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 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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Lesson 34 – I’ll be back: The language of Return Period

The average time between Arnold Schwarzenegger’s being back is the return period of his stunt.

I have to wait 5 minutes for my next bus. Some days I wait for 1 minute; some days, I wait for 15 minutes. The wait time is variable → random variable 😉 The average of these wait times is the return period of my bus.

Your recent vocabulary may include “100-year event” (happening more often), (drainage system designed for) “10-year storm,” and so on, courtesy mainstream media and news outlets.

Houston drainage grid ‘so obsolete it’s just unbelievable’

What exactly is this return period business?

Does a 10-year return period event occur diligently every ten years?

Can a 100-year event occur three times in a row?

THE LOGIC

Let’s visit annual maximum rainfall for Houston. If we take the daily rainfall data for each year from January 1 to December 31 and choose the maximum rainfall among these days, we call it annual maximum rainfall for that year. So this is the rainfall for the wettest day of the year. Likewise, if we do this for all the years that we have data for, we get a data series (also called time series since we are recording this in time units).

You can get the data from here if you like. You may have to register using your email, but its free. We have 79 years of recorded data from 1939 to 2017. 79 data points, one number per year as the rainfall for the wettest day in that year. You will see that five years are missing between 1942 and 1946.

I want you to understand that these numbers represent a random variable X. Each number (outcome) is assumed to be independent, i.e., the occurrence of one event in one year does not influence the occurrence of the subsequent event. In other words, 2017 rainfall does not depend on 2016 rainfall.

Now, I want you to see Brays Bayou, the lake that detains excess rainfall in Houston. Let us assume that it can store up to eight inches of rainfall on any day. If it rains more than eight inches in a day, the Bayou will overflow and cause flood — as we saw in Houston during hurricane Harvey.

So, if the rainfall is greater than eight inches, we define this as an event. Let us call him Bob. The first time we see Bob was in 1949. We started recording data in 1939. Bob happened after 11 years. The wait time for Bob (1949) is 11 years.

Then we get on with our lives, 11 years passed, Bob is not back, 22 years passed, no sign of Bob. Suddenly, after 30 years of waiting from 1949, Bob Strikes Back (1979).

Two years after this event happened, Bob wanted to greet the Millenials, so he came back in 1981. This time, the waiting period is only two years.

Then, in 1989, for no particular reason, Bob returns. The return of Bob (1989) is after eight years.

You must be thinking: “I don’t see any pattern here.” Yes, that is because there is none.

Years pass, Bob seems to be resting. At the turn of the century, Bob decided to come back. So Bob Meets the 21st Century in 2001 after 12 years since his prior occurrence. Bob re-occurs. Recurrence.

During the first decade of the 21st century, Bob re-occurs two times, once in 2006 as the Restless Bob (5-year wait time) and again in 2008 as Miss Me Yet, Bob (2-year wait time).

We all know what happened after that. Vengeant Bob (2017), aka Harvey, happened after nine years.

Now, let’s summarize all Bobs along with their recurrence times. We started with the assumption that the maximum rainfall events represent a random variable X. Let us define T as another random variable that measures the time between the event Bob (wait time or time to the next event or time to the first event since the previous event).

The return period of the event Bob, (X > 8 inches) is the expected value of T, i.e., E[T], its average measured over a large number of such occurrences.

As you can see here, in the table, the return period of Bob is approximately ten years. Bob is a 10-year return period event.

Another way of thinking about this: Since there are eight Bob events in 79 years, they occur at an average rate of 79/8. Approximately, once in 10 years. Hence originated the 10-year event concept.

Remember, they don’t happen cyclically every ten years. If we average the wait times of a lot of events, we will get approximately ten years.

Just like when you wait for the bus, you wait for short time or a long time, but you think of the average time you wait for a bus everyday, you can see events happening in a cluster or spaced out, but all average to an nyear return period.

The relation to Geometric distribution

Last week when we learned Geometric distribution, I told you that we would relate the expected value of the Geometric distribution to return period of an event. Let’s see how Bob relates to Geometric distribution.

I want you to convert the maximum rainfall data series into a series of independent Bernoulli trials of 0s and 1s. 0 if the rainfall is < eight inches (No Bob), 1 if the rainfall is > eight inches (Yes Bob). The 1s can occur with some probability of occurrence p. In our example, since we have 8 Bobs in 79 years the probability of occurrence p = 8/79 = 0.101.

Now, assume T to be a random variable that measures the number of trials (years) it takes to see the first success (event), or the next event from each such event. For the first event, Bob (1949), it took 11 years to occur. The probability that T = 11, P(T = 11) is (1 – p)^10*p. Similarly, the next Bob happened after 30 years and so on. T is the time to first success (next success) → Geometrically distributed.

We can derive the expected value of T using the expectation operation we learned in lesson 24.

Now, recall from your math classes that the expression inside the parenthesis looks like a power series. Ponder over it and confirm that the whole expression will reduce to

E[T] = 1/p

The expected value of the wait time that is Geometrically distributed is the inverse of the probability of the event. Since the probability of Bob is 0.101, the return period (expected value of the wait times) is 1/0.101 ~ ten years. A 10-year return period event.

The Question

We measured the probability over 79 years; n = 79. We assumed that the probability is constant over all the trials.

In other words, we are assuming that we know p and it does not change.

If I were writing this lesson last year, the probability would have been 7/78 = 0.089. Since Harvey (The Vengeant Bob), the probability became 8/79 = 0.101. There are also five missing years.

Perhaps we do not know the true value of p, and perhaps it is not constant.

How then, can you estimate the risk of anything? How then, can you predict anything? How then, can you design anything? 

If I haven’t confused you enough, let me end with one of my favorite quotes from Nicholas Taleb’s book Antifragile: Things that gain from disorder.

“It is hard to explain to naive data-driven people that risk is in the future, not in the past.”

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Lesson 33 – Trials to first success: The language of Geometric distribution

So goes the legendary story: Bruce, Robert I, the King of Scotland, defeated the English armies on his seventh trial. He bore six successive defeats before that.

 

Giants are yet to win their first game of the season. Their record: L, L, … I wonder how many games until their first win.

 

The last deadly hurricane that hit New York City is Sandy in 2012. We are now five years through without such a dangerous event.

 

The common thread tying the three examples is the number of trials to the first success. This is the language of the 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.

We can get the success on the first trial, in which case X will be 1. We can see the success on the second trial, in which case the sequence will be 01, and X will be 2. We can see the success on the third trial; the sequence will be 001 and X will be 3 and so forth. As you can guess, X = {1, 2, 3, … }, positive integers.

There is some probability that X can take any integer value. We should also figure out this probability, i.e., P(X = 1), P(X = 2), P(X = 3), and so on.

Let us take the example of a coin toss. The outcomes are head or tail. Binary outcomes → Bernoulli trial. The probability p of a head or tail is 0.5. In other words, if you toss a coin a large number of times, say 100, roughly 50 of them will be heads, and 50 of them will be tails. Let’s play. Heads you win, tails you lose.

 

Great, you win on the first trial. The probability of seeing head is 0.5. Hence, P(X = 1) = 0.5.

 

Let’s play again.

 

Ah, this time the first outcome is a tail and the second outcome is a head. You lose on the first trial but win on the second. It took two trials to wins. X = 2. P(X=2) is P(tail on the first toss)*P(head on the second toss) = 0.5*0.5 = 0.25. Why did we multiply? What is P(A and B) for independent events?

One more time.

  

Now it took three trials to win. X = 3, and P(X = 3) = P(tail on the first toss)*P(tail on the second toss)*P(head on the third toss) = 0.5*0.5*0.5 = 0.125.

For X = 4, it will be P(tail on the first toss)*P(tail on the second toss)*P(tail on the third toss)*P(head on the fourth toss) = 0.5*0.5*0.5*0.5 = 0.0625.

If we now plot X and P(X = k), k being 1, 2, 3, 4, …, we get a probability distribution like this.

The height of the line at X = 2 is 0.5 times the height of the line at X = 1. In the same way, the height of the line at X = 3 is 0.5 times the height of the line at X = 2 and so on. P(X=k) decreases in a geometric progression. Hence the name Geometric distribution.

We can generalize this for any probability p. In our game, we estimated P(X = 1) as 0.5, i.e., the probability of seeing a head p. P(X=2) is 0.5*0.5, i.e., (1-p)*p. P(X = 10) = (1-p)^9*p. First success on the tenth toss is nine tails followed by a head.

More generally,

We can derive the expected value and variance of X as: 

The expected value of a Geometric distribution relates to a special concept called return period → we will look at it next week.

Meanwhile, here are some more geometric probability distributions with different values of p.

p = 0.1

p = 0.3

p = 0.5

p = 0.7

p = 0.9

Notice how the shape changes with changing values of p. p is the parameter that controls the shape of the distribution. The greater the value for p, the steeper the fall.

If the probability of success is close to 1, the odds of winning in the first few trials is high → notice the height of the line for p = 0.9. If the probability of success is close to 0, it takes several trials to get to greater odds of winning overall.

Have you now conceptualized the idea of geometric distribution?

Let me challenge you to a bet then.

I have a coin toss game where I give you two times your bet if you win; you get nothing if you lose. Assume we have a fair coin, would you play the game with me and bet your money? If you will, then what is your strategy, assuming you are in it to win.

Since I challenged you to a bet, I also looked into some lottery games myself at nylottery.ny.gov.

First observation

The odds of winning first prize in any of the games is next to 0. So if you plan to keep buying the tickets until you win the first time, and then retire, you now know that you will keep buying forever.

The chances of winning per game get better for lower prize levels. For example, in the Mega Million, if you want to win the ninth prize, the odds are 1 in 21. Still low, and will take a long time to win.

But, who wants to win the ninth prize. It is like saying “America Ninth.”

Second observation

I keep wondering why on earth is New York Government running a lottery business … only to reconcile that “bread and circuses” have always been up the state’s sleeves to expand.

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Lesson 32 – Exactly k successes: The language of Binomial distribution

 

I may be in one of those transit buses. Since I moved to New Jersey, I am going through this mess every day.

 

 

Well, you wanted to enjoy Manhattan skyline. It has a price tag.

 

 

D, glad you are here. It’s been a while. In our last meeting, we were discussing the concepts of variance operation and its properties. I continue to read your lessons every week. As I paused and reflected on all the lessons, I noticed that there is a systematic approach to them. You started with the basics of sets and probability, introduced lessons on visualizing, summarizing and comparing data using various statistics, then extended those ideas into random variables and probability distributions. The readership seems to have grown considerably, and people are tweeting about our classroom. Have you reached 25000 pageviews yet?

We are at 24350 pageviews now. We will certainly hit the 25k mark today 😉 I am thankful to all the readers for their time. Special thanks to all those who are spreading the word. Our classroom is a great resource for anyone starting data analysis.

 

 

So, whats on the show today?

 

 

As you correctly pointed out, we are now slowly getting into various types of probability distributions. I mentioned in lesson 31 that we would learn several discrete probability distributions that are based on Bernoulli trials. We start this adventure with Binomial distribution.

 

Great. Let me refresh my memory of probability distributions before we get started. We discussed the basics of probability distribution in lesson 23. Let’s assume X is a random variable, and P(X = x) is the probability that this random variable takes any value x (i.e., an outcome). Then, the distribution of these probabilities on a number line, i.e., the probability graph is called the probability distribution function f(x) for a random variable. We are now looking at various mathematical forms for this f(x).

 

Fantastic. Now imagine you have a Bernoulli sequence of yes or no.

 

 

Sure. It is a sequence of 0s and 1s with a probability p; 0 if the trial yields a no (failure, or event not happening) and 1 if the trial yields a yes (success, or event happening). Something like this: 00101001101

 

From this sequence, 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 probability that this random variable X takes any value k, i.e., the probability of exactly k successes in n trials is:

 

 

 

The expected value of this random variable, E[X] = np, and the variance V[X] = np(1-p).

 

 

😯 Wow, that’s a fastball. Can we parse through the lingo?

 

 

Oops… Okay, let us take the example of your daily commute. Imagine buses and cars pass through the tunnel each morning. Can you guesstimate the probability of buses?

 

 

Yeah, I usually see more buses than cars in the morning. Let’s say the likelihood of seeing a bus is p=0.7.

 

 

Now let us imagine that buses and cars come in a Bernoulli sequence. Assign a 1 if it is a bus, and 0 if it is a car.

 

That is reasonable. The vehicle passage is usually random. If we take that as a Bernoulli sequence, there will be some 1s and some 0s with a 0.7 probability of occurrence. In the long run, you will have 70% buses and 30% cars in any order.

 

 

Correct. Now think about this. In the next four vehicles that pass through the tunnel, how many of them will be buses?

 

 

Since there is randomness in the sequence, in the next four vehicles, I can say, all of them may be buses, or none of them will be buses or any number in between.

 

Exactly. The number of buses in a sequence of 4 vehicles can be 0, 1, 2, 3 or 4. These are the random variables represented by X. In other words, if X is the number of buses in 4 vehicles coming at random, then X can take 0, 1, 2, 3 or 4 as the outcomes. The probability distribution of X is binomial.

 

I understand how we came up with X. Why is the probability distribution of X called binomial?

 

It originates from the idea of the binomial coefficient that you may have learned in an elementary math/combinations class. Let us continue with our logical deduction to see how the probability is derived, and you will see why.

 

Sure. We have X as 0, 1, 2, 3 and 4. We should calculate the probability P(X = 0), P(X = 1), P(X = 2), P(X = 3) and P(X = 4). This will give us the distribution of the probabilities.

 

Take an example, say 2. Let us compute P(X = 2). The probability of seeing exactly two buses in 4 vehicles. The probability of exactly k successes in n trials. If the buses and cars come in a Bernoulli sequence (1 for bus and 0 for a car) with a probability p, in how many ways can you see two buses out of 4 vehicles?

 

Ah, I see where we are going with this. Let me list out the possibilities. Two buses in four vehicles can occur in six ways. 0011, 0101, 1100, 1010, 1001, 0110. In each of these six possible sequences, there will be exactly two buses among four vehicles. I remember from my combinations class that this is four choose two. Four factorial divided by the product of two factorial and (four minus two) factorial. 4C2 = 4!/4!(4-2)!

For each possibility, the probability of that sequence can also be written down. Let me make a table like this:

 

 

 

 

 

 

 

You can see from the table that there are six possibilities. Any of the possibilities, 1 or 2 or 3 or 4 or 5 or 6 can occur. Hence, the probability of seeing two in four is the sum of these probabilities. Remember P(A or B) = P(A) + P(B). If you follow through this, you will get, 6*p*p*(1-p)*(1-p). = 6*p^2*(1-p)^(4-2). Can you see where the formula for binomial distribution comes from?

 

 

 

Absolutely. For each outcome of X, i.e., 0, 1, 2, 3 and 4, we should apply this logic/formula and compute the probability of the outcome. Let me finish it and make a plot.

 

Very nicely done. Let me jump in here and show you another plot with a different n and p. If p = 0.5 (equal probability) and n = 100; this is how the binomial distribution looks like.

 

Nice. It looks like an inverted bell centered around 50.

 

 

Yeah. You noticed that the distribution is centered around 50. It is the expected value of the distribution. Remember E[X] is the central tendency of the distribution. For binomial, you can derive it as np = 100 (0.5) = 50. In the same way, the variance, i.e. spread of the function around this center is np(1-p) = 100(0.5)(0.5) = 25. Or standard deviation is 5. You can see that the distribution is spread out within three standard deviations from the center. Can you now imagine how the distribution will look like for p = 0.3 or p = 0.7?

 

Following the same logic, those distributions will be centered on 100*0.3 = 30 and 100*0.7 = 70 with their variance. Now it all makes sense.

 

You see how easy it is when you go through the logic. We started with Bernoulli sequence. When we are interested in the random variable that is the number of successes in so many trials, it follows a binomial distribution. Exactly k successes” is the language of Binomial distribution. Can you think of any other examples that can be modeled as a binomial distribution?

 

Probability that Derek Jeter, with a batting average of 0.3, gets three hits out of the three times he comes to bat 😆  This is fun. I am glad I learned some useful concepts out of the messy commute experience. By the way, Exactly one landfall in the next four hurricanes is also binomial. With Jose coming up, I wonder if we can compute the probability of damage for New York City based on the probability of landfall.

 

Don’t worry Joe. Our Mayor is graciously implementing his comprehensive $20 billion resiliency plan. NYC is safe now. Forget probability of damage. You need to worry about the probability of bankruptcy.

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Lesson 31 – Yes or No: The language of Bernoulli trials

Downtown Miami will be flooded due to hurricane Irma.

Your vehicle will pass the inspection test this year.          

Each toss of a coin results in either a head or a tail.          

Did you notice that I am looking for an answer, an outcome that is “yes” or “no.” We often summarize data as the occurrence (or non-occurrence) of an event in a sequence of trials. For example, if you are designing dikes for flood control in Miami, you may want to look at the sequence of floods over several years to analyze the number of events, and the rate at which they occur.

There are two possibilities, a hit (event occurred – success) or miss (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 itself are assumed to be independent, i.e., the occurrence of one event does not influence the occurrence of the subsequent event.

Now, imagine these outcomes, 0’s or 1’s can be represented using a random variable X. In other words, X is a random variable that can take 0 or 1 with a probability p. If in Miami, there were ten extreme flood events in the last 100 years, the sequence will have 90 0’s and 10 1’s in some order. The probability of the event is hence 0.1. If the probability is 0.5, then, in a sequence of 100 trials (coin tosses for example), you will see 50 heads on average. We can derive the expected value of X and the variance of X as follows:

Since the Bernoulli trials are independent, the probability of a sequence of events happening will be equal to the product of the probability of each event. For instance, the probability of observing a sequence of No Flood, No Flood, No Flood and Flood over the last four years is 0.9*0.9*0.9*0.1 = 0.072 (assuming p = 0.1).

Bernoulli trials form the basis for deriving several discrete probability distributions that we will learn over the next few weeks.

While you ponder over what these distributions are, their mathematical forms, and how they represent the variation in the data, I will leave you with this image of the daily rainfall data from Miami International Airport. An approximate 6.38 inches of rain (~160mm/day) is forecasted for Sunday. Notice how you can remap the data into a sequence of 0’s (if rain is less than 160) and 1’s (if rain is greater than 160).

After tomorrow, when you hear “unprecedented rains” in the news, keep in mind that we seek the historical sequence data like this precisely because our memory is weak.

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Lesson 30 – Pause and rewind

Why am I doing this?

What you just read is my blog post on January 1, 2017. It is what happens on a new year day when you don’t have a social life.

Over the last few years, I have concluded that the largest crowd I can ever reach, if I continue delivering lectures in the usual University setting is 80 per semester. I am sure some of them are not there by choice. Since I believe that I can distill ideas into easily understandable forms, I felt the urge to spread my voice to a larger audience. Hence our Data Analysis Classroom.

I usually emphasize the importance of collecting more data for understanding the uncertainties in the system and using them in the final decision-making process. Practicing the preachings, after 29 lessons, I want to pause, reflect on the readership data and rewind what we learned.

We started this journey on the fifth day of February 2017. After 202 days, i.e., six months and 21 days, the monthly readership data is

We more than made up for the slump in July. The total page views in this time are 19051 with average monthly page views of 2721. Being very conservative, I would like to think that the blog may only capture the interest of 0.5% of the readers. That will be approximately 95 more people I could reach in the six month period. As I said, that is easily more than who I can reach through a class in a semester. I just hope these 95 don’t include the same folks in the class who are already captivated!

Here is a map of the readership. If you are in one of the blue countries, thank you for your attention. Let’s get more people on board. They will like it too.

What did we learn so far?

Uncertainties surround us. Understanding where they arise from and recognizing their extent is fundamental to improving our knowledge about anything.

We started off with the fact that one needs to observe the system to understand it better. Ask for data. The more, the merrier.

Lesson 1: When you see something, say data. A total of 412 page views.

Data may be grouped in sets → collection of elements. We have various forms to visualize the sets. Unions, intersections, subsets, and their properties.

Lesson 3: The Setup. 321 page views.
Lesson 4: I am a visual person. 301 page views.

We defined probability and understood that there are some rules (axioms) of probability.

Lesson 6: It’s probably me. 377 page views.
Lesson 7: The nervousness axiom – fight or flight. 253 page views.

We learned conditional events, independent events and how the probability plays out under these events.

Lesson 9: The necessary condition for Vegas. 813 page views.
Lesson 10: The fight for independence. 641 page views.

We now know the law of total probability and the Bayes theorem.

Lesson 12: Total recall. 471 page views.
Lesson 13: Dear Mr. Bayes. 2280 page views — most popular lesson so far.

In lessons 14 through 20, you can learn the basics of exploratory data analysis; visualization techniques, and summary statistics.

Lesson 14: The time has come; execute order statistics. A lesson to explain the concept of percentiles and boxplots. 327 page views.

Lesson 15: Another brick in the wall — for building histograms. 297 page views.

Lesson 16: Joe meets the average. Explains the mean of the data. 218 page views.

Lesson 17: We who deviate from the norm. Explains the idea of variance and standard deviation. 159 page views.

Lesson 18: Data comes in all shapes. A short lesson explaining the concept of skewness. 116 page views.

Lesson 19: Voice of the outliers. Outliers are significant. As Nicholas Taleb puts it: “Don’t be a turkey” by removing them. 87 page views.

Lesson 20: Compared to what?. Use the coefficient of variation to compare data. 115 page views.

Lessons 22 to 28 introduce the idea of random variables and probability distribution.

Lesson 22: You are so random. The basic idea of discrete and continuous random variables. 109 page views.

Lesson 23: Let’s distribution the probability. This lesson will teach the concept of the probability distribution. 126 page views.

Lesson 24: What else did you expect? 130 page views.
Lesson 25: More expectation. 95 page views. These two lessons go through the expected value of a random variable.

Lesson 26: The variety in consumption. 175 page views.
Lesson 27: More variety. 568 page views. These two lessons are for understanding the variance of a random variable concept.

Lesson 28: Apples and Oranges. How to standardize the data for comparison. 325 page views.

Lessons in R

As I mentioned in lesson 2, R is your companion in this journey through data. Computer programming is very very (cannot emphasize enough “very”) essential for data analysis. Wherever required, I have provided brief lessons on how to use R for data analysis.

Lesson 2: R is your companion. The very first lesson to get going with R. 434 page views.

Lesson 5: Let us Review. Reading data files and more fun stuff. 228 page views.

Lesson 8: The search for wifi. Learning for and if else statements in R. 194 page views.

Lesson 11: Fran, the functionin R-bot. The essentials of writing functions in R. 332 page views.

Lesson 21: Beginners guide to summarize data in R. A step-by-step exploratory data analysis. 1786 page views.

Lesson 29: Large number games in R. 933 page views.

Where do we go from here?

A long way.

While you pause and reflect on these lessons, I will pause and come back with lesson 31 on the ninth day of September 2017. Help spread the word as we build this knowledge platform one lesson at a time while the university system becomes obsolete.

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