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The Comer Lectures on Random Variables and Signals

Slectures by Maliha Hossain


Topic 14: Joint Expectation



Joint Expectation

Given random variables X and Y, let Z = g(X,Y) for some g:R$ _2 $→R. Then E[Z] can be computed using f$ _Z $(z) or p$ _Z $(z) in the original definition of E[ ]. Or, we can use

$ E[Z] = \int\int_{\mathbb R^2}g(x,y)f_{XY}(x,y)dxdy $

or

$ E[Z]=\sum_{x\in\mathcal R_x}\sum_{y\in\mathcal R_y}g(x,y)p_{XY}(x,y) $

The proof is in Papoulis.

We will use joint expectation to define some important moments that help characterize the joint behavior of X and Y.

Note that joint expectation is a linear operator, so if g$ _1 $,...,g$ _n $ are n functions from R$ ^2 $ to R and a$ _1 $,...,a$ _n $ are constants, then

$ E\left[\sum_{i=1}^na_ig_i(X,Y)\right]=\sum_{i=1}^na_iE[g_i(X,Y)] $



Important Moments for X,Y

We still need $ \mu_X $, $ \mu_Y $, $ \sigma_X $$ ^2 $, $ \sigma_X $$ ^2 $, the means and variances of X and Y. Other moments that are of great interest are:

  • The correlation between X and Y is
$ \mbox{Corr}(X,Y)\equiv E[XY] $
  • The covariance of X and Y
$ \mbox{Cov}(X,Y)\equiv E[(X-\mu_X)(Y-\mu_Y)] $
  • The correlation coefficient of X and Y is
$ r_{XY}\equiv\frac{\mbox{Cov}(X,Y)}{\sigma_X\sigma_Y} $
  • Note:
    • |r$ _{XY} $| ≤ 1 (proof)
    • If X and Y are independent, then r$ _{XY} $ = 0. The converse is not true in general.
  • If r$ _{XY} $=0, then X and Y are said to be uncorrelated. It can be shown that
    • X and Yare uncorrelated iff Cov(X,Y)=0 (proof).
    • X and Y are uncorrelated iff E[XY] = $ \mu_X\mu_Y $ (proof).
  • X and Y are orthogonal if E[XY]=0.



The Cauchy Schwarz Inequality

For random variables X and Y,

$ |E[XY]|\leq\sqrt{E[X^2]E[Y^2]} $

with equality iff Y = a$ _0 $X with probability 1, where a$ _0 $ is a constant. Note that "equality with probability 1" will be defined later.

Proof: $ \qquad $ We start by considering

$ 0\leq E[(aX-Y)^2]=a^2E[X^2]-2aE[XY]+E[Y^2] $

which is a quadratic function of a ∈ R. Consider two cases:

(i)$ \; $ E[(aX-Y)^2] > 0
(ii) E[(aX-Y)^2] = 0

Case (i):

$ E[X^2]a^2-2E[XY]a+E[Y^2]>0 \ $


Fig 1: A possible depiction of the quadratic when the discriminant is greater than zero.

Since this quadratic is greater than 0 for all a, there are no real roots. There must be two complex roots. From the quadratic equation, this means that

$ E[XY] < \sqrt{E[X^2]E[Y^2]} $

Case (ii):

$ E[(aX-Y)^2]=0 \ $

In this case

$ E[X^2]a^2-2E[XY]a+E[Y^2]=0 \ $


Fig 2: A possible depiction of the quadratic when the discriminant is equal to zero.


This means that ∃a ∈ R such that

$ E[(a_0X-Y)^2] = 0 \ $

It can be shown that if a random variable X has E[X$ ^2 $]=0, then X=0 except possibly on a set of probability 0. Note that previously, we have defined equality between random variables X and Y to mean

$ X=Y\;\mbox{iff}\;X(\omega)=Y(\omega)\;\forall\omega\in\mathcal S $

With the same notion of equality, we would have that

$ X=0\;\mbox{iff}\;X(\omega)=0\;\forall\omega\in\mathcal S $

However, the result for the case E[X$ ^2 $]=0 is that X=0 with probability 1, which means that there is a set A = {$ \omega $S: X($ \omega $) = 0} with P(A) = 1.

Returning to the case E[(a$ _0 $X-Y)$ ^2 $]=0, we get that Y = a$ _0 $X for some a$ _0 $R, with probability 1.
Thus we have shown that

$ E[XY]=\sqrt{E[X^2]E[Y^2]} $

with equality iff Y = a$ _0 $X with probability 1.



The Correlation Coefficient for Jointly Gaussian Random Variables X and Y

We have previously seen that the joint density function for Gaussian X and Y contains a parameter r. It can be shown that this r is the correlation coeffecient of X and Y:

$ r = \frac{E[XY]-\mu_X\mu_Y}{\sigma_X\sigma_Y} $

Note that one way to show this is to use a concept called iterated expectation, which we will cover later.



Joint Moments

Definition $ \qquad $ The joint moments of X and Y are

$ \mu_{jk}\equiv E[X^jY^k] $

and the joint central moments are

$ \sigma_{jk}=E[(X-\overline{X})^j(Y-\overline{Y})^k] $

for j = 0,1,...; k = 0,1,...



Joint Characteristic Function

Definition $ \qquad $ The joint characteristic function of X and Y is

$ \Phi_{XY}(\omega_1,\omega_2)\equiv E[e^{i(\omega_1X+\omega_2Y)}] $

for $ \omega_1,\omega_2 $R.

If X and Y are continuous, we write this as

$ \Phi_{XY}(\omega_1,\omega_2)=\int\int_{\mathbf R^2}e^{i(\omega_1x+\omega_2y)}f_{XY}(x,y)dxdy $

If X and Y are discrete, we use

$ \Phi_{XY}(\omega_1,\omega_2)=\sum_{x\in\mathcal R_x}\sum_{y\in\mathcal R_y}e^{i(\omega_1x+\omega_2y)}p_{XY}(x,y) $

Note:

$ \bullet\; \Phi_X(\omega)=\Phi_{XY}(\omega,0);\quad\Phi_Y(\omega)=\Phi_{XY}(0,\omega) $
(proof)
$ \bullet\; Z = aX+bY \;\Rightarrow\;\Phi_Z(\omega) =\Phi_{XY}(a\omega,b\omega) $
(proof)
$ \bullet\; X \perp\!\!\!\perp Y \;\mbox{iff}\;\Phi_{XY}(\omega_1,\omega_2)=\Phi_X(\omega_1)\Phi_Y(\omega_2) $
(proof)
$ \bullet\; X \perp\!\!\!\perp Y \;\mbox{and}\;Z=X+Y\;\Rightarrow\Phi_{Z}(\omega)=\Phi_X(\omega)\Phi_Y(\omega) $
(proof)



Moment Generating Function

The joint moment generating function of X and Y is

$ \phi_{XY}(s_1,s_2)=E[e^{s_1X+s_2Y}];\qquad s_1,s_2\in\mathbb C $

Then it can be shown that

$ \begin{align} \mu_{jk}&=E[X^jY^k] \\ &=\frac{\partial^j\partial^k}{\partial s_1^j\partial s_2^k}\phi_{XY}(s_1,s_2)|_{s_1=0,s_2=0} \end{align} $



References



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