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 view this blog post beneath the fold to watch the video on this blog.  You can also watch it directly on YouTube.

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Sorting correlation coefficients by their magnitudes in a SAS macro

Theoretical Background

Many statisticians and data scientists use the correlation coefficient to study the relationship between 2 variables.  For 2 random variables, X and Y, the correlation coefficient between them is defined as their covariance scaled by the product of their standard deviations.  Algebraically, this can be expressed as

\rho_{X, Y} = \frac{Cov(X, Y)}{\sigma_X \sigma_Y} = \frac{E[(X - \mu_X)(Y - \mu_Y)]}{\sigma_X \sigma_Y}.

In real life, you can never know what the true correlation coefficient is, but you can estimate it from data.  The most common estimator for \rho is the Pearson correlation coefficient, which is defined as the sample covariance between X and Y divided by the product of their sample standard deviations.  Since there is a common factor of

\frac{1}{n - 1}

in the numerator and the denominator, they cancel out each other, so the formula simplifies to

r_P = \frac{\sum_{i = 1}^{n}(x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_i - \bar{x})^2 \sum_{i = 1}^{n}(y_i - \bar{y})^2}} .

 

In predictive modelling, you may want to find the covariates that are most correlated with the response variable before building a regression model.  You can do this by

  1. computing the correlation coefficients
  2. obtaining their absolute values
  3. sorting them by their absolute values.

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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)

odds of all 15 home teams winning on same day

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 – Basu’s Theorem

Today’s Statistics Lesson of the Day will discuss Basu’s theorem, which connects the previously discussed concepts of minimally sufficient statistics, complete statistics and ancillary statistics.  As before, I will begin with the following set-up.

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) or probability mass function (PMF) for X_1, X_2, ..., X_n.

Let

t = T(\mathbf{X})

be a statistics based on \textbf{X}.

Basu’s theorem states that, if T(\textbf{X}) is a complete and minimal sufficient statistic, then T(\textbf{X}) is independent of every ancillary statistic.

Establishing the independence between 2 random variables can be very difficult if their joint distribution is hard to obtain.  This theorem allows the independence between minimally sufficient statistic and every ancillary statistic to be established without their joint distribution – and this is the great utility of Basu’s theorem.

However, establishing that a statistic is complete can be a difficult task.  In a later lesson, I will discuss another theorem that will make this task easier for certain cases.

Mathematical Statistics Lesson of the Day – An Example of An Ancillary Statistic

Consider 2 random variables, X_1 and X_2, from the normal distribution \text{Normal}(\mu, \sigma^2), where \mu is unknown.  Then the statistic

D = X_1 - X_2

has the distribution

\text{Normal}(0, 2\sigma^2).

The distribution of D does not depend on \mu, so D is an ancillary statistic for \mu.

Note that, if \sigma^2 is unknown, then D is not ancillary for \sigma^2.

Mathematical Statistics Lesson of the Day – Ancillary Statistics

The set-up for today’s post mirrors my earlier Statistics Lessons of the Day on sufficient statistics and complete 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) or probability mass function (PMF) for X_1, X_2, ..., X_n.

Let

a = A(\mathbf{X})

be a statistics based on \textbf{X}.

If the distribution of A(\textbf{X}) does NOT depend on \theta, then A(\textbf{X}) is called an ancillary statistic.

An ancillary statistic contains no information about \theta; its distribution is fixed and known without any relation to \theta.  Why, then, would we care about A(\textbf{X})  I will address this question in later Statistics Lessons of the Day, and I will connect ancillary statistics to sufficient statistics, minimally sufficient statistics and complete statistics.

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.

Christian Robert Shows that the Sample Median Cannot Be a Sufficient Statistic

I am grateful to Christian Robert (Xi’an) for commenting on my recent Mathematical Statistics Lessons of the Day on sufficient statistics and minimally sufficient statistics.

In one of my earlier posts, he wisely commented that the sample median cannot be a sufficient statistic.  He has supplemented this by writing on his own blog to show that the median cannot be a sufficient statistic.

Thank you, Christian, for your continuing readership and contribution.  It’s a pleasure to learn from you!

Mathematical Statistics Lesson of the Day – Minimally Sufficient Statistics

In using a statistic to estimate a parameter in a probability distribution, it is important to remember that there can be multiple sufficient statistics for the same parameter.  Indeed, the entire data set, X_1, X_2, ..., X_n, can be a sufficient statistic – it certainly contains all of the information that is needed to estimate the parameter.  However, using all n variables is not very satisfying as a sufficient statistic, because it doesn’t reduce the information in any meaningful way – and a more compact, concise statistic is better than a complicated, multi-dimensional statistic.  If we can use a lower-dimensional statistic that still contains all necessary information for estimating the parameter, then we have truly reduced our data set without stripping any value from it.

Our saviour for this problem is a minimally sufficient statistic.  This is defined as a statistic, T(\textbf{X}), such that

  1. T(\textbf{X}) is a sufficient statistic
  2. if U(\textbf{X}) is any other sufficient statistic, then there exists a function g such that

T(\textbf{X}) = g[U(\textbf{X})].

Note that, if there exists a one-to-one function h such that

T(\textbf{X}) = h[U(\textbf{X})],

then T(\textbf{X}) and U(\textbf{X}) are equivalent.

Mathematical Statistics Lesson of the Day – Sufficient Statistics

*Update on 2014-11-06: Thanks to Christian Robert’s comment, I have removed the sample median as an example of a sufficient statistic.

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}.  Let g_\theta(t) be the PDF for T(X).

If the conditional PDF

h_\theta(\mathbf{X}) = f_\theta(x) \div g_\theta[T(\mathbf{X})]

is independent of \theta, then T(\mathbf{X}) is a sufficient statistic for \theta.  In other words,

h_\theta(\mathbf{X}) = h(\mathbf{X}),

and \theta does not appear in h(\mathbf{X}).

Intuitively, this means that T(\mathbf{X}) contains everything you need to estimate \theta, so knowing T(\mathbf{X}) (i.e. conditioning f_\theta(x) on T(\mathbf{X})) is sufficient for estimating \theta.

Often, the sufficient statistic for \theta is a summary statistic of X_1, X_2, ..., X_n, such as their

  • sample mean
  • sample median – removed thanks to comment by Christian Robert (Xi’an)
  • sample minimum
  • sample maximum

If such a summary statistic is sufficient for \theta, then knowing this one statistic is just as useful as knowing all n data for estimating \theta.

*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 further and further from its “central value”, we would expect those distant-from-the-centre vales 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.

Mathematical and Applied Statistics Lesson of the Day – Don’t Use the Terms “Independent Variable” and “Dependent Variable” in Regression

In math and science, we learn the equation of a line as

y = mx + b,

with y being called the dependent variable and x being called the independent variable.  This terminology holds true for more complicated functions with multiple variables, such as in polynomial regression.

I highly discourage the use of “independent” and “dependent” in the context of statistics and regression, because these terms have other meanings in statistics.  In probability, 2 random variables X_1 and X_2 are independent if their joint distribution is simply a product of their marginal distributions, and they are dependent if otherwise.  Thus, the usage of “independent variable” for a regression model with 2 predictors becomes problematic if the model assumes that the predictors are random variables; a random effects model is an example with such an assumption.  An obvious question for such models is whether or not the independent variables are independent, which is a rather confusing question with 2 uses of the word “independent”.  A better way to phrase that question is whether or not the predictors are independent.

Thus, in a statistical regression model, I strongly encourage the use of the terms “response variable” or “target variable” (or just “response” and “target”) for Y and the terms “explanatory variables”, “predictor variables”, “predictors”, “covariates”, or “factors” for x_1, x_2, .., x_p.

(I have encountered some statisticians who prefer to reserve “covariate” for continuous predictors and “factor” for categorical predictors.)

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!

You can also watch this new video below the fold!

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Video Tutorial – Rolling 2 Dice: An Intuitive Explanation of The Central Limit Theorem

According to the central limit theorem, if

  • n random variables, X_1, ..., X_n, are independent and identically distributed,
  • n is sufficiently large,

then the distribution of their sample mean, \bar{X_n}, is approximately normal, and this approximation is better as n increases.

One of the most remarkable aspects of the central limit theorem (CLT) is its validity for any parent distribution of X_1, ..., X_n.  In my new Youtube channel, you will find a video tutorial that provides an intuitive explanation of why this is true by considering a thought experiment of rolling 2 dice.  This video focuses on the intuition rather than the mathematics of the CLT.  In a later video, I will discuss the technical details of the CLT and how it applies to this example.

You can also watch the video below the fold!

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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.