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.

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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.

Mathematics and Applied Statistics Lesson of the Day – Contrasts

A contrast is a linear combination of a set of variables such that the sum of the coefficients is equal to zero.  Notationally, consider a set of variables

\mu_1, \mu_2, ..., \mu_n.

Then the linear combination

c_1 \mu_1 + c_2 \mu_2 + ... + c_n \mu_n

is a contrast if

c_1 + c_2 + ... + c_n = 0.

There is a reason for why I chose to use \mu as the symbol for the variables in the above notation – in statistics, contrasts provide a very useful framework for comparing multiple population means in hypothesis testing.  In a later Statistics Lesson of the Day, I will illustrate some examples of contrasts, especially in the context of experimental design.

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.

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.

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.)

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}

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).