Difference between revisions of "ECE PhD QE CNSIP 2006 Problem1.2" - Rhea

Latest revision as of 09:36, 10 March 2015

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). $