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.

A: I think there are two answers to this question because it is slightly ambiguous. I'm not sure whether you mean you wish to calculate:
$Pr(-a \le X \le a)$ 
or you wish to calculate 
$E(X | -a \le X \le a)$
In either case you would in general have different answers. In either case the answer would not be $\theta$ unless $\theta = 0 $. Simply imagine the case where $\theta = 110$ and we choose $a = 100$, we can imagine conceptually that the expectation would not hold. 

Now, in above you might be confused to why I said "do you wish to calculate $Pr(-a \le X \le a)$", because this is what I gather you were trying to calculate from your question. This is because we cannot simply just figure out the expectation by bounding it. Instead we will have to renormalise the data first before calculating the expected value. 
Regardless, we should renormalise the distribution to our $Z \sim N(0,1)$.
Since we have $X \sim N(\theta, 1)$, we then have:
$ Z = X - \theta \implies X = Z+\theta$ 
For the first case, we have:
$Pr(-a \le X \le a) = Pr(-a \le Z + \theta \le a) $ 
$ = Pr(- (a+\theta) \le Z \le a - \theta) = \Phi(a-\theta) - \Phi(-(a+\theta))$
In the 2nd case, we will have to renormalise the distribution to calculate the required expectation. In that instance we can simply refer to wikipedia for the truncated normal distribution, 
$E(X | -a \le X \le a) = \theta + \frac{\phi(-a-\theta)-\phi(a-\theta)}{\Phi(a-\theta)-\Phi(-a-\theta)}$
