Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I try to resolve this problem, but I have some difficulties to get a clear result.

The problem :

Let X be a normal random variable with mean 0 and variance 1 (ie. $X\sim \mathcal{N}(0,1)$).

Let Y be a normal random variable with mean $m$ and variance $\sigma^{2}$ (ie. $Y\sim \mathcal{N}(m,\sigma^{2})$).

X and Y are independent random variables.

What I want is to compute $I=\mathbb{E}[\Phi(Y)]$ where $\Phi$ is the the cumulative distribution function (CDF) of $X$.

*What I done is wrong * Sorry for my english :)

share|improve this question
1  
I might be missing something, but don't quite get why are you considering $X$ as a random variable. $\phi(\cdot)$ is just a function like any other. If you want to compute $E(g(Y))$,that's just $\int g(y) f_Y(y) dy$ –  leonbloy Jan 11 '13 at 15:01

1 Answer 1

As leonbloy says, you just need to use $$E[\Phi(Y)] = \int_{-\infty}^\infty \Phi(y)\frac{1}{\sigma \sqrt{2\pi}}\exp\left(-\frac{(y-\mu)^2}{2\sigma^2}\right)\,\mathrm dy.$$ But, rather than trying to evaluate the integral directly, consider that we can write it as $$\begin{align*} E[\Phi(Y)] &= \int_{-\infty}^\infty \Phi(y)\frac{1}{\sigma \sqrt{2\pi}}\exp\left(-\frac{(y-\mu)^2}{2\sigma^2}\right)\,\mathrm dy\\ &= \int_{-\infty}^\infty P\{X \leq y\}f_Y(y)\,\mathrm dy\\ &= \int_{-\infty}^\infty P\{X \leq y \mid Y = y\}f_Y(y)\,\mathrm dy\\ &= P\{X \leq Y\}\\ &= P\{X-Y\leq 0\}. \end{align*}$$ What kind of random variable is $X-Y$? Can you find its mean and variance without doing any integrations? Can you write an expression for this probability in terms of $\Phi(\cdot)$?

share|improve this answer
1  
Thanks a lot ! $X-Y\sim \mathbb{N}(-m,1+\sigma^{2})$ because $X$ and $Y$ are independent. Then, $\mathbb{P}(X-Y\le 0) = \Phi\left(\frac{m}{\sqrt{1+\sigma^{2}}}\right)$. Your result is correct because I check it for simples values : If $m=0$ and $\sigma=1$, then, $\Phi(Y)\sim \mathcal{U}(0,1)$, $\mathbb{E}[\Phi(Y)]=\Phi(0)$. –  Gauss Jan 11 '13 at 16:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.