Q:

Expected Value Normal Distribution over an interval

The mean of a Normal distribution is $\theta$ and variance is 1. I know that $\text{E}(X)=\theta$. Then, if I compute the integral I would use to find $\text{E}(X)$ but instead I only take the integral from $(-a,a)$. How would I find the value of this integral over the specified interval. Is it also $\theta$? 

A:

$$\begin{align}
\int_{-a}^a x\cdot\frac{1}{\sqrt{2\pi}}
\exp\left(-\frac{(x-\theta)^2}{2}\right)\,\mathrm dx
&= \int_{-a}^a (x-\theta)\cdot\frac{1}{\sqrt{2\pi}}
\exp\left(-\frac{(x-\theta)^2}{2}\right)\,\mathrm dx\\
&\qquad + \int_{-a}^a \theta\cdot\frac{1}{\sqrt{2\pi}}
\exp\left(-\frac{(x-\theta)^2}{2}\right)\,\mathrm dx
\end{align}$$
An antiderivative of the integrand of the first integral
is $-\exp\left(-\frac{(x-\theta)^2}{2}\right)$ (work out the
derivative and check it if you don't believe this assertion)
and so is easily evaluated. The second integral can be expressed
in terms of $\Phi(\cdot)$, the cumulative probability distribution
function of the standard normal random variable.

The answer is not $\theta$ in general.

