence of n trials, then this X follows a binomial distribution. The probability that this random variable X takes any value k, i.e., the probability of exactly k successes in n trials is:

Review lesson 33.

If we consider independent Bernoulli trials of 0s and 1s with some probability of occurrence p and assume X to be a random variable that measures the number of trials it takes to see the first success, then, X is said to be Geometrically distributed. The probability of first success in the kth trial is:

Review lesson 36.

The number of times an event occurs (counts) in an interval follows a Poisson distribution. The probability that X can take any particular value P(X = k) is:

The characteristic feature in all these distributions is that the random variable X is discrete. The possible outcomes are distinct numbers, which is why we called them discrete probability distributions.

Have you asked yourself, “what if the random variable X is continuous?” What is the probability that X can take any particular value x on the real number line which has infinite possibilities?

Let’s start with the most cliched thought experiment.

Yes, your guess is correct.

I am going to ask you to draw a ball at random from a box of ten balls. I am also going to ask you “what is the probability of selecting any particular ball?

Your answer will include, “not again,” and “since the balls are all identical, and there are ten in the box, the probability of selecting any particular ball is one-tenth (1/10); P(X = any ball out of ten balls) = 1/10.”

As I am about to ask my next question, you will interrupt me and give me the answer. “And if there are 20 balls, the probability will be 1/20.” You might also say, “spare your next question, because the answer is 1/100, and the visual for increasing number of balls looks like this.”

I am sure you have recognized the pattern here. As the sample size (n) becomes large, the probability of any one value approaches zero. For a continuous random variable, the number of possible outcomes is infinite, hence,

P(X = x) = 0.

For continuous random variables, the probability is defined in an interval between two values. It is computed using continuous probability distribution functions.

If you go back to lesson 15, you will recall how we made frequency plots. We partitioned the real number space into intervals or groups, recorded the number of observations (values) that fall into each group and used this grouping to build stacks.

Based on the number of observations in each interval, we can compute the probability that the random variable will occur in that intervals. For example, if there are ten observations out of 100 observations in a group, we estimate the probability that the variable occurs in this group as 10/100.

For continuous random variables, the proportion of observations in the group approaches the probability of being in the group, and the size of the group (interval range) approaches zero. For a large n, we can imagine a large number of very small intervals.

Is it too abstract?

If so, let’s take some data and observe this behavior.

We will use the same data that we used last week — daily temperature data for New York City. We have this data from 1869 to 2017, a large sample of 54227 values. We can assume that temperature data is a continuous random variable that has infinite possible values on the real number line.

I will take 500 data points at a time and place them on the number line. If there are two or more observations with the same temperature value, I will stack them. Recall that this is how we create histograms, the only difference is that I am not grouping. Each value is independent.

Observe this animation.

As the number of data points (sample size) increases, the stacks get denser and denser with overlaps. The final compact histogram can be approximated using a smooth function – a continuous probability distribution function.

Since for continuous random variables, the proportion of observations in the group approaches the probability of being in the group, the area of the interval block or the area under the curve of the smooth function is the probability that X is in that interval.

Finally, from calculus, you can see that the probability of a continuous variable in an interval a and b is:

An example area computation between -1 and 2 is shown.

The continuous probability distribution functions should obey the property of unit probability.

The limits of the integral are negative to positive infinity.

Now, we can integrate this function, f(x) up to any value x to get the cumulative distribution function (F(x)).

Since cumulative distribution function is the area under the curve up to a value of x, we are essentially computing .

Having this cumulative function is handy for computing the percentiles of the random variable.

Do you remember the concept of percentiles?

We learned in lesson 14 that percentiles are order statistics that can be used to summarize the data. A 75th percentile is that value of x which has 75% of the data less than this number. In other words, F(x) = 0.75.

Can you see how the cumulative distribution function, F(x) =  can be used to compute the percentiles?

Over the next few weeks, we will learn some special types of continuous distribution functions. Since you’ve been struck by smooth functions today, I will invite you to solve this.

If X is a random variable with a probability distribution function defined as

  • What is the median of X?
  • What is the probability that X is between 0.2 and 0.3?
  • What is the probability that X will exceed 0.9?

Post your answers in the comments section. The first correct answer will be highlighted next week.

No medal for solving it, but you don’t have to feel bad if you cannot. Many “modern engineering Ph.D. students” cannot solve this. Basic mathematics is no more a requirement in the new world with diverse backgrounds.

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Lesson 40 – Discrete distributions in R: Part II

The scarfs and gloves come out of the closet.

The neighborhood Starbucks coffee cups change red.

It’s a reminder that autumn is ending.

It’s a reminder that 4 pm is 8 pm.

It’s a reminder that winter is coming.

Today’s temperature in New York is below 30F – a cold November day.

  1. Do you want to know what the probability of a cold November day is?
  2. Do you want to know what the return period of such an event is?
  3. Do you want to know how many such events happened in the last five years?

Get yourself some warm tea. Let the room heater crackle. We’ll dive into rest of the discrete distributions in R.

Get the Data

The National Center for Environmental Information (NCEI) archives weather day for most of the United States. I requested temperature data for Central Park, NYC. Anyone can go online and submit requests for data. They will deliver it to your email in your preferred file format. I filtered the data for our lesson. You can get it from here.

Preliminary Analysis

You can use the following line to read the data file into R workspace.

# Read the temperature data file #
temperature_data = read.table("temperature_data_CentralPark.txt",header=T)

The data has six columns. The first three columns indicate the year, month and day of the record. The fourth, fifth and sixth columns provide the data for average daily temperature, maximum daily temperature, and minimum daily temperature. We will work with the average daily temperature data.

Next, I want to choose the coldest day in November for all the years in the record. For this, I will look through each year’s November data, identify the day with lowest average daily temperature and store it in a matrix. You can use the following lines to get this subset data.

# Years #
years = unique(temperature_data$Year) # Identifying unique years in the data
nyrs = length(years) # number of years of data

# November Coldest Day #
november_coldtemp = matrix(NA,nrow=nyrs,ncol=1)

for (i in 1:nyrs)
{
 days = which((temperature_data$Year==years[i]) & (temperature_data$Month==11)) # index to find november days in each year
 november_coldtemp[i,1] = min(temperature_data$TAVG[days]) # computing the minimum of these days
}

Notice how I am using the which command to find values.

When I plot the data, I notice that there is a long-term trend in the temperature data. In later lessons, we will learn about identifying trends and their causes. For now, let’s take recent data from 1982 – 2016 to avoid the issues that come with the trend.

# Plot the time series #
plot(years, november_coldtemp,type="o")
# There is trend in the data #

# Take a subset of data from recent years to avoid issues with trend (for now)-- # 
# 1982 - 2016
november_recent_coldtemp = november_coldtemp[114:148]
plot(1982:2016,november_recent_coldtemp,type="o")
Geometric Distribution

In lesson 33, we learned that the number of trials to the first success is Geometric distribution.

If we consider independent Bernoulli trials of 0s and 1s with some probability of occurrence p and assume X to be a random variable that measures the number of trials it takes to see the first success, then, X is said to be Geometrically distributed.

In our example, the independent Bernoulli trials are years. Each year can have a cold November day (if the lowest November temperature in that year is less than 30F) or not.

The probability of occurrence is the probability of experiencing a cold November day. A simple estimate of this probability can be obtained by counting the number of years that had a temperature < 30F and dividing this number by the total sample size. In our restricted example, we chose 35 years of data (1982 – 2016) in which we see ten years with lowest November temperature less than 30F. You can see them in the following table.

Success (Cold November) can happen in the first year, in which case X will be 1. We can see the success in the second year, in which case the sequence will be 01, and X will be 2 and so on.

In R, we can compute the probability P(X=1), P(X=2), etc., using the command dgeom. The Inputs are X and p. Try the following lines to create a visual of the distribution.

######################### GEOMETRIC DISTRIBUTION #########################

# The real data case # 
n = length(november_recent_coldtemp)

cold_years = which(november_recent_coldtemp <= 30)

ncold = length(cold_years)

p = ncold/n

x = 0:n
px = dgeom(x,p)
plot((x+1),px,type="h",xlab="Random Variable X (Cold on kth year)",ylab="Probability P(X=k)",font=2,font.lab=2)
abline(v = (1/p),col="red",lwd=2)
txt1 = paste("probability of cold year (p) = ",round(p,2),sep="")
txt2 = paste("Return period of cold years = E[X] = 1/p ~ 3.5 years",sep="")
text(20,0.2,txt1,col="red",cex=1)
text(20,0.1,txt2,col="red",cex=1)

Notice the geometric decay in the distribution. It can take X years to see the first success (or the next success from the current success). You must have seen that I have a thick red line at 3.5 years. This is the expected value of the geometric distribution. In lesson 34, we learned that the expected value of the geometric distribution is the return period of the event. On average, how many years does it take before we see the cold year again?

Did we answer the first two questions?

  1. Do you want to know what the probability of a cold November day is?
  2. Do you want to know what the return period of such an event is?

Suppose we want to compute the probability that the first success will occur within the next five years, we can use the command pgeom for this purpose.

pgeom computes P(X < 5) as P(X = 1) + P(X = 2) + P(X=3) + P(X = 4). Try it for yourself and verify that they both match.

Suppose the probability is higher or lower, how do you think the distribution will change?

For this, I created an animation of the geometric distribution with changing values of p. See how the distribution is wider for smaller values of p and steeper for larger values of p. A high value of p (probability of the cold November year) indicates that the event will occur more often; hence the trials to success are less in number. On the other hand, a smaller value for p suggests that the event will occur less frequently. The number of trials it takes to see the first/next success is more; creating a wider distribution.

Here is the code for creating the animation. We used similar code last week for animating the binomial distribution.

######## Animation (varying p) #########
# Create png files for Geometric distribution #

png(file="geometric%02d.png", width=600, height=300)

n = 35 # to mimic the sample size for november cold 
x = 0:n

p = 0.1
for (i in 1:5)
{
 px = dgeom(x,p)
 
 plot(x,px,type="h",xlab="Random Variable X (First Success on kth trial)",ylab="Probability P(X=k)",font=2,font.lab=2)
 txt = paste("p=",p,sep="")
 text(20,0.04,txt,col="red",cex=2)
 p = p+0.2
}
dev.off()

# Combine the png files saved in the folder into a GIF #

library(magick)

geometric_png1 <- image_read("geometric01.png","x150")
geometric_png2 <- image_read("geometric02.png","x150")
geometric_png3 <- image_read("geometric03.png","x150")
geometric_png4 <- image_read("geometric04.png","x150")
geometric_png5 <- image_read("geometric05.png","x150")

frames <- image_morph(c(geometric_png1, geometric_png2, geometric_png3, geometric_png4, geometric_png5), frames = 15)
animation <- image_animate(frames)

image_write(animation, "geometric.gif")

Negative Binomial Distribution

In lesson 35, we learned that the number of trials it takes to see the second success is Negative Binomial distribution. The number of trials it takes to see the third success is Negative Binomial distribution. More generally, the number of trials it takes to see the ‘r’th success is Negative Binomial distribution.

We can think of a similar situation where we ask the question, how many years does it take to see the third cold year from the current cold year. It can happen in year3, year 4, year 5, and so on, following a probability distribution.

You can set this up in R using the following lines of code.

################ Negative Binomial DISTRIBUTION #########################
require(combinat)

comb = function(n, x) {
 return(factorial(n) / (factorial(x) * factorial(n-x)))
}

# The real data case # 
n = length(november_recent_coldtemp)

cold_years = which(november_recent_coldtemp <= 30)

ncold = length(cold_years)

p = ncold/n

r = 3 # third cold year

x = r:n

px = NA

for (i in r:n)
{
 dum = comb((i-1),(r-1))*p^r*(1-p)^(i-r)
 px = c(px,dum)
}

px = px[2:length(px)]

plot(x,px,type="h",xlab="Random Variable X (Third Cold year on kth trial)",ylab="Probability P(X=k)",font=2,font.lab=2)

There is an inbuilt command in R for Negative Binomial distribution (dnbinom). I chose to write the function myself using the logic of the negative binomial distribution for a change.

The distribution has a mean of 10.5 years. The third cold year can occur approximately on the 10th year on average.

If you are comfortable so far, think about the following questions:

What happens to the distribution if you change r.

What is the probability that the third cold year will occur within seven years?

Poisson Distribution

Now let’s address the question: how many such events happened in the last five years?”

In lesson 36, Able and Mumble taught us about the Poisson distribution. We now know that counts, i.e., the number of times an event occurs in an interval follows a Poisson distribution. In our example, we are counting events that occur in time, and the interval is five years. Observe the data table and start counting how many events (red color rows) are there in each five-year span starting from 1982.

From 1982 – 1986, there is one event; 1987 – 1991, there are two events; 1992 – 1996, there is one event; 1997 – 2001, there is one event; 2002 – 2006, there are two events; 2007 – 2011, there is one event; 2012 – 2016, there are two events.

These counts (1, 2, 1, 1, 2, 1, 2) follow a Poisson distribution with an average rate of occurrence of 1.43 per five-years.

The probability that X can take any particular value P(X = k) can be computed using the dpois command in R.

Before we create the probability distribution, here are a few tricks to prepare the data.

Data Rearrangement

We have the data in a single vector. If we want to rearrange the data into a matrix form with seven columns of five years each, we can use the array command.

# rearrange the data into columns of 5 years #
data_rearrange = array(november_recent_coldtemp,c(5,7))

This rearrangement will help in computing the number of events for each column.

Counting the number of events

We can write a for loop to count the number of years with a temperature less than 30F for each column. But, R has a convenient function called apply that will perform this same analysis faster.

The apply command can be used to perform any function on the data row-wise, column-wise or both. The user can define the function.

For example, we can count the number of years with November temperature less than 30F for each column using the following one line code.

# count the number of years in each column with temp < 30
counts = apply(data_rearrange,2,function(x) length(which(x <= 30)))

The first argument is the data matrix; the second argument “2” indicates that the function has to be applied for the columns (1 for rows); the third argument is the definition of the function. In this case, we are counting the number of values with a temperature less than 30F. This one line code will count the number of events.

The rate of occurrence is the average of these numbers = 1.43 per five-year period.

We are now ready to plot the distribution for counts assuming they follow a Poisson distribution. Use the following line:

plot(0:5,dpois(0:5,mean(counts)),type="h",xlab="Random Variable X (Cold events in 5 years)",ylab="Probability P(X=k)",font=2,font.lab=2)

You can now tune the knobs and see what happens to the distribution. Remember that the tuning parameter for Poisson distribution is

I will leave you this week with these thoughts.

If we know the function f(x), we can find out the probability of any possible event from it. If the outcomes are discrete (as we see so far), the function is also discrete for every outcome.

What if the outcomes are continuous?

How does the probability distribution function look if the random variable is continuous where the possibilities are infinite?

Like the various types of discrete distributions, are there different types of continuous distributions?

I reminded you at the beginning that the autumn is ending. I am reminding you now that continuous distributions are coming.

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You can also follow me on Twitter @realDevineni for updates on new lessons.

Lesson 39 – Discrete distributions in R: Part I

It happened again. I got a ticket for parking close to the fire hydrant. Since my first hydrant violation ticket, I have been carrying a measuring tape. I may have miscalculated the curb length this time, or the enforcer’s tape is different than mine. Either way, today’s entry for the Department of Finance’s account is +$115, and my account is -$115.

We can’t win this game. Most hydrants in New York City don’t have painted curbs. It is up to the parking enforcer’s expert judgment — also called our fate. We park and hope that our number does not get picked in the lucky draw.

I want to research the fire hydrant violation tickets in my locality. Since we are learning discrete probability distributions, the violation tickets data can serve as a neat example. New York City Open Data has this information. I will use a subset of this data: parking violations on Broadway in precinct 24 in 2017.

I am also only analyzing those unfortunate souls whose vehicles are not registered in New York and who are parked at least seven feet from the hydrant. No excuse for those who park on the hydrant. Here is a look at the 27 instances under the given criteria.

Today’s lesson includes a journey through Bernoulli trials and Binomial distribution in R and how this example fits the description. Let’s start.

First Steps

Step 1: Get the data
You can download the filtered data file here.

Step 2: Create a new folder on your computer
Let us call this folder “lesson39”. Save the data file in this folder.

Step 3: Create a new code in R
Create a new code for this lesson. “File >> New >> R script”. Save the code in the same folder “lesson39” using the “save” button or by using “Ctrl+S”. Use .R as the extension — “lesson39_code.R”. Now your R code and the data file are in the same folder.

Step 4: Choose your working directory
“lesson39” is the folder where we stored the data file. Use “setwd(“path”)” to set the path to this folder. Execute the line by clicking the “Run” button on the top right.

setwd("path to your folder")

Step 5: Read the data into R workspace
I have the filtered data in the file named “parking_data.txt“. It only contains the date when the ticket is issued. Type the following line in your code and execute it. Use header=TRUE in the command.

# Read the data to the workspace #
parking_violations = read.table("parking_data.txt",header=T)
Bernoulli Trials

There are two possibilities each day, a ticket (event occurred – success) or no ticket (event did not occur – failure). A yes or a no. These events can be represented as a sequence of 0’s and 1’s (0001100101001000) called Bernoulli trials with a probability of occurrence of p. This probability is constant over all the trials, and the trials are assumed to be independent, i.e., the occurrence of one event does not influence the occurrence of the subsequent event.

In R, we can use the command “rbinom” to create as many outcomes as we require, assuming each trial is independent.

The input arguments are the number of observations we want, the trial (1 in the case of Bernoulli) and the probability p of observing 1s.

#### Bernoulli Trials ####

# The generalized Case #
p = 0.5 # probability of success -- user defined (using 0.5 here)
rbinom(1,1,p) # create 1 random Bernoulli trial
rbinom(10,1,p) # create 10 random Bernoulli trials
rbinom(100,1,p) # create 100 random Bernoulli trials

For this example, based on our data, there are 27 parking violation tickets issues in 181 days from January 1, 2017, to June 30, 2017.

Let us first create a binary coding 0 and 1 from the data. Use the following lines to convert the day into a 1 or a 0.

# Create a Date Series #
days = seq(from=as.Date('2017/1/1'), to=as.Date('2017/6/30'), by="day")

y = as.numeric(format(days,"%y"))
m = as.numeric(format(days,"%m"))
d = as.numeric(format(days,"%d"))

binary_code = matrix(0,nrow=length(m),ncol=1)

for (i in 1:nrow(parking_violations))
{
 dummy = which(m == parking_violations[i,1] & d == parking_violations[i,2])
 
 binary_code[dummy,1] = 1
}

plot(days,binary_code,type="h",font=2,font.lab=2,xlab="Days",ylab="(0,1)")

Assuming each day is a trial where the outcome can be getting a ticket or not getting a ticket, we can estimate the probability of getting a ticket on any day as 27/181 = 0.15.

So, with p = 0.15, we can simulate 181 outcomes (0 if the ticket is not issued or 1 if the ticket is issued) using the rbinom command. An example sequence:

# For the example case #
n = 1 # it is the number of trials 
p = 0.15 # probability of the event 
nobs = 181

rbinom(nobs,n,p)

0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0
0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0
1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 0 
0 0 0 1 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 0 
0 0 0 0 0 0

plot(rbinom(nobs,n,p),type="h",font=2,font.lab=2,xlab="Days",ylab="(0,1)")

If you simulate such sequences multiple times, on the average, you will see 27 ones.

Run this command multiple times and see how the plot changes. Each run is a new simulation of possible tickets during the 181 days. It occurs randomly, with a probability of 0.15.

Slightly advanced plotting using animation and GIF (for those more familiar with R)

In R, you can create animation and GIF based on these simulations. For example, I want to run the “rbinom” command five times and visualize the changing plots as an animation.

We first save the plots as “.png” files and then combine them into a GIF. You will need to install the “animation” and “magick” packages for this. Try the following lines to create a GIF for the changing Bernoulli plot.

######## Animation #########
# Create png files for Bernoulli sequence #
library(animation)

png(file="bernoulli%02d.png", width=600, height=300)
for (i in 1:5)
 {
 plot(rbinom(nobs,n,p),type="h",font=2,font.lab=2,xlab="Days",ylab="(0,1)")
 }
dev.off()

# Combine the png files saved in the folder into a GIF #

library(magick)

bernoulli_png1 <- image_read("bernoulli01.png","x150")
bernoulli_png2 <- image_read("bernoulli02.png","x150")
bernoulli_png3 <- image_read("bernoulli03.png","x150")
bernoulli_png4 <- image_read("bernoulli04.png","x150")
bernoulli_png5 <- image_read("bernoulli05.png","x150")

frames <- image_morph(c(bernoulli_png1, bernoulli_png2, bernoulli_png3, bernoulli_png4, bernoulli_png5), frames = 10)
animation <- image_animate(frames)

image_write(animation, "bernoulli.gif")

This is how the final GIF looks. Each frame is a simulation from the Bernoulli distribution. Depending on what version of R you are using, there may be more straightforward functions to create GIF.

Binomial Distribution

From the above, if you are interested in the number of successes (1s) in n trials, this number follows a Binomial distribution. If you assume X is a random variable that represents the number of successes in a Bernoulli sequence of n trials, then this X should follow a binomial distribution.

The number of trials is n = 181. The number of successes (getting a ticket) can be between 0 (if there are no tickets in all the 181 days) or 181
(if there is a ticket issued every day).

In R, the probability that this random variable X takes any value k (between 0 and 181), i.e., the probability of exactly k successes in n trials is computed using the command “dbinom.”

For computing the probability of exactly ten tickets in 181 days we can input:

px = dbinom(10,181,p)

For computing the probability of exactly 20 tickets in 181 days we can input:

px = dbinom(20,181,p)

To compute this probability for all possible k‘s and visualizing the probability distribution, we can use the following lines:

n = 181 # define the number of trials 
p = 0.15 # probability of the event 
x = 0:181 # number of successes varying from 0 to 181

px = dbinom(x,n,p)

plot(x,px,type="h",xlab="Random Variable X (Number of tickets in 181 days)",ylab="Probability P(X=k)",font=2,font.lab=2)

Do you know why the probability distribution is centered on 27? What is the expected value of a Binomial distribution?

If we want to compute the probability of getting more than five tickets in one month (30 days), we first calculate the probability for k = 6 to 30 (i.e., for exactly 6 tickets in 30 days to exactly 30 tickets in 30 days) with n = 30 to represent 30 trials.

n = 30 # define the number of trials 
p = 0.15 # probability of the event 
x = 6:30 # number of successes varying from 6 to 30

px = dbinom(x,n,p)
sum(px)

We add all these probability since
P(More than 5 in 30) = P(6 in 30) or P(7 in 30 ) or P(8 in 30) …

P(X > 5 in 30) = P(X = 6 in 30) + P(X=7 in 30) + ... + P(30 in 30).
Slightly advanced plotting using animation and GIF (for those more familiar with R)

I want to experiment with different values of p and check how the probability distribution changes.

I can use the animation and GIF trick to create this visualization.

Run the following lines in your code and see for yourself.

######## Animation #########
# Create png files for Binomial distribution #

png(file="binomial%02d.png", width=600, height=300)

n = 181
x = 0:181

p = 0.1
for (i in 1:5)
{
 px = dbinom(x,n,p)
 
 plot(x,px,type="h",xlab="Random Variable X (Number of tickets in 181 days)",ylab="Probability P(X=k)",font=2,font.lab=2)
 txt = paste("p=",p,sep="")
 text(150,0.04,txt,col="red",cex=2)
 p = p+0.2
}
dev.off()

# Combine the png files saved in the folder into a GIF #

library(magick)

binomial_png1 <- image_read("binomial01.png","x150")
binomial_png2 <- image_read("binomial02.png","x150")
binomial_png3 <- image_read("binomial03.png","x150")
binomial_png4 <- image_read("binomial04.png","x150")
binomial_png5 <- image_read("binomial05.png","x150")

frames <- image_morph(c(binomial_png1, binomial_png2, binomial_png3, binomial_png4, binomial_png5), frames = 15)
animation <- image_animate(frames)

image_write(animation, "binomial.gif")

You can also try changing the values for n and animate those plots.

We will return next week with more R games for the Geometric distribution, Negative Binomial distribution, and the Poisson distribution.

Meanwhile, did you know that RocketMan owes NYC $156,000 for unpaid parking tickets?

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Lesson 38 – Correct guesses: The language of Hypergeometric distribution

Now that John prefers Pepsi let’s put him on the spot and ask him to choose it correctly. We will tell him that there is one Pepsi drink. We’ll ask him to chose it correctly when given three drinks.

These are his possible outcomes. He will pick the first, second or third can as Pepsi. If he chooses the first can, his guess is correct. So the probability that John is correctly guessing the one Pepsi can is 1/3. The probability that his guess is wrong is 2/3.

Since he is only choosing one correct Pepsi can, his outcomes are 0 or 1 right guess.

P(X = 0) = 2/3
P(X = 1) = 1/3

Let’s say that he liked Pepsi so much that he guessed it correctly in our first test.

We will put him through another test. This time, we will ask him to choose Pepsi correctly, when there are two cans in three.

He is to choose two cans, 1-2, 1-3, or 2-3. His guess outcomes are selecting a combination of two cans that will yield one Pepsi or both Pepsis. The only way to correctly guess both Pepsis is to choose 1-2. The probability is 1/3 (one option among three combinations).

There are two other ways (1-3, 2-3) for him to choose one Pepsi correctly. So the probability is 2/3.

P(X = 1) = 2/3
P(X = 2) = 1/3

John is good at this. He correctly guessed both the Pepsi cans. Let’s troll him by saying he got lucky so that he will take on the next test.

This time, he has to choose Pepsi correctly where there are two cans in four.

His picks can be 1-2, 1-3, 1-4, 2-3, 2-4, and 3-4. Six possibilities, since he is choosing two from four – 4C2. Based on his pick, he can get zero Pepsi, one Pepsi or both Pepsis. X = 0, 1, 2.

There is one possibility of both Pepsis; the combination 1-2. The probability that he chooses this combination among all others is 1/6.

P(X = 2) = 1/6

Similarly, there is one possibility of no Pepsi, the combination 3-4. The probability that he chooses these cans among all his options is 1/6.

P(X = 0) = 1/6

The other guess is choosing one Pepsi correctly. There are two Pepsi cans; he can pick either the first can or the second can as his choice, and then his second can will be the third or the fourth can.

1 with 3
1 with 4

2 with 3
2 with 4

There are 2C1 ways of choosing one Pepsi can among the two Pepsi cans. There are 2C1 ways of picking one Coke can from the two Coke cans.
The total possibilities are 2C1*2C1.

P(X = 1) = 4/6

Fortune favors the brave. John again correctly picked 1-2. Ask him to take one more test, and he will ask us to get out.

If you are like me, you want to play out one more scenario to understand the patterns and where we are going with this. So, let’s leave John alone and do this thought experiment ourselves. I will pretend you are John.

I will ask you to correctly pick Pepsis when there are three Pepsi cans among five.

When you pick three cans out of five, only one of them can be Pepsi, two of them can be Pepsi, or all three of them can be Pepsi. So X = 1, 2, 3.
You have a total of 10 combinations; pick three correctly from five. 5C3 ways as shown in this table.

The probability of picking all three Pepsis is 1/10.

P(X = 3) = 1/10

Now, let’s work out how many options we have for picking exactly one Pepsi in three cans. It has to be one Pepsi (one correct guess) with two Cokes (two wrong guesses). One Pepsi from three Pepsi cans can be selected in 3C1 ways. Two Cokes from two Cokes can be selected in 2C2 ways making it a total of 3C1*2C2 = 3 options. We can identify these options from our table as 1-4-5, 2-4-5 and 3-4-5. So,

P(X = 1) = 3/10

By the same logic, picking exactly two Pepsis in three cans can be done in 3C2*2C1 ways. Two Pepsi cans selected from three and one Coke chosen from two Coke cans. Six ways. Can you identify them in the table?

P(X = 2) = 6/10

Let us generalize.

If there are R Pepsi cans in a total of N cans (N-R Cokes) and we are asked to identify them correctly, in our choice selection of R Pepsis, we can get k = 0, 1, 2, … R Pepsis. The probability of correctly selecting k Pepsis is

X, the number of correct guesses (0, 1, 2, …, R) assumes a hypergeometric distribution.

The control parameters of the hypergeometric distribution are N and R.

The probability distribution for N = 5, and R = 2 is given.

When N =5 and R = 3.

In more generalized terms, if there are R Pepsi cans in a total of N cans (N-R Cokes) and we randomly select n cans from this lot of N and define X to be the number of Pepsis in our sample of n, then the distribution of X is hypergeometric distribution. P(X=k) in this sample is then:

The denominator NCn is the number of ways you can select n cans out of a total of N. A sample of n from N. The first term in the numerator is selecting k (correct guesses) out of R Pepsis. The second term is selecting (n-k) (wrong guesses) from the remaining (N-R) Cokes.

I want you to start visualizing how the probability distribution changes for different values of N, R, and n.

Next week, we will learn how to work with all discrete distributions in our computer programming tool R.

Hypergeometric distribution is typically used in quality control analysis for estimating the probability of defective items out of a selected lot.

The Pepsi-Coke marketing analysis is another example application. Companies can analyze the preferences of one product to other among a subset of customers in their region.

Now think of this:

There are R Republican leaning voters in a population of N. For simplicity (and for all practical purposes since LP and GP will not win); the other N-R voters are leaning Democratic. If you select a random sample of n voters, what is the probability that you will have more than k of them voting Republican? You know that the control parameters for this “election forecast” model are N, R, and n. What if you underestimated the number of R leaning voters? What if your sample of n voters is not random?

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Lesson 37 – Still counting: Poisson distribution

The conference table was arranged neatly with a notebook and pen at each chair. Mumble’s Macbook Air is hooked up to the projector at the other end.

He peeps over the misty window. A hazy and rainy outside did not elevate his senses. He will meet Lani from the risk team of California builders. Able invited him to New York to discuss a potential deal on earthquake insurance products.

Mumble goes over the talking points in his mind. He is to discuss the technical details with Able and Lani in the morning. Later, he will present his analysis of product lines and market studies to the executive officers.

Still facing the window, Mumble takes a deep breath, feels confident about his opening jokes, and turns around to greet Able and Lani who just entered the conference room.

Lani initiates the conversation and talks about teamwork and success. Mumble was not impressed. He has heard this teamwork mantra many a time now. Lani came across as an all talk no action guy.

Lani nods and picks up the pen to take some notes.

Able’s eyes rolled over towards Lani. He expected him to interject. Afterall, Lani’s bio says that he is a mathematician.

Lani did not interrupt. He was busy writing down points on his notepad.

Able was again expecting Lani to latch onto the equations.

“This is very good. Spending time on these details is essential. I am looking forward to our successful collaboration Mumble. I love Math too.”

Able and Mumble glanced at each other and politely concealed their emotions.

“Definitely. California Builders are at the forefront in Earthquake risk for housing projects. We can collect data for tremors in California. We work on several models for the benefit of our clients. Our team consists of many mathematicians and engineers. Mumble, I agree with Mr. Able. You did a fantastic job. We can work together on this project to produce risk models.”

Mumble has now confirmed that Lani is all talk no substance. He is just looking to delegate work and take credit.

Able has little patience for these platitudes. He is beginning to realize that Lani is a paper mathematician who may have taken one extra Math course in college and just calls himself a “mathematician” since it sounds “Einsteinian.” The deal with California Builders just became “no deal.”

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Lesson 36 – Counts: The language of Poisson distribution

I want you to meet my friends, Mr. Able and Mr. Mumble.

Unlike me who talks about risk and risk management for a living, they take and manage risk and make a killing.

True to his name, Mr. Mumble is soft-spoken and humble. He is the go-to person for numbers, data, and models. Like many of his contemporaries, he checked off the bucket list given to him and stepped on the ladder to success.

Some of you may be familiar with this bucket list.

Mumble was a “High School STEM intern,” was part of a “High School to College Bridge Program for STEM,” has a “STEM degree,” and completed his “STEM CAREER Development Program.” Heck, he even was called a “STEM Coder” during his brief stint with a computer programming learning center. A few more “STEMs,” and he’d be ready for stem cell research.

These “STEM initiatives” have landed him a risk officer job for an assets insurance company in the financial district.

On the contrary, Mr. Able is your quintessential American pride who has a standard high school education, was able to pay for his college through odd jobs, worked for a real estate company for some time, learned the trade and branched off to start his own business that insures properties against catastrophic risk.

He is very astute and understands the technical aspects involved in his business. Let’s say you cannot throw in some lingo into your presentation and get away without his questions. He is not your “I don’t get the equations stuff” person. He is the BOSS.

Last week, while you were learning the negative binomial distribution, Able and Mumble were planning a new hurricane insurance product. Their company would sell insurance against hurricane damages. Property owners will pay an annual premium to collect payouts in case of damages.

As you know, the planning phase involves discussion about available data and hurricane and damage probabilities. The meeting is in their 61st-floor conference room that oversees the Brooklyn Bridge.

Mumble, is there an update on the hurricane data? Do you have any thoughts on how we can compute the probabilities of a certain number of hurricanes per year?

 

Mr. Able, the National Oceanic and Atmospheric Administration’s (NOAA) National Hurricane Center archives the data on hurricanes and tropical storms. I could find historical information on each storm, their track history, meteorological statistics like wind speeds, pressures, etc. They also have information on the casualties and damages.

That is excellent. A good starting point. Have you crunched the numbers yet? There must be a lot of these hurricanes this year. I keep hearing they are unprecedented.

 

Counting Ophelia, we had ten hurricanes this year. Take a look at this table from their website. I am counting hurricanes of all categories. I recall from our last meeting that Hurin will cover all categories. By the way, I never liked the name Hurin for hurricane insurance. It sounds like aspirin.

Don’t worry about the name. Our marketing team has it covered. Funny name branding has its influence. You will learn when you rotate through the sales and marketing team. Tell me about the counts for 2016, 2015, etc. Did you count the number of hurricanes for all the previous years?

 

Yes. Here are the table and a plot showing the counts for each year from 1996 to 2017.

 

Based on this 22-year data, we see that the lowest number per year is two hurricanes and the highest number is 15 hurricanes. When we are designing the payout structure, we should have this in mind. Our claim applications will be a function of the number of hurricanes. Can we compute the probability of having more than 15 hurricanes in a year using some distribution?

Absolutely. If we assume hurricane events are independent (the occurrence of one event does not affect the probability that a second event will occur), then the counts per year can be assumed a random variable that follows a probability distribution. Counts, i.e., the number of times an event occurs in an interval follows a Poisson distributionIn our case, we are counting events that occur in time, and the interval is one year.

Let’s say the random variable is X and it can be any value zero hurricanes, one hurricane, two hurricanes, ….. What is the probability that X can take any particular value P(X = k)? What are the control parameters?

 

Poisson distribution has one control parameter 

It is the rate of occurrence; the average number of hurricanes per year. Based on our data, lambda is 7.18 hurricanes per year. The probability P(X = k) for a unit time interval t is

The expected value and the variance of this distribution are both

We can compute the probability of having more than 15 hurricanes in a year by adding P(X = 16) + P(X = 17) + P(X = 18) and so on. Since 15 happened to be in the extreme, the probability will be small, but our risk planning should include it. Extreme events will create a catastrophic damage. I see you have more slides on your deck. Do you also have the probability distribution plotted?

 

Yes, I have them. I computed the P(X = k) for k = 0, 1, 2, …, 20 and plotted the distribution. It looks like this for = 7.18.

Let me show you one more probability distribution. This one is for storms originating in the western Pacific. They are reported to the Joint Typhoon Warning Center. Since we also insure assets in Asia, this data and the probability estimates will be useful to design premiums and payouts there. The rate of events is higher in Asia; an average of 14.95 typhoons per year. The maximum number of typhoons is 21.

 

Very impressive Mumble. You have the foresight to consider different scenarios.

 

As the meeting comes to closure, Mr. Able is busy checking his emails on the phone. A visibly jubilant Mumble sits in his chair and collects the papers from the table. He is happy for having completed a meeting with Mr. Able without many questions. He is already thinking of his evening drink.

The next meeting is in one week. Just as Mr. Able gets up to leave the conference room, he pauses and looks at Mumble.

“Why is it called Poisson distribution? How is this probability distribution different from the Binomial distribution? Didn’t you say in a previous meeting that exactly one landfall in the next four hurricanes is binomial?”

Mumble gets cold feet. His mind already switched over to the drinks after the last slide; he couldn’t come up with a quick answer. As he begins to mumble, Able gets sidetracked with a phone call. “See you next week Mumble.” He leaves the room.

Mumble gets up and watches over the window — bright sunny afternoon. He refills his coffee mug, takes a sip and reflects on the meeting and the question.

To be continued…

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Lesson 35 – Trials to ‘r’th success: The language of Negative Binomial distribution

We all know that the trials to the first success is a Geometric distribution. It can take one trial, two trials, three trials, etc., to see the first success. These trials are assumed a random variable X = {1, 2, 3, … }; they have a probability, i.e., P(X = 1), P(X = 2), P(X = 3), and so on.

However, Mr. Gardner needs more success. He has already sold his first “time machine” (aka bone density scanner). He’d have to sell more. He is looking for his second success, third success and so forth to get through.

The number of trials it takes to see the second success is Negative Binomial distribution.

The number of trials it takes to see the third success is Negative Binomial distribution.

The number of trials it takes to see the ‘r’th success is Negative Binomial distribution.

Are you paying attention to the pattern here? Negative Binomial distribution is Geometric distribution if r is 1 (trials to first success).

The Geometric distribution has one control parameter, p, the probability of success.

Since we are interested in more than the first success, r is another parameter in the Negative Binomial distribution. Together, p and r determine how the distribution looks.

Let’s take the example of Mr. Gardner. “I’d have to sell one more.” Each of his visit to a doctor’s office is a trial. They will buy the bone density scanner or show him the door. So Mr. Gardner is working with a probability of success, p. Let’s say that p is 0.25, i.e., there is a 25% chance that he will succeed in selling it.

Let’s assume that he sold one machine. He is looking for his second sale. r = 2.

His second success can occur in the second trial, the third trial, the fourth trial, etc. When r = 2, X, the random variable of the number of trials will be X = {2, 3, 4, …}.

When r = 3, X = {3, 4, 5, …}.

You correctly guessed my next line. When r = 1, i.e., for Geometric distribution, X = {1, 2, 3, …}.

There is a pattern. We are learning a distribution which is an extension of Geometric distribution.

Now let us compute the probability that X can take any integer value.

P(X = 2) is the probability that he makes his second sale (second success) on the second trial.

Remember the trials are independent. The second doctor’s decision is not dependent on what happened before. He buys or not with a 0.25 probability.

So, P(X = 2) is 0.25*0.25 = 0.0625. The first trial is a success and the second trial is a success.

P(X = 3) is the probability that he makes his second sale on the third trial. It means he must have made his first sale in either the first trial or the second trial, and then he makes his second sale on the third trial.

The probability of succeeding second time on the third trial is the probability of succeeding once in two trials, and the third trial is a success.

P(1 success in 2 trials) * P(3rd is a success)

The probability of the one success in two trials is computed using the Binomial distribution.

2C1*p^1*(1-p)^2-1

This probability is multiplied by p, the probability of success in the third trial.

P(X = 3) = 2C1*p^1*(1-p)^2-1*p

If this is your expression now, 😕 let’s take another case to clear up the concept.

Suppose we want P(X = 6), the probability of making the second sale on the sixth trial.

This will happen in the following way.

Mr. Gardner has to sell one machine in five trials. P(1 in 5), one success in five trials → Binomial.

5C1*p^1(1-p)^(5-1)

Then, the sixth trial is a sell. So we multiply the above binomial probability with p, the hit in the sixth.

You should have noticed the origin of the name → Negative Binomial.

To generalize,

I computed these probabilities for X ranging from 2 to 50. Here is the probability distribution. Remember r = 2 and p = 0.25.

Now I change the value for r to 3 and 5 to see how the Negative Binomial distribution looks. r = 3 means Mr. Gardner will sell his third machine.

r = 3

r = 5

Notice how the probability distribution shifts with increasing values of r.

The control parameters are r and p.

You can try different values of p and see what happens. For a fixed value of p and changing values of r, the tail is getting bigger and bigger. What does it mean regarding the number trials and their probability for Mr. Gardner?

Think about this. If he sets a target of selling three machines in the day, what is the probability that he can achieve his goal within 20 doctor visits?

What if he needs to sell five machines within this 20 visits to make his ends meet. Can he make it?

If you try out the probability distribution plots for p = 0.5, you will see that his chance of selling the fifth machine within 20 visits goes up tremendously. So perhaps he should learn the six principles of influence and persuasion to increase p, the probability of saying yes by the doctors.

People mostly prefer to say yes to the request of someone they know and like. I want our blog to have the 9000th user within nine months. See, I am requesting in the language of Negative Binomial distribution.

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