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Random Variables and Signals

Topic 9: Expectation

Thus far, we have learned how to represent the probabilistic behavior or random variables X using the density function f $$ _X $$ or the mass function p $$ _X $$ .
Sometimes, we want to describe X probabilistically using only a small number of parameters. The expectation is often used to do this.

Definition $\qquad$ the expected value of continuous random variable X is defined as

E[X) = \int_{-\infty}^{\infty}xf_X(x)dx

Definition $\qquad$ the expected value of discrete random variable X is defined as

E[X] = \sum_{x\in\mathcal R_X}xp_X(x)

where $$ R_X $$ is the range space of X.

Note:

E[X] is also known as the mean of X. Other notation for E[X] include:

EX,\;\overline{X},\;m_X,\;\mu_X

The equation defining E[X] for discrete X could have been derived from the continuous X, using the density function f $$ _X $$ containing $\delta$ -functions.

Example $\qquad$ X is an exponential random variable. find E[X].

f_X(x) = \lambda e^{-\lambda x}u(x) \

\begin{align} E[X] &= \int_{-\infty}^{\infty}xf_X(x)dx \\ &= \int_{0}^{\infty}x\lambda e^{-\lambda x}dx \\ &= \frac{1}{\lambda} \end{align}

Let $\mu = 1/\lambda$ . We often write

f_X(x) = \frac{1}{\mu} e^{-\frac{1}{\mu}x}u(x) \

Example $\qquad$ X is a uniform discrete random varibable with $$ R_X $$ = {1,...,n}. Then,

\begin{align} E[X]&=\sum_{k=1}^n\frac{k}{n} \\ &=\frac{1}{n}\sum_{k=1}^n k \\ \\ &= \frac{1}{n}(\frac{1}{2})(n)(n+1) \\ \\ &=\frac{n+1}{2} \end{align}

Having defined E[X], we will now consider more general E[g(X)] for a function g:R → R.

Let Y = g(X). What is E[Y]? From previous definitions:

E[Y]=\int_{-\infty}^{\infty}yf_Y(y)dy

or

E[Y] = \sum_{y\in\mathcal R_Y}yp_Y(y)

We can find this by first finding f $$ _Y $$ or p $$ _Y $$ in terms of g and f $$ _X $$ or p $$ _X $$ . Alternatively, it can be shown that

E[Y]=E[g(X)]=\int_{-\infty}^{\infty}g(x)f_X(x)dx

or

E[Y] = E[g(X)]=\sum_{y\in\mathcal R_X}g(x)p_X(x)

See Papoulis for the proof of the above.

Two important cases or functions g:

g(x) = x. Then E[g(X)] = E[X]
g(x) = (x - $\mu_X)^2$ . Then E[g(X)] = E[(X - $\mu_X)^2$ ]

E[g(X)] = \int_{-\infty}^{\infty}(x-\mu_X)^2f_X(x)dx

or

E[g(X)] = \sum_{x\in\mathcal R_x}(x-\mu_X)^2p_X(x)

Note: $\qquad$ E[(X - $\mu_X)^2$ ] is called the variance of X and is often denoted $\sigma_X$ $$ ^2 $$ . $\sigma_X$ is called the standard deviation of X.

Important property of E[]:
Let g $$ _1 $$ :R → R; g $$ _2 $$ :R → R; $\alpha,\beta$ ∈ R, Then

E[\alpha g_1(X) +\beta g_2(X)] = \alpha E[g_1(X)]+\beta E[g_2(X)] \

So E[] is a linear operator. The proof follows from the linearity of integration.

Important property of Var():

Var(X) = E[X^2]-\mu_X^2

Proof:

\begin{align} E[(X-\mu)^2]&=E[X^2-2X\mu_X+\mu_X^2] \\ &=E[X^2]-2\mu_XE[X]+E[\mu_X^2] \\ &=E[X^2]-2\mu_X^2+\mu_X^2 \\ &=E[X^2]-\mu_X^2 \end{align}

Example $\qquad$ X is Gaussian N( $\mu,\sigma^2$ ). Find E[X} and Var(X).

E[X] = \int_{-\infty}^{\infty}\frac{x}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx

Let r = x - $\mu$ . Then

E[X] = \int_{-\infty}^{\infty}\frac{r}{\sqrt{2\pi}\sigma}e^{-\frac{r^2}{2\sigma^2}}dr\;+\; \mu\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{r^2}{2\sigma^2}}dr

First term: Integrating an odd function over (-∞,∞) ⇒ first term is 0.
Second term: Integrating a Gaussian pdf over (-∞,∞) gives one ⇒ second term is $\mu$ .
So E[X] = $\mu$

E[X^2] = \int_{-\infty}^{\infty}\frac{x^2}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx

Using integration by parts, we see that this integral evaluates to $\sigma^2+\mu^2$ . So,

Var(X) = \sigma^2+\mu^2-\mu^2 = \sigma^2

Example $\qquad$ X is Poisson with parameter $\lambda$ . Find E[X] and Var(X).

\begin{align} E[X] &= \sum_{k=0}^{\infty}k\frac{e^{-\lambda}\lambda^k}{k!} \\ &= \sum_{k=1}^{\infty}\frac{e^{-\lambda}\lambda^k}{(k-1)!} \\ &= \lambda\sum_{k=0}^{\infty}e^{-\lambda}\frac{\lambda^k}{k!} \\ &= \lambda \end{align}

\begin{align} E[X^2] &= \sum_{k=0}^{\infty}k^2\frac{e^{-\lambda}\lambda^k}{k!} \\ &= \sum_{k=0}^{\infty}(k+1)\frac{e^{-\lambda}\lambda^{k+1}}{k!} \\ &= \lambda\sum_{k=0}^{\infty}\frac{ke^{-\lambda}\lambda^k}{k!}\;+\;\lambda\sum_{k=0}^{\infty}\frac{e^{-\lambda}\lambda^k}{k!} \\ \\ &= \lambda E[X] + \lambda(1) \\ &= \lambda^2+\lambda \end{align}

So,
$E[X^2] = \lambda^2 +\lambda \$
$\Rightarrow Var(X) = \lambda^2 +\lambda - \lambda^2 = \lambda \$

Moments

Moments generalize mean and variance to nth order expectations.

Definition $\qquad$ the nth order moment of random variable X is

\mu_n\equiv E[X^n]=\int_{-\infty}^{\infty}x^nf_X(x)dx\quad n=1,2,...

and the nth central moment of X is

v_n\equiv E[(X-\mu_X)^n] = \int_{-\infty}^{\infty}(x-\mu_X)^nf_X(x)dx\qquad n = 2,3,...

So

$\mu_1$ = E[X] mean
$\mu_2$ = E[X $$ ^2 $$ ] mean-square
v $$ _2 $$ = Var(X) variance

Conditional Expectation

For an event M ∈ F with P(M) > 0.

E[g(X)|M] = \int_{-\infty}^{\infty}g(x)f_X(x|M)dx

or

E[g(X)|M] = \sum_{x\in\mathcal R_x}g(x)p_X(x|M)dx

Example $\qquad$ X is an exponential random variable. Let M = {X > $\mu$ }. Find E[X|M]. Note that P(M) = P(X > $\mu$ ) and since $\mu$ > 0,

P(M) = P(X>\mu) =\int_{\mu}^{\infty}\frac{1}{\mu}e^{-\frac{x}{\mu}}dx \;>\;0

It can be shown that

f_X(x|X>\mu) = \frac{1}{\mu}e^{-\frac{x-\mu}{\mu}}u(x-\mu)

Then,

\begin{align} E[X|X>\mu] &=\int_{\mu}^{\infty}\frac{x}{\mu}e^{-\frac{x-\mu}{\mu}} \\ &=2\mu \end{align}

Fig 1: Conditional Expectation; X is exponential.

References

M. Comer. ECE 600. Class Lecture. Random Variables and Signals. Faculty of Electrical Engineering, Purdue University. Fall 2013.

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