## 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.)

## Getting Ready for Mathematical Classes in the New Semester – Guest-Blogging on SFU’s Career Services Informer

The following blog post was slightly condensed for editorial brevity and then published on the Career Services Informer, the official blog of the Career Services Centre at my undergraduate alma mater, Simon Fraser University

As a new Fall semester begins, many students start courses such as math, physics, computing science, engineering and statistics.  These can be tough classes with a rapid progression in workload and difficulty, but steady preparation can mount a strong defense to the inevitable pressure and stress.  Here are some tips to help you to get ready for those classes.

## 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.

## The Chi-Squared Test of Independence – An Example in Both R and SAS

#### Introduction

The chi-squared test of independence is one of the most basic and common hypothesis tests in the statistical analysis of categorical data.  Given 2 categorical random variables, $X$ and $Y$, the chi-squared test of independence determines whether or not there exists a statistical dependence between them.  Formally, it is a hypothesis test with the following null and alternative hypotheses:

$H_0: X \perp Y \ \ \ \ \ \text{vs.} \ \ \ \ \ H_a: X \not \perp Y$

If you’re not familiar with probabilistic independence and how it manifests in categorical random variables, watch my video on calculating expected counts in contingency tables using joint and marginal probabilities.  For your convenience, here is another video that gives a gentler and more practical understanding of calculating expected counts using marginal proportions and marginal totals.

Today, I will continue from those 2 videos and illustrate how the chi-squared test of independence can be implemented in both R and SAS with the same example.

## Applied Statistics Lesson of the Day – The Coefficient of Variation

In my statistics classes, I learned to use the variance or the standard deviation to measure the variability or dispersion of a data set.  However, consider the following 2 hypothetical cases:

1. the standard deviation for the incomes of households in Canada is $2,000 2. the standard deviation for the incomes of the 5 major banks in Canada is$2,000

Even though this measure of dispersion has the same value for both sets of income data, $2,000 is a significant amount for a household, whereas$2,000 is not a lot of money for one of the “Big Five” banks.  Thus, the standard deviation alone does not give a fully accurate sense of the relative variability between the 2 data sets.  One way to overcome this limitation is to take the mean of the data sets into account.

A useful statistic for measuring the variability of a data set while scaling by the mean is the sample coefficient of variation:

$\text{Sample Coefficient of Variation (} \bar{c_v} \text{)} \ = \ s \ \div \ \bar{x},$

where $s$ is the sample standard deviation and $\bar{x}$ is the sample mean.

Analogously, the coefficient of variation for a random variable is

$\text{Coefficient of Variation} \ (c_v) \ = \ \sigma \div \ \mu,$

where $\sigma$ is the random variable’s standard deviation and $\mu$ is the random variable’s expected value.

The coefficient of variation is a very useful statistic that I, unfortunately, never learned in my introductory statistics classes.  I hope that all new statistics students get to learn this alternative measure of dispersion.

## Using Your Vacation to Develop Your Career – Guest Blogging on Simon Fraser University’s Career Services Informer

The following post was originally published on the Career Services Informer.

I recently took a vacation from my former role as a statistician at the BC Centre for Excellence in HIV/AIDS. I did not plan a trip out of town – the spring weather was beautiful in Vancouver, and I wanted to spend time on the things that I like to do in this city. Many obvious things came to mind – walking along beaches, practicing Python programming and catching up with friends – just to name a few.

Yes, Python programming was one of the obvious things on my vacation to-do list, and I understand how ridiculous this may seem to some people. Why tax my brain during a time that is meant for mental relaxation, especially when the weather is great?

## Machine Learning and Applied Statistics Lesson of the Day – Positive Predictive Value and Negative Predictive Value

For a binary classifier,

• its positive predictive value (PPV) is the proportion of positively classified cases that were truly positive.

$\text{PPV} = \text{(Number of True Positives)} \ \div \ \text{(Number of True Positives} \ + \ \text{Number of False Positives)}$

• its negative predictive value (NPV) is the proportion of negatively classified cases that were truly negative.

$\text{NPV} = \text{(Number of True Negatives)} \ \div \ \text{(Number of True Negatives} \ + \ \text{Number of False Negatives)}$

In a later Statistics and Machine Learning Lesson of the Day, I will discuss the differences between PPV/NPV and sensitivity/specificity in assessing the predictive accuracy of a binary classifier.

(Recall that sensitivity and specificity can also be used to evaluate the performance of a binary classifier.  Based on those 2 statistics, we can construct receiver operating characteristic (ROC) curves to assess the predictive accuracy of the classifier, and a minimum standard for a good ROC curve is being better than the line of no discrimination.)

## 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.

You can also watch the video below the fold!

## Video Tutorial – Calculating Expected Counts in Contingency Tables Using Marginal Proportions and Marginal Totals

A common task in statistics and biostatistics is performing hypothesis tests of independence between 2 categorical random variables.  The data for such tests are best organized in contingency tables, which allow expected counts to be calculated easily.  In this video tutorial in my Youtube channel, I demonstrate how to calculate expected counts using marginal proportions and marginal totals.  In a later video, I will introduce a second method for calculating expected counts using joint probabilities and marginal probabilities.

In a later tutorial, I will illustrate how to implement the chi-squared test of independence on the same data set in R and SAS – stay tuned!

You can also watch the video below the fold!

## Mathematics and Applied Statistics Lesson of the Day – The Geometric Mean

Suppose that you invested in a stock 3 years ago, and the annual rates of return for each of the 3 years were

• 5% in the 1st year
• 10% in the 2nd year
• 15% in the 3rd year

What is the average rate of return in those 3 years?

It’s tempting to use the arithmetic mean, since we are so used to using it when trying to estimate the “centre” of our data.  However, the arithmetic mean is not appropriate in this case, because the annual rate of return implies a multiplicative growth of your investment by a factor of $1 + r$, where $r$ is the rate of return in each year.  In contrast, the arithmetic mean is appropriate for quantities that are additive in nature; for example, your average annual salary from the past 3 years is the sum of last 3 annual salaries divided by 3.

If the arithmetic mean is not appropriate, then what can we use instead?  Our saviour is the geometric mean, $G$.  The average factor of growth from the 3 years is

$G = [(1 + r_1)(1 + r_2) ... (1 + r_n)]^{1/n}$,

where $r_i$ is the rate of return in year $i$, $i = 1, 2, 3, ..., n$.  The average annual rate of return is $G - 1$.  Note that the geometric mean is NOT applied to the annual rates of return, but the annual factors of growth.

Returning to our example, our average factor of growth is

$G = [(1 + 0.05) \times (1 + 0.10) \times (1 + 0.15)]^{1/3} = 1.099242$.

Thus, our annual rate of return is $G - 1 = 1.099242 - 1 = 0.099242 = 9.9242\%$.

Here is a good way to think about the difference between the arithmetic mean and the geometric mean.  Suppose that there are 2 sets of numbers.

1. The first set, $S_1$, consists of your data $x_1, x_2, ..., x_n$, and this set has a sample size of $n$.
2. The second, $S_2$,  set also has a sample size of $n$, but all $n$ values are the same – let’s call this common value $y$.
• What number must $y$ be such that the sums in $S_1$ and $S_2$ are equal?  This value of $y$ is the arithmetic mean of the first set.
• What number must $y$ be such that the products in $S_1$ and $S_2$ are equal?  This value of $y$ is the geometric mean of the first set.

Note that the geometric means is only applicable to positive numbers.

## Mathematics and Applied Statistics Lesson of the Day – The Weighted Harmonic Mean

In a previous Statistics Lesson of the Day on the harmonic mean, I used an example of a car travelling at 2 different speeds – 60 km/hr and 40 km/hr.  In that example, the car travelled 120 km at both speeds, so the 2 speeds had equal weight in calculating the harmonic mean of the speeds.

What if the cars travelled different distances at those speeds?  In that case, we can modify the calculation to allow the weight of each datum to be different.  This results in the weighted harmonic mean, which has the formula

$H = \sum_{i = 1}^{n} w_i \ \ \div \ \ \sum_{i = 1}^{n}(w_i \ \div \ x_i)$.

For example, consider a car travelling for 240 kilometres at 2 different speeds and for 2 different distances:

1. 60 km/hr for 100 km
2. 40 km/hr for another 140 km

Then the weighted harmonic mean of the speeds (i.e. the average speed of the whole trip) is

$(100 \text{ km} \ + \ 140 \text{ km}) \ \div \ [(100 \text{ km} \ \div \ 60 \text{ km/hr}) \ + \ (140 \text{ km} \ \div \ 40 \text{ km/hr})]$

$= 46.45 \text{ km/hr}$

Notice that this is exactly the same calculation that we would use if we wanted to calculate the average speed of the whole trip by the formula from kinematics:

$\text{Average Speed} = \Delta \text{Distance} \div \Delta \text{Time}$

## Vancouver Machine Learning and Data Science Meetup – NLP to Find User Archetypes for Search & Matching

I will attend the following seminar by Thomas Levi in the next R/Machine Learning/Data Science Meetup in Vancouver on Wednesday, June 25.  If you will also attend this event, please come up and say “Hello”!  I would be glad to meet you!

To register, sign up for an account on Meetup, and RSVP in the R Users Group, the Machine Learning group or the Data Science group.

Title: NLP to Find User Archetypes for Search & Matching

Speaker: Thomas Levi, Plenty of Fish

Location: HootSuite, 5 East 8th Avenue, Vancouver, BC

Time and Date: 6-8 pm, Wednesday, June 25, 2014

Abstract

As the world’s largest free dating site, Plenty Of Fish would like to be able to match with and allow users to search for people with similar interests. However, we allow our users to enter their interests as free text on their profiles. This presents a difficult problem in clustering, search and machine learning if we want to move beyond simple ‘exact match’ solutions to a deeper archetypal user profiling and thematic search system. Some of the common issues that arise are misspellings, synonyms (e.g. biking, cycling and bicycling) and similar interests (e.g. snowboarding and skiing) on a several million user scale. In this talk I will demonstrate how we built a system utilizing topic modelling with Latent Dirichlet Allocation (LDA) on a several hundred thousand word vocabulary over ten million+ North American users and explore its applications at POF.

Bio

Thomas Levi started out with a doctorate in Theoretical Physics and String Theory from the University of Pennsylvania in 2006. His post-doctoral studies in cosmology and string theory, where he wrote 19 papers garnering 650+ citations, then took him to NYU and finally UBC.  In 2012, he decided to move into industry, and took on the role of Senior Data Scientist at POF. Thomas has been involved in diverse projects such as behaviour analysis, social network analysis, scam detection, Bot detection, matching algorithms, topic modelling and semantic analysis.

Schedule
• 6:00PM Doors are open, feel free to mingle
• 6:30 Presentations start
• 8:00 Off to a nearby watering hole (Mr. Brownstone?) for a pint, food, and/or breakout discussions

## 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.)

## Applied Statistics Lesson of the Day – Polynomial Regression is Actually Just Linear Regression

Continuing from my previous Statistics Lesson of the Day on what “linear” really means in “linear regression”, I want to highlight a common example involving this nomenclature that can mislead non-statisticians.  Polynomial regression is a commonly used multiple regression technique; it models the systematic component of the regression model as a $p\text{th}$-order polynomial relationship between the response variable $Y$ and the explanatory variable $x$.

$Y = \beta_0 + \beta_1 x + \beta_2 x^2 + ... + \beta_p x^p + \varepsilon$

However, this model is still a linear regression model, because the response variable is still a linear combination of the regression coefficients.  The regression coefficients would still be estimated using linear algebra through the method of least squares.

Remember: the “linear” in linear regression refers to the linearity between the response variable and the regression coefficients, NOT between the response variable and the explanatory variable(s).

## Mathematics and Applied Statistics Lesson of the Day – The Harmonic Mean

The harmonic mean, H, for $n$ positive real numbers $x_1, x_2, ..., x_n$ is defined as

$H = n \div (1/x_1 + 1/x_2 + .. + 1/x_n) = n \div \sum_{i = 1}^{n}x_i^{-1}$.

This type of mean is useful for measuring the average of rates.  For example, consider a car travelling for 240 kilometres at 2 different speeds:

1. 60 km/hr for 120 km
2. 40 km/hr for another 120 km

Then its average speed for this trip is

$S_{avg} = 2 \div (1/60 + 1/40) = 48 \text{ km/hr}$

Notice that the speed for the 2 trips have equal weight in the calculation of the harmonic mean – this is valid because of the equal distance travelled at the 2 speeds.  If the distances were not equal, then use a weighted harmonic mean instead – I will cover this in a later lesson.

To confirm the formulaic calculation above, let’s use the definition of average speed from physics.  The average speed is defined as

$S_{avg} = \Delta \text{distance} \div \Delta \text{time}$

We already have the elapsed distance – it’s 240 km.  Let’s find the time elapsed for this trip.

$\Delta \text{ time} = 120 \text{ km} \times (1 \text{ hr}/60 \text{ km}) + 120 \text{ km} \times (1 \text{ hr}/40 \text{ km})$

$\Delta \text{time} = 5 \text{ hours}$

Thus,

$S_{avg} = 240 \text{ km} \div 5 \text{ hours} = 48 \text { km/hr}$

Notice that this explicit calculation of the average speed by the definition from kinematics is the same as the average speed that we calculated from the harmonic mean!

## 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!

## 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!