Expected Value of a Function (Standard Normal)

[rp2000ap]

Homework Statement

Let Z be a standard normal random variable, f is a measurable function and a is a constant.
Show E[Zf(e^aZ)] = aE[e^aZ]E[f'(e^(a(Z+a))]

Homework Equations

The Attempt at a Solution

I set up an integral from negative to positive infinity of

z*f(e^aZ)*e^((-z^2)/2)*1/sqrt(2pie)*dz

My 2 questions are can I combine the z from the pdf of the normal distrbution function into f(e^aZ) and does the z in the beginning of my equation require integration by parts to evaluation the integral.

Thanks!

 

[mighty2000]

Hello!

Let's start by defining some terms and understanding what the problem is asking for. A standard normal random variable, Z, is a random variable with a mean of 0 and a standard deviation of 1. This means that the probability density function (pdf) of Z is given by f(z)=1/sqrt(2π)e^(-z^2/2). A measurable function, f, is a function that maps measurable sets to measurable sets. And a is a constant.

Now, the problem is asking us to show that the expected value of the random variable Z multiplied by the function f(e^aZ) is equal to the product of the expected value of e^aZ and the derivative of the function f evaluated at e^(a(Z+a)). In other words, we need to show that E[Zf(e^aZ)] = aE[e^aZ]E[f'(e^(a(Z+a)))]

To do this, we can start by writing out the expected value of Zf(e^aZ) in terms of the pdf of Z. This is given by:

E[Zf(e^aZ)] = ∫-∞∞ zf(e^aZ)f(z)dz

Now, we can use a change of variable to simplify this integral. Let u=e^aZ, then du=a*e^aZ*dZ. This means that dZ=du/(a*e^aZ) and we can rewrite our integral as:

E[Zf(e^aZ)] = ∫-∞∞ f(u)du/(a*e^aZ)

Next, we can use the fact that Z is a standard normal random variable, and substitute in the pdf of Z. This gives us:

E[Zf(e^aZ)] = ∫-∞∞ f(u)du/(a*e^aZ) = ∫-∞∞ f(u)e^(-u^2/2)/a*e^aZ du

Now, we can simplify this even further by using the fact that e^aZ is a constant with respect to the integral and can be pulled out. This gives us:

E[Zf(e^aZ)] = e^a ∫-∞∞ f(u)e^(-u^2/2)/a du

Next, we can use another change of variable, let v=u+a,