[Home]

Table of contents

Random variables

What they are

Suppose that I toss a fair coin, and offer you Rs 10 for a head, and demand $Rs 20$ for a tail. In other words, your gain (in Rs) from this deal is $10$ for head and $-20$ for tail. Both $10$ and $-20$ are constants, but since you do not know which of these two constants you are going to get, you gain is a variable. Since it varies with chance, we call it a random variable.

Notice that here we have a function from $\{head,~tail\}$ to $\{10,-20\}$ defined as $$\begin{eqnarray*} head & \mapsto & 10,\\ tail & \mapsto & -20. \end{eqnarray*}$$ There is nothing random about this function. The randomness comes from mechanism that decides what gois into this: head or tail?

We use this idea to define random variables mathematically. We start with a random experiment which is the provider of the randomness. Then any function defined on its sample space is called a random variable. To be precise, it is the function (which is not at all random) that is called the random variable. Thus, if in the above coin toss example, we replace the fair coin with a biased coin, but keep the payment rules the same, then we still have the same random variable.

Beginners often find it odd: a random variable is neither random nor a variable!

However, it is not as unnatural as it sounds. In calculus also we write $y = x^2$ and say $y$ is a variable as well as $y$ is a function of $x.$

EXAMPLE: In the coin tossing example with a fair coin, let your gain be denoted by $X.$ (or sometimes $X(w)$, if you want to emphasize that it is a function). Find $P(X=10).$

SOLUTION: The immediate answer is $\frac 12.$ Let's see the steps that led to this answer. $P(X=10)$ is the probability that $X$ is $10,$ i.e., the probability that the coin toss has produced an outcome for which the function $X$ takes the value $10.$ Thus $$ P(X=10) = P\big\{w\in\{head,tail\}~:~X(w)=10\big\}. $$ Now $\big\{w\in\{head,tail\}~:~X(w)=10\big\} = \{head\},$ and so the problem now reduces to finding $P(\{head\}),$ which is $\frac 12.$

The general case, then, looks like this: We have a random experiment with sample space $\Omega.$ A random variable $X$ is a function $X:\Omega\rightarrow S$ where $S$ is any codomain of our choice. If some one gives us some $A\subseteq S$ and asks us to find $P(X\in A),$ we are to actually find $$ P\big(\big\{w\in\Omega~:~X(w)\in A\big\}\big). $$ Remember that this is the definition of $P(X\in A).$ The complicated looking set $\big\{w\in\Omega~:~X(w)\in A\big\}$ is often abbreviated to $\{X\in A\}$ or $X ^{-1} (A).$
Earlier we had talked about "good" sets and "bad" sets. What if someone asks us to find $P(X\in A),$ where $X ^{-1} (A)$ is a "bad" subset of $\Omega?$ Well, the answer is: We shall simply refuse to find $P(X\in A)$ for such an $A.$ We shall call such an $A$ a "bad" subset of $S$ (w.r.t. this $X$). A subset $A\subseteq S$ is "good" or "bad" according as $X ^{-1} (A)$ is "good" or "bad" in $\Omega.$

EXAMPLE: A fair die is rolled. I shall pay you Rs 10 if the die shows an even number, you'll pay me Rs 20 otherwise. Again, let's denote by $X$ your gain (in Rs). Express $X$ as a function (as codomain you can take ${\mathbb R}$ or ${\mathbb C}$ or ${\mathbb Z}$ or $\{10,-20\}$ or any other superset of $\{10,-20\}$). Let $A = \{10\}.$ Find $X ^{-1} (A)$ and using it find $P(X\in A).$

SOLUTION: Here $X ^{-1}(A) = \{2,4,6\}.$ So $P(X=10) = P(\{2,4,6\}) = \frac 16+\frac 16+\frac 16 = \frac 12.$

In each of these examples we had a random variable $X$ that took only two values $10$ and $-20.$ Which $X$ do you think is more profitable for you? Well, both are actually the same so far as profit goes. Understand this carefully: the two different $X$'s are completely different as functions (their domains are also different), but in terms of the behaviour of the output of the functions they are identical. This behaviour is called the distribution of the random variable. It is the distribution which we care about mostly in real applications. So we often start a discussion as
Let $X$ be a random variable taking values $10$ and $-20$ each with probability $\frac 12.$
We understand implicitly that there is some random experiment (say the coin toss experiment or the die roll experiment or something similar) and some function from its sample space to $\{10,-20\}$ such that the distribution is as specified. In this course, we shall often omit the sample space or the function.

Combining different random variables

Sometimes we need to combine the values of two or more random variables. Say $X,Y$ are both random variables and we want to compute $X+Y.$ Since random variables are actually functions, so this sum can be formed only when $X$ and $Y$ have the same domain. This simple point sometimes needs careful handling as the following example shows.

EXAMPLE: I am playing against two gamblers simultaneosly. One gambler tosses a fair coin and pays Rs 10 for a head and takes Rs 20 for a tail. The other gambler takes Rs 3 from me, rolls a fair die and pays me as many rupees as the outcome. What is my total gain?

SOLUTION: If I call the gain from the first gambler $X,$ then $X$ is a function from $\{head,tail\}$ to ${\mathbb R},$ while the gain from the second gambler is a function $Y:\{1,2,3,4,5,6\}\rightarrow{\mathbb R}.$ Obviously, $X+Y$ does not make any sense here. We need to first combine the two random experiments to get the product sample space: $\{head,tail\}\times\{1,2,3,4,5,6\}$ and then consider $X,Y$ both as functions from $\Omega$ to ${\mathbb R}.$ For example, $X(head,4) = 10$ and $Y(head,4) = 4-3 = 1.$

Now it is meaningful to talk about $X+Y.$

Different types

Depending on the distribution, a random variable may be of 3 types: In this course we shall focus on discrete random variables only.

The distribution of a discrete random variable is completely specified by the countable set of values it can take, and the probability with which it takes each of those values. These two specifications together are called the probability mass function (PMF) of the rv.

Functions of a random variable

Any function of a random variable is again a random variable. This is immediate from the definition of a random variable (since composition of two functions is again a function).

Notice that any function of a discrete random variable must again be a discrete random variable.

CDF

Definition: Let $X$ be any real valued random variable. Then its (cumulative) distribution function (CDF) is defined as the function $F:{\mathbb R}\rightarrow[0,1]$ where $\forall x\in{\mathbb R}~~F(x) = P(X\leq x).$
This definition holds for all real-valued random variables (not just the discrete ones).

The following properties are easy to prove.
Theorem Let $F(x)$ be the CDF of some rv $X.$ Then
  1. $F(x)$ must be nondecreasing, i.e., $\forall x < y\in{\mathbb R}~F(x)\leq F(y).$
  2. $\lim_{x\rightarrow-\infty} F(x) = 0.$
  3. $\lim_{x\rightarrow\infty} F(x) = 1.$
  4. $F(x)$ must be right continuous, i.e., $\forall a\in{\mathbb R}~~F(a+)=F(a).$

Proof:

  1. Since $\{X\leq x\} \subseteq \{X\leq y\},$ hence $P(\{X\leq x\}) \leq P(\{X\leq y\}),$ i.e., $F(x)\leq F(y).$
  2. Shall show $$ \forall \epsilon>0 ~~ \exists M \in{\mathbb R} ~~ \forall x < M~~ |F(x)-0| < \epsilon. $$ (Actually we may drop the absolute value sign around $F(x)$ since it is anyway $\geq 0$).

    Take any $\epsilon>0.$

    Let $A_n$ be the event that $\{X \leq -n\}$ for $n\in{\mathbb N}.$ Then $F(-n) = P(A_n).$ Clearly, $A_1\supseteq A_2\supseteq A_3\supseteq\cdots$ and $\cap A_n=\phi.$

    So $P(A_n)\rightarrow 0,$ i.e., $F(-n)\rightarrow 0.$

    So $N\in{\mathbb N} ~~F(-N)<\epsilon.$

    Choose $M = -N.$

    Take any $x < M.$

    Then $0\leq F(x) \leq F(M)<\epsilon,$ since $F(\cdot)$ is nondecreasing.

    So $|F(x)-0| < \epsilon,$ as required.
  3. Very similar. Try yourself first, before clicking here.
  4. Shall show: $$ \forall a\in{\mathbb R}~~\forall \epsilon>0~~\exists \delta>0~~ \forall x\in (a,a+\delta)~~|F(x)-F(a)| < \epsilon. $$ Take any $a\in{\mathbb R}$ and any $\epsilon>0.$

    Let $A_n$ be the event that $\left\{X\leq a+\frac 1n\right\}$ for $n\in{\mathbb N}.$ Also let $A$ be the event that $\{X\leq a\}.$

    Try the rest yourself.
[QED]

A rather nontrivial theorem is that the converse is also true. This converse is called the fundamental theorem of probability.
Fundamental theorem of probability Let $F:{\mathbb R}\rightarrow[0,1]$ be any function satisfying the properties listed in the last theorem. Then there must exist a real-valued rv $X$ with this $F(x)$ as its CDF.

Proof:Too technical for this course.[QED]

Theorem Any nondecreasing function bounded from above (and hence all CDF's) must have finite left hand limit at each point.

Proof: Let $F:{\mathbb R}\rightarrow{\mathbb R}$ be nondecreasing and bounded from above.

Take any $a\in {\mathbb R}.$

We shall show that $\lim_{x\rightarrow a-} F(x)$ exists as a finite number, i.e., $$ \exists\ell\in{\mathbb R}~~\forall \epsilon>0~~\exists \delta>0~~\forall x\in(a-\delta,a)~~|F(x)-\ell|\leq\epsilon. $$ Consider the set $A=\{F(x)~:~x < a\}.$ Then $A\neq\phi$ and bounded from above (by $F(a)$).

So $\sup(A)\in{\mathbb R}.$

Choose $\ell = \sup(A).$

Take any $\epsilon>0.$

Then $\exists y < a~~F(y) > \ell-\epsilon.$

Choose $\delta = a-y > 0.$

Take any $x\in(a-\delta,a) = (y,a).$

Then $F(y)\leq F(x) \leq \ell,$ or, in other words, $\ell-\epsilon\leq F(x)\leq \ell.$

So $|F(x)-\ell|\leq \epsilon,$ as required. [QED]

Theorem Let $X$ have CDF $F.$ Then $$ \forall a\in{\mathbb R}~~F(a-) = P(X < a). $$

Proof: Take any $a\in{\mathbb R}.$

Let $A = \{X < a\}$ and let $A_n = \left\{X \leq a-\frac 1n\right\}$ for $n\in{\mathbb N}.$

Then $A_n\nearrow A.$

Hence $P(A_n)\rightarrow P(A).$

So $F\left(a-\frac 1n\right)\rightarrow P(A).$

But $F\left(a-\frac 1n\right)\rightarrow F(a-),$ since $F(a-)$ exists.

Hence $P(X < a) = F(a-),$ as required. [QED]

Theorem Let $X$ have CDF $F.$ Then $$ \forall a\in{\mathbb R}~~P(X=a) = F(a)-F(a-). $$

Proof: $P(X=a) = P(X\leq a)-P(X < a).$ [QED]

The following theorem justifies the adjective "continuous" for a random variable.
Theorem A random variable is continuous if and only if its CDF is continuous everywhere.

Proof: Obvious from the last theorem. [QED]

PMF

Definition: Probability Mass Function (PMF) Let $X$ be a discrete random variable taking values $x_1,x_2,...$ with probabilities $p_1,p_2,...$. Then the probability mass function (PMF) of $X$ is defined as $p(x)$ where $$ \forall i~~p(x_i) = p_i. $$
There is no clear guideline as to the choice of the domain of the PMF, except that it must be a superset of $\{x_1,x_2,...\}.$ If you take the domain to be a strict superset, then you define the PMF as $$ p(x) = \left\{\begin{array}{ll}p_i&\text{if }x=x_i\\0&\text{otherwise.}\end{array}\right.. $$ Clearly, $\sum p_i = 1$ and $\forall i~~p_i\geq 0.$ A consequence of the fundamental theorem of probability is that for any countable set $\{x_1,x_2,...\}$ and for any sequence $(p_i)_i,$ for which $\forall i~~p_i\geq 0$ and $\sum p_i=1,$ there is a (discrete) random variable of which the PMF is $$ p(x) = \left\{\begin{array}{ll}p_i&\text{if }x=x_i\\0&\text{otherwise.}\end{array}\right.. $$ The CDF of a discrete random variable is like a step function.

Expectation

For many random variables we see a striking example of statistical regularity. 
w = sample(6,1000,rep=T)
profit =c(-20,10,-20,10,-20,10)
X = profit[w]
avgX = cumsum(X)/(1:1000)
plot(avgX,ty='l')
In fact, it is this phenomenon that first let man to study probability. If you run a gambling game a large number of time the average profit becomes more and more stable. Gamblers wanted to guess this stable value beforehand. They argued as follows:
If I play this game a large number of times (say $n$ times), then approximately $\frac n2$ times I should get $10$ and the remaining $\frac n2$ times I should get $-20.$ So approximately my total gain would be approximately $$ \frac n2\times 10 + \frac n2\times (-20). $$ So the average would be approximately this divided by $n,$ i.e., $$ \frac 12\times 10 + \frac 12\times (-20) = -5. $$
Indeed, this simple argument turns out to be remarkably accurate. Gamblers could not understand why it becomes so accurate as $n$ becomes large. But they used this formula to find out what they could expect the random variable to do in the long run.
There is a wierd point in the definition of $E(X).$ We are working with $\sum p_i x_i.$ But then why is the condition on $\sum |p_i x_i|$? The reason for taking this absolute value should be clear from our crash course on infinite series.
Definition: Expectation Let a (discrete) random variable $X$ take values $x_1,x_2,...$ with probabilities $p_1,p_2,...$. If $\sum |p_i x_i| <\infty,$ then we define the expectation of $X$ as $$ E(X) = \sum p_i x_i. $$ If $\sum |p_i x_i| =\infty,$ then

Expectation of a function

EXAMPLE: Suppose I have a rv that takes values $-1,0$ and $1$ with probabilities $0.1, 0.5$ and $0.4,$ respectively. What is $E(X^2)?$

SOLUTION: Here $X^2$ is a new rv. Call it $Y,$ say. Then $Y$ takes values $0$ and $1$ with probabilities $0.5$ each.

So $E(Y) = \frac 12.$

Here is another technique to arrive at the same result. $$ E(X^2) = 0.1\times (-1)^2 + 0.5\times 0^2 + 0.4\times 1^2 = 0.5. $$ This technique is often easier because here we do not need to find the distribution of $Y=X^2$ first. Both these techniques will always give the same answer.
Theorem Let a (discrete) random variable $X$ take values $x_1,x_2,...$ with probabilities $p_1,p_2,...$. Let $h(\cdot)$ be any function defined on the set $\{x_1,x_2,...\}.$ If $\sum |p_i h(x_i)| <\infty,$ then we must have $$ E(h(X)) = \sum p_i h(x_i). $$ Also, if $\sum|p_i h(x_i)|=\infty$ and all but finitely many $h(x_i)$'s are $>0$ (resp, $<0$), then $E(h(X))=\infty $(resp, $-\infty$).

Proof: If $X$ takes only finitely many values, then the result follows from distributivity of multiplication over addition.

For the countably infinite case, the result follows from rearrangement property of absolutely convergent series. [QED]

Properties of expectation

Relation of $E(X)$ with values of $X$

Theorem If $X$ is a degenerate rv (i.e., takes only one value with probability 1), then $E(X)$ equals that value.
Here $X\leq Y$ means $$ \forall w\in\Omega~~X(w)\leq Y(w). $$
Theorem Let $X$ be a discrete random variable. Let $f(x), g(x)$ be functions such that $P(f(X)\leq g(X))=1$ and both $E(f(X))$ and $E(g(X))$ exist. Then $E(f(X))\leq E(g(X)).$

A similar result holds if $P(X\geq f(X))=1.$
An immediate consequence of the above theorems is the following theorem.
Theorem If $X$ always takes values in $[a,b],$ then $E(X)$ must exist finitely, and lie in $[a,b].$
The condition "$X$ always lies in $[a,b]$" may be written as $P(X\in[a,b])=1.$
Theorem Let $a\in{\mathbb R}$ be any number. If $P(X\leq a)=1,$ then $E(X)=a$ if and only if $X$ is degenerate at $a.$
However, if $a\in{\mathbb R}$ is replaced by $\infty,$ then the result fails, i.e., it is possible to have a random variable $X$ that is always finite (any real-valued random variable will do, since $\infty\not\in{\mathbb R}$) such that $E(X)=\infty.$

EXAMPLE: It is a standard fact that $\sum\frac{1}{n^2}<\infty.$ Let the sum be $c.$ (The exact value of $c$ which is $\frac{\pi^2}{6},$ is of no importance here).

Then consider a random variable $X$ that takes values in ${\mathbb N}$ and $P(X=n)=\frac{1}{cn^2}.$

Then $E(X) = \frac 1c\sum\frac 1n=\infty.$

By the way, if $X$ can take values $x_1,x_2,...$, there is no guaranty that $E(X)$ will equal any of the $x_i$'s. For example, if the $X$ is the outcome of a fair die, then $E(X)=3.5,$ which is not a possible outcome.

Transformation properties

Theorem Let $X$ be a discrete rv such that $E(X)$ is defined. If $a,b$ are constants, then $E(a+bX) = a+bE(X).$

Proof: Trivial. [QED]

EXERCISE: If $E(X) = \mu,$ then what is $E(X-\mu)?$

Theorem Let $X$ be a random variable and $f,g$ be two functions such that $E(f(x))$ and $E(g(x))$ both exist finitely.

Then $E(f(X)+ g(X))$ also exists finitely and we have $$ E(f(X)+ g(X)) = E(f(X))+ E(g(X)). $$
Next we shall need a new concept, that of a convex function. Graphically, $f(x)$ is a convex function if its graph is like a bowl opening upwards (possibly slanted). Some examples are shown below.
Mathematically we may define a convex function as follows.
While this definition is graphically quite intuitive, you may have seen other definitions of convexity elsewhere. Read here to learn more about equivalences between different definitions of convexity.
Definition: Convex function A function $f:{\mathbb R}\rightarrow{\mathbb R}$ is called convex if $\forall a\in{\mathbb R}$ there is a line $y = \ell_a(x)$ through $(a,f(a))$ that lies on or below the graph of $f(x),$ i.e., $$ \forall x\in{\mathbb R}~~ \ell_a(x) \leq f(x). $$
In the following diagram the blue line is $\ell_a.$ Both the red lines are candidates for $\ell_b.$
Jensen's inequality Let $X$ be a discrete rv such that $E(X)$ is defined. If $f:{\mathbb R}\rightarrow{\mathbb R}$ is any convex function for which $E(f(X))$ exists, then $f(E(X))\leq E(f(X)).$

Proof: Let $\mu = E(X).$ Consider $\ell_\mu(x)$ as mentioned in the definition of convexity.

Since the graph of $\ell_\mu(x)$ is a straight line passing through $(\mu,f(\mu)),$ hence it must be of the form $$ \ell_\mu(x) = f(\mu)+m(x-\mu),~~x\in{\mathbb R}. $$ So $$ E(f(X)) \leq E(\ell_\mu(X)) = E(f(\mu))+mE(X-\mu) = f(\mu)+0 = f(E(X)), $$ as required. [QED]

EXERCISE:Which is larger $(E(X))^2$ or $E(X^2)?$ Assume that both exist finitely.

Problems for practice

  1. What is $E(X)$ if $X$ takes the values $1,2,...,n$ with probability $\frac 1n$ each.

Comments

To post an anonymous comment, click on the "Name" field. This will bring up an option saying "I'd rather post as a guest."