Difference between revisions of "ECE600 F13 Joint Expectation mhossain" - Rhea

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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

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

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
  

• The covariance of X and Y
  

• The correlation coefficient of X and Y is
  

• 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,

with equality iff Y = a$ _0 $ 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

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):

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

Case (ii):

In this case

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

This means that ∃a ∈ R such that

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

With the same notion of equality, we would have that

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

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:

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

and the joint central moments are

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

Joint Characteristic Function

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

for $ \omega_1,\omega_2 $ ∈ R.

If X and Y are continuous, we write this as

If X and Y are discrete, we use

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

Then it can be shown that

References

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

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