# Expectation of a function of a normally distributed random variable

Consider that I have to produce this result:

$$E[u(W_0+r(\theta))] = u(W_0)+\theta-\frac 12\rho\sigma^2$$

From this:

$$E[u(W_0+r(\theta))] = \int_{-\infty}^\infty u(w_0+r) \frac{1}{\sigma \sqrt{2\pi}} \exp\left(-\frac{(r-\theta)^2}{2 \sigma^2}\right) dr$$

And:

$$\int_{-\infty}^\infty \frac{1}{\sigma \sqrt{2\pi}} \exp\left(-\frac{(r-\theta)^2}{2 \sigma^2}\right) dr = 1$$

Also: $u(w) = -\exp(-\rho w)$

And: $r(\theta) \sim N(\theta, \sigma^2)$

I have reached the point where I removed the constants, and realize that I have both a positive exponent and negative exponent, and I believe no way to combine them.

Help is much appreciated. Thanks.

• there is a ) missing in the first equation on the left side, where does it go? Commented Jan 26, 2016 at 17:19
• Woops. Added the bracket. It goes next to the r(theta) Commented Jan 26, 2016 at 17:39
• notice that the integrals on the RHS of eqn2 and lhs of eqn3 are the same. is that the intent? Commented Jan 26, 2016 at 17:48
• notation $r(\theta)$ is problematic - it denotes a function on the lhs of eqns 1 and 2, while inside the integrals you compute $(r-\theta)^2$ while integrating $dr$ -- what is the connection? Commented Jan 26, 2016 at 17:50
• Made edit to equation 2. I left out the $u(w_0+r)$ by accident. My apologies. Commented Jan 26, 2016 at 18:07

Hint: Complete the square of the exponent in the integral by using $$2\rho\sigma^2 r + r^2-2r\theta = (r+\rho\sigma^2-\theta)^2-(\rho\sigma^2-\theta)^2$$
An "alternative" way to see the solution without explicitly doing the integral, since you are given a Gaussian random variable $R\sim\mathcal N(\theta, \sigma^2)$, is to use its moment generating function (m.g.f) which is well known, so that for some constant $w_0$, $$\mathbb E[u(w_0+R)]=\mathbb E[(-1)e^{-\rho(w_0+R)}]=-\mathbb E[e^{-\rho(w_0+R)}]\\[3ex]=-e^{-\rho w_0}\mathbb E[e^{-\rho R}]\color{blue}{\stackrel{m.g.f.}{=}}-e^{-\rho w_0}e^{-\rho\theta+1/2\rho^2\sigma^2}=-e^{-\rho(w_0+\theta-1/2\rho\sigma^2)}=\color{red}{u(w_0+\theta-1/2\rho\sigma^2)}$$ The answer, in red, is what I think your answer should be.
Random variables should be written in capital letters and non-random variables in lower case. When computing an expectation with respect to some random variable, the variable of integration should be lower case. So, for instance, I interpret what you wrote as $$\mathbb E[u(w_0+R)]=\frac1{\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty}u(w_0+r)e^{-\frac12(r-\theta)^2/\sigma^2}\mathrm dr$$ for $R\sim\mathcal N(\theta, \sigma^2)$.