m (Protected "ECE600 F13 Expectation mhossain" [edit=sysop:move=sysop])
 
(3 intermediate revisions by 2 users not shown)
Line 5: Line 5:
 
[[ECE600_F13_rv_Functions_of_random_variable_mhossain|Previous Topic: Functions of a Random Variable]]<br/>
 
[[ECE600_F13_rv_Functions_of_random_variable_mhossain|Previous Topic: Functions of a Random Variable]]<br/>
 
[[ECE600_F13_Characteristic_Functions_mhossain|Next Topic: Characteristic Functions]]
 
[[ECE600_F13_Characteristic_Functions_mhossain|Next Topic: Characteristic Functions]]
 
+
----
 +
[[Category:ECE600]]
 +
[[Category:probability]]
 +
[[Category:lecture notes]]
 +
[[Category:slecture]]
  
 
<center><font size= 4>
 
<center><font size= 4>
'''Random Variables and Signals'''
+
[[ECE600_F13_notes_mhossain|'''The Comer Lectures on Random Variables and Signals''']]
 
</font size>
 
</font size>
 +
 +
[https://www.projectrhea.org/learning/slectures.php Slectures] by [[user:Mhossain | Maliha Hossain]]
 +
  
 
<font size= 3> Topic 9: Expectation</font size>
 
<font size= 3> Topic 9: Expectation</font size>
 
</center>
 
</center>
 
  
 
----
 
----
 
+
----
  
 
Thus far, we have learned how to represent the probabilistic behavior of a random variable X using the density function f<math>_X</math> or the mass function p<math>_X</math>. <br/>
 
Thus far, we have learned how to represent the probabilistic behavior of a random variable X using the density function f<math>_X</math> or the mass function p<math>_X</math>. <br/>

Latest revision as of 11:12, 21 May 2014

Back to all ECE 600 notes
Previous Topic: Functions of a Random Variable
Next Topic: Characteristic Functions


The Comer Lectures on Random Variables and Signals

Slectures by Maliha Hossain


Topic 9: Expectation



Thus far, we have learned how to represent the probabilistic behavior of a random variable 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 that for 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}\left(\frac{1}{2}\right)(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:RR.

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 the expectation of Y 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 a 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 $. The positive square root, denoted $ \sigma_X $, is called the standard deviation of X.

Important property of E[]:
Let g$ _1 $:RR; g$ _2 $:RR; $ \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_X)^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 (proof), 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 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



Questions and comments

If you have any questions, comments, etc. please post them on this page



Back to all ECE 600 notes
Previous Topic: Functions of a Random Variable
Next Topic: Characteristic Functions

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett