Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

August 2006

2

Let $\Phi$ be the standard normal distribution, i.e., the distribution function of a zero-mean, unit-variance Gaussian random variable. Let $\mathbf{X}$ be a normal random variable with mean $\mu$ and variance 1 . We want to find $E\left[\Phi\left(\mathbf{X}\right)\right]$ .

(a) (10 points)

First show that $E\left[\Phi\left(\mathbf{X}\right)\right]=P\left(\mathbf{Z}\leq\mathbf{X}\right)$ , where $\mathbf{Z}$ is a standard normal random variable independent of $\mathbf{X}$ . Hint: Use an intermediate random variable $\mathbf{I}$ defined as

$\mathbf{I}=\left\{ \begin{array}{lll} 1 & & \text{if }\mathbf{Z}\leq\mathbf{X}\\ 0 & & \text{if }\mathbf{Z}>\mathbf{X}. \end{array}\right.$

(b) (10 points)

Now use the result from Part (a) to show that $E\left[\Phi\left(\mathbf{X}\right)\right]=\Phi\left(\frac{\mu}{\sqrt{2}}\right)$ .

Let $\mathbf{Y}=\mathbf{Z}-\mathbf{X}$ . Since $\mathbf{Z}$ and $\mathbf{X}$ are Gaussian random variables, $\mathbf{Y}$ is also a Gaussian random variable.

$E\left[\mathbf{Y}\right]=E\left[\mathbf{Z}\right]-E\left[\mathbf{X}\right]=-\mu.$

$Var\left[\mathbf{Y}\right]=E\left[\left(\mathbf{Y}-E\left[\mathbf{Y}\right]\right)^{2}\right]=E\left[\left(\mathbf{Z}-\left(\mathbf{X}-\mu\right)\right)^{2}\right]=E\left[\mathbf{Z}^{2}\right]+E\left[\left(\mathbf{X}-\mu\right)^{2}\right]-2E\left[\mathbf{Z}\right]E\left[\mathbf{X}-\mu\right] $$=E\left[\mathbf{Z}^{2}\right]-E\left[\mathbf{Z}\right]E\left[\mathbf{X}-\mu\right]+E\left[\left(\mathbf{X}-\mu\right)^{2}\right]-E\left[\mathbf{Z}\right]E\left[\mathbf{X}-\mu\right]$$ =E\left[\mathbf{Z}^{2}\right]-\left(E\left[\mathbf{Z}\right]\right)^{2}+E\left[\left(\mathbf{X}-\mu\right)^{2}\right]-\left(E\left[\mathbf{X}-\mu\right]\right)^{2}=Var\left[\mathbf{Z}\right]+Var\left[\mathbf{X}\right]=2.$

$E\left[\Phi\left(\mathbf{X}\right)\right]=P\left(\left\{ \mathbf{Z}\leq\mathbf{X}\right\} \right)=P\left(\left\{ \mathbf{Y}\leq0\right\} \right)=\Phi\left(\frac{0-\left(-\mu\right)}{\sqrt{2}}\right)=\Phi\left(\frac{\mu}{\sqrt{2}}\right).$

## Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood