## Arnab Chakraborty on The Monty Hall Problem and Bayes’ Theorem – The Central Equilibrium – Episode 6

I am pleased to welcome Arnab Chakraborty back to my talk show, “The Central Equilibrium“, to talk about the Monty Hall Problem and Bayes’ theorem.  In this episode, he shows 2 solutions to this classic puzzle in probability, and invokes Bayes’ Theorem for the second solution.

If you have not watched Arnab’s first episode on Bayes’ theorem, then I encourage you to do that first.

Marilyn Vos Savant provided a solution to this problem in PARADE Magazine in 1990-1991.  Thousands of readers disagreed with her solution and criticized her vehemently (and incorrectly) for her error.  Some of these critics were mathematicians!  She included some of those replies and provided alternative perspectives that led to the same conclusion.  Although I am dismayed by the disrespect that some people showed in their letters to her, I am glad that a magazine column on probability was able to attract so much readership and interest.  Arnab and I referred to one of her solutions in our episode.  Thank you, Marilyn!

Enjoy this episode of “The Central Equilibrium“!

## Video Tutorial – Obtaining the Expected Value of the Exponential Distribution Using the Moment Generating Function

In this video tutorial on YouTube, I use the exponential distribution’s moment generating function (MGF) to obtain the expected value of this distribution.  Visit my YouTube channel to watch more video tutorials!

## Video Tutorial – The Moment Generating Function of the Exponential Distribution

In this video tutorial on YouTube, I derive the moment generating function (MGF) of the exponential distribution.  Visit my YouTube channel to watch more video tutorials!

## Arnab Chakraborty on Bayes’ Theorem – The Central Equilibrium – Episode 3

Arnab Chakraborty kindly came to my new talk show, “The Central Equilibrium”, to talk about Bayes’ theorem.  He introduced the concept of conditional probability, stated Bayes’ theorem in its simple and general forms, and showed an example of how to use it in a calculation.

Check it out!

## Christopher Salahub on Markov Chains – The Central Equilibrium – Episode 2

It was a great pleasure to talk to Christopher Salahub about Markov chains in the second episode of my new talk show, The Central Equilibrium!  Chris graduated from the University of Waterloo with a Bachelor of Mathematics degree in statistics.  He just finished an internship in data development at Environics Analytics, and he is starting a Master’s program in statistics at ETH Zurich in Switzerland.

Chris recommends “Introduction to Probability Models” by Sheldon Ross to learn more about probability theory and Markov chains.

The Central Equilibrium is my new talk show about math, science, and economics. It focuses on technical topics that involve explanations with formulas, equations, graphs, and diagrams.  Stay tuned for more episodes in the coming weeks!

You can watch all of my videos on my YouTube channel!

Please watch the video on this blog.  You can also watch it directly on YouTube.

## Odds and Probability: Commonly Misused Terms in Statistics – An Illustrative Example in Baseball

Yesterday, all 15 home teams in Major League Baseball won on the same day – the first such occurrence in history.  CTV News published an article written by Mike Fitzpatrick from The Associated Press that reported on this event.  The article states, “Viewing every game as a 50-50 proposition independent of all others, STATS figured the odds of a home sweep on a night with a full major league schedule was 1 in 32,768.”  (Emphases added)

Screenshot captured at 5:35 pm Vancouver time on Wednesday, August 12, 2015.

Out of curiosity, I wanted to reproduce this result.  This event is an intersection of 15 independent Bernoulli random variables, all with the probability of the home team winning being 0.5.

$P[(\text{Winner}_1 = \text{Home Team}_1) \cap (\text{Winner}_2 = \text{Home Team}_2) \cap \ldots \cap (\text{Winner}_{15}= \text{Home Team}_{15})]$

Since all 15 games are assumed to be mutually independent, the probability of all 15 home teams winning is just

$P(\text{All 15 Home Teams Win}) = \prod_{n = 1}^{15} P(\text{Winner}_i = \text{Home Team}_i)$

$P(\text{All 15 Home Teams Win}) = 0.5^{15} = 0.00003051757$

Now, let’s connect this probability to odds.

It is important to note that

• odds is only applicable to Bernoulli random variables (i.e. binary events)
• odds is the ratio of the probability of success to the probability of failure

For our example,

$\text{Odds}(\text{All 15 Home Teams Win}) = P(\text{All 15 Home Teams Win}) \ \div \ P(\text{At least 1 Home Team Loses})$

$\text{Odds}(\text{All 15 Home Teams Win}) = 0.00003051757 \div (1 - 0.00003051757)$

$\text{Odds}(\text{All 15 Home Teams Win}) = 0.0000305185$

The above article states that the odds is 1 in 32,768.  The fraction 1/32768 is equal to 0.00003051757, which is NOT the odds as I just calculated.  Instead, 0.00003051757 is the probability of all 15 home teams winning.  Thus, the article incorrectly states 0.00003051757 as the odds rather than the probability.

This is an example of a common confusion between probability and odds that the media and the general public often make.  Probability and odds are two different concepts and are calculated differently, and my calculations above illustrate their differences.  Thus, exercise caution when reading statements about probability and odds, and make sure that the communicator of such statements knows exactly how they are calculated and which one is more applicable.

## Mathematical Statistics Lesson of the Day – Complete Statistics

The set-up for today’s post mirrors my earlier Statistics Lesson of the Day on sufficient statistics.

Suppose that you collected data

$\mathbf{X} = X_1, X_2, ..., X_n$

in order to estimate a parameter $\theta$.  Let $f_\theta(x)$ be the probability density function (PDF)* for $X_1, X_2, ..., X_n$.

Let

$t = T(\mathbf{X})$

be a statistic based on $\mathbf{X}$.

If

$E_\theta \{g[T(\mathbf{X})]\} = 0, \ \ \forall \ \theta,$

implies that

$P \{g[T(\mathbf{X})]\} = 0] = 1,$

then $T(\mathbf{X})$ is said to be complete.  To deconstruct this esoteric mathematical statement,

1. let $g(t)$ be a measurable function
2. if you want to use $g[T(\mathbf{X})]$ to form an unbiased estimator of the zero function,
3. and if the only such function is almost surely equal to the zero function,
4. then $T(\mathbf{X})$ is a complete statistic.

I will discuss the intuition behind this bizarre definition in a later Statistics Lesson of the Day.

*This above definition holds for discrete and continuous random variables.

## Mathematics and Mathematical Statistics Lesson of the Day – Convex Functions and Jensen’s Inequality

Consider a real-valued function $f(x)$ that is continuous on the interval $[x_1, x_2]$, where $x_1$ and $x_2$ are any 2 points in the domain of $f(x)$.  Let

$x_m = 0.5x_1 + 0.5x_2$

be the midpoint of $x_1$ and $x_2$.  Then, if

$f(x_m) \leq 0.5f(x_1) + 0.5f(x_2),$

then $f(x)$ is defined to be midpoint convex.

More generally, let’s consider any point within the interval $[x_1, x_2]$.  We can denote this arbitrary point as

$x_\lambda = \lambda x_1 + (1 - \lambda)x_2,$ where $0 < \lambda < 1$.

Then, if

$f(x_\lambda) \leq \lambda f(x_1) + (1 - \lambda) f(x_2),$

then $f(x)$ is defined to be convex.  If

$f(x_\lambda) < \lambda f(x_1) + (1 - \lambda) f(x_2),$

then $f(x)$ is defined to be strictly convex.

There is a very elegant and powerful relationship about convex functions in mathematics and in mathematical statistics called Jensen’s inequality.  It states that, for any random variable $Y$ with a finite expected value and for any convex function $g(y)$,

$E[g(Y)] \geq g[E(Y)]$.

A function $f(x)$ is defined to be concave if $-f(x)$ is convex.  Thus, Jensen’s inequality can also be stated for concave functions.  For any random variable $Z$ with a finite expected value and for any concave function $h(z)$,

$E[h(Z)] \leq h[E(Z)]$.

In future Statistics Lessons of the Day, I will prove Jensen’s inequality and discuss some of its implications in mathematical statistics.

## Mathematical Statistics Lesson of the Day – The Glivenko-Cantelli Theorem

In 2 earlier tutorials that focused on exploratory data analysis in statistics, I introduced

There is actually an elegant theorem that provides a rigorous basis for using empirical CDFs to estimate the true CDF – and this is true for any probability distribution.  It is called the Glivenko-Cantelli theorem, and here is what it states:

Given a sequence of $n$ independent and identically distributed random variables, $X_1, X_2, ..., X_n$,

$P[\lim_{n \to \infty} \sup_{x \epsilon \mathbb{R}} |\hat{F}_n(x) - F_X(x)| = 0] = 1.$

In other words, the empirical CDF of $X_1, X_2, ..., X_n$ converges uniformly to the true CDF.

My mathematical statistics professor at the University of Toronto, Keith Knight, told my class that this is often referred to as “The First Theorem of Statistics” or the “The Fundamental Theorem of Statistics”.  I think that this is a rather subjective title – the central limit theorem is likely more useful and important – but Page 261 of John Taylor’s An introduction to measure and probability (Springer, 1997) recognizes this attribution to the Glivenko-Cantelli theorem, too.

## Mathematical and Applied Statistics Lesson of the Day – The Motivation and Intuition Behind Chebyshev’s Inequality

In 2 recent Statistics Lessons of the Day, I

Chebyshev’s inequality is just a special version of Markov’s inequality; thus, their motivations and intuitions are similar.

$P[|X - \mu| \geq k \sigma] \leq 1 \div k^2$

Markov’s inequality roughly says that a random variable $X$ is most frequently observed near its expected value, $\mu$.  Remarkably, it quantifies just how often $X$ is far away from $\mu$.  Chebyshev’s inequality goes one step further and quantifies that distance between $X$ and $\mu$ in terms of the number of standard deviations away from $\mu$.  It roughly says that the probability of $X$ being $k$ standard deviations away from $\mu$ is at most $k^{-2}$.  Notice that this upper bound decreases as $k$ increases – confirming our intuition that it is highly improbable for $X$ to be far away from $\mu$.

As with Markov’s inequality, Chebyshev’s inequality applies to any random variable $X$, as long as $E(X)$ and $V(X)$ are finite.  (Markov’s inequality requires only $E(X)$ to be finite.)  This is quite a marvelous result!

## Mathematical Statistics Lesson of the Day – Chebyshev’s Inequality

The variance of a random variable $X$ is just an expected value of a function of $X$.  Specifically,

$V(X) = E[(X - \mu)^2], \ \text{where} \ \mu = E(X)$.

Let’s substitute $(X - \mu)^2$ into Markov’s inequality and see what happens.  For convenience and without loss of generality, I will replace the constant $c$ with another constant, $b^2$.

$\text{Let} \ b^2 = c, \ b > 0. \ \ \text{Then,}$

$P[(X - \mu)^2 \geq b^2] \leq E[(X - \mu)^2] \div b^2$

$P[ (X - \mu) \leq -b \ \ \text{or} \ \ (X - \mu) \geq b] \leq V(X) \div b^2$

$P[|X - \mu| \geq b] \leq V(X) \div b^2$

Now, let’s substitute $b$ with $k \sigma$, where $\sigma$ is the standard deviation of $X$.  (I can make this substitution, because $\sigma$ is just another constant.)

$\text{Let} \ k \sigma = b. \ \ \text{Then,}$

$P[|X - \mu| \geq k \sigma] \leq V(X) \div k^2 \sigma^2$

$P[|X - \mu| \geq k \sigma] \leq 1 \div k^2$

This last inequality is known as Chebyshev’s inequality, and it is just a special version of Markov’s inequality.  In a later Statistics Lesson of the Day, I will discuss the motivation and intuition behind it.  (Hint: Read my earlier lesson on the motivation and intuition behind Markov’s inequality.)

## Mathematical and Applied Statistics Lesson of the Day – The Motivation and Intuition Behind Markov’s Inequality

Markov’s inequality may seem like a rather arbitrary pair of mathematical expressions that are coincidentally related to each other by an inequality sign:

$P(X \geq c) \leq E(X) \div c,$ where $c > 0$.

However, there is a practical motivation behind Markov’s inequality, and it can be posed in the form of a simple question: How often is the random variable $X$ “far” away from its “centre” or “central value”?

Intuitively, the “central value” of $X$ is the value that of $X$ that is most commonly (or most frequently) observed.  Thus, as $X$ deviates farther and farther from its “central value”, we would expect those distant-from-the-centre values to be less frequently observed.

Recall that the expected value, $E(X)$, is a measure of the “centre” of $X$.  Thus, we would expect that the probability of $X$ being very far away from $E(X)$ is very low.  Indeed, Markov’s inequality rigorously confirms this intuition; here is its rough translation:

As $c$ becomes really far away from $E(X)$, the event $X \geq c$ becomes less probable.

You can confirm this by substituting several key values of $c$.

• If $c = E(X)$, then $P[X \geq E(X)] \leq 1$; this is the highest upper bound that $P(X \geq c)$ can get.  This makes intuitive sense; $X$ is going to be frequently observed near its own expected value.

• If $c \rightarrow \infty$, then $P(X \geq \infty) \leq 0$.  By Kolmogorov’s axioms of probability, any probability must be inclusively between $0$ and $1$, so $P(X \geq \infty) = 0$.  This makes intuitive sense; there is no possible way that $X$ can be bigger than positive infinity.

## Mathematical Statistics Lesson of the Day – Markov’s Inequality

Markov’s inequality is an elegant and very useful inequality that relates the probability of an event concerning a non-negative random variable, $X$, with the expected value of $X$.  It states that

$P(X \geq c) \leq E(X) \div c,$

where $c > 0$.

I find Markov’s inequality to be beautiful for 2 reasons:

1. It applies to both continuous and discrete random variables.
2. It applies to any non-negative random variable from any distribution with a finite expected value.

In a later lesson, I will discuss the motivation and intuition behind Markov’s inequality, which has useful implications for understanding a data set.

## Video Tutorial – Allelic Frequencies Remain Constant From Generation to Generation Under the Hardy-Weinberg Equilibrium

The Hardy-Weinberg law is a fundamental principle in statistical genetics.  If its 7 assumptions are fulfilled, then it predicts that the allelic frequency of a genetic trait will remain constant from generation to generation.  In this new video tutorial in my Youtube channel, I explain the math behind the Hardy-Weinberg theorem.  In particular, I clarify the origin of the connection between allelic frequencies and genotyopic frequencies in the second generation – I have not found a single textbook or web site on this topic that explains this calculation, so I hope that my explanation is helpful to you.

## Mathematical and Applied Statistics Lesson of the Day – The Central Limit Theorem Can Apply to the Sum

The central limit theorem (CLT) is often stated in terms of the sample mean of independent and identically distributed random variables.  An often unnoticed or forgotten aspect of the CLT is its applicability to the sample sum of those variables.  Since $n$, the sample size, is just a constant, it can be multiplied to $\bar{X}$ to obtain $\sum_{i = 1}^{n} X_i$.  For a sufficiently large $n$, this new statistic still has an approximately normal distribution, just with a new expected value and a new variance.

$\sum_{i = 1}^{n} X_i \overset{approx.}{\sim} \text{Normal} (n\mu, n\sigma^2)$

## Video Tutorial – Useful Relationships Between Any Pair of h(t), f(t) and S(t)

I first started my video tutorial series on survival analysis by defining the hazard function.  I then explained how this definition leads to the elegant relationship of

$h(t) = f(t) \div S(t)$.

In my new video, I derive 6 useful mathematical relationships that exist between any 2 of the 3 quantities in the above equation.  Each relationship allows one quantity to be written as a function of the other.

I am excited to continue adding to my Youtube channel‘s collection of video tutorials.  Please stay tuned for more!

## Presentation on Statistical Genetics at Vancouver SAS User Group – Wednesday, May 28, 2014

I am excited and delighted to be invited to present at the Vancouver SAS User Group‘s next meeting.  I will provide an introduction to statistical genetics; specifically, I will

• define basic terminology in genetics
• explain the Hardy-Weinberg equilibrium in detail
• illustrate how Pearson’s chi-squared goodness-of-fit test can be used in PROC FREQ in SAS to check the Hardy-Weinberg equilibrium
• illustrate how the Newton-Raphson algorithm can be used for maximum likelihood estimation in PROC IML in SAS

You can register for this meeting here.  The meeting’s coordinates are

9:00am – 3:00pm

Wednesday, May 28th, 2014

BC Cancer Agency Research Centre

675 West 10th Avenue.

Vancouver, BC

If you will attend this meeting, please feel free to come up and say “Hello!”.  I look forward to meeting you!

## Mathematical and Applied Statistics Lesson of the Day – The Central Limit Theorem Applies to the Sample Mean

Having taught and tutored introductory statistics numerous times, I often hear students misinterpret the Central Limit Theorem by saying that, as the sample size gets bigger, the distribution of the data approaches a normal distribution.  This is not true.  If your data come from a non-normal distribution, their distribution stays the same regardless of the sample size.

Remember: The Central Limit Theorem says that, if $X_1, X_2, ..., X_n$ is an independent and identically distributed sample of random variables, then the distribution of their sample mean is approximately normal, and this approximation gets better as the sample size gets bigger.

## Video Tutorial – The Hazard Function is the Probability Density Function Divided by the Survival Function

In an earlier video, I introduced the definition of the hazard function and broke it down into its mathematical components.  Recall that the definition of the hazard function for events defined on a continuous time scale is

$h(t) = \lim_{\Delta t \rightarrow 0} [P(t < X \leq t + \Delta t \ | \ X > t) \ \div \ \Delta t]$.

Did you know that the hazard function can be expressed as the probability density function (PDF) divided by the survival function?

$h(t) = f(t) \div S(t)$

In my new Youtube video, I prove how this relationship can be obtained from the definition of the hazard function!  I am very excited to post this second video in my new Youtube channel.

## Video Tutorial: Breaking Down the Definition of the Hazard Function

The hazard function is a fundamental quantity in survival analysis.  For an event occurring at some time on a continuous time scale, the hazard function, $h(t)$, for that event is defined as

$h(t) = \lim_{\Delta t \rightarrow 0} [P(t < X \leq t + \Delta t \ | \ X > t) \ \div \ \Delta t]$,

where

• $t$ is the time,
• $X$ is the time of the occurrence of the event.

However, what does this actually mean?  In this Youtube video, I break down the mathematics of this definition into its individual components and explain the intuition behind each component.

I am very excited about the release of this first video in my new Youtube channel!  This is yet another mode of expansion of The Chemical Statistician since the beginning of 2014.  As always, your comments are most appreciated!