Expectation of a Noisy Truncated Normal Let $X_1$ and $X_2$ be independent normally distributed random variables.  Does this conditional expectation have a closed form.
$$
\mathbb{E}\left[X_1 \mid \alpha < X_1 + X_2 < \beta \right]
$$
This expression corresponds to the expectation of a random variable which is imperfectly truncated.  As the variance of $X_2$ decreases, it converges to the standard truncated normal distribution.
I tried putting it into Mathematica as:
Expectation[
 x \[Conditioned] \[Alpha] < x + e < \[Beta], {x, e} \[Distributed] 
  MultinormalDistribution[{\[Mu], 
    0}, {{\[Sigma]x^2, 0}, {0, \[Sigma]e^2}}]]

but got nothing.  There's an old post here that characterizes the sum $X_1'$ and $X_2'$ where $X_1'$ is truncated and $X_2'$ is independent of $X_1'$.  In my case, the truncation is two-sided, and $X_2'$ is correlated with $X_1' \equiv X_1 + X_2$.  However, it's possible that MathStatica still produces an answer -- I don't have a copy to try.
EDIT:
I worked on this a bit more and got an expression for the the case of standard normals $X_1$ and $X_2$: $$\mathbb{E}\left[X_1 * \mathbb{1}_{\alpha < X_1 + X_2 < \beta}\right] = \frac{1}{2 \sqrt{\pi}} \left(\exp\left(-\frac{\alpha^2}{4}\right) - \exp \left(-\frac{\beta^2}{4}\right)\right).$$ However, the expectation of the indicator function $(\mathbb{E}\left[\mathbb{1}_{\alpha < X_1 + X_2 < \beta}\right])$ can't be evaluated.  This was a bit surprising, but that's what I've got thus far.
 A: Alright I ended up working it out myself this morning.  Here I'm using $X$ and $Y$ for the random variables and $\left(\mu_i,\sigma_i\right)$ for the parameters of distribution for random variable $i$. I found it helpful to think about a change of variables $X = \mu_X + \sigma_X V$ where $V \sim \mathcal{N}\left(0, 1\right)$
The expectation $$E\left[X \mid \alpha < X + Y < \beta\right]$$ then becomes $$\mu_X + \sigma_X E \left[V \mid \alpha < \mu_X + \sigma_X V + Y < \beta\right].$$ This normalized expectation is given by $$\frac{\int_{\mathbb{R}^2} v\,f_{VY}\left(v, y\right) \mathbb{1}_{\left[\alpha - \mu_X,\beta-\mu_X\right)}\left(\sigma_X v+y\right) dv\,dy}{\int_{\mathbb{R}^2} f_{VY}\left(v, y\right) \mathbb{1}_{\left[\alpha-\mu_X,\beta-\mu X\right)}\left(\sigma_X v+y\right) dv\,dy} \tag{1}$$
In the case where $V$ and $Y$ are independent, the denominator of (1), which is the probability that $\alpha - \mu_X < \sigma_X V +Y < \beta - \mu_X$, is $$ F_Z\left(\beta\right) - F_Z\left(\alpha\right) $$ where $Z$ is a normal distribution with parameters $\mu_Z \equiv \mu_X + \mu_Y$ and $\sigma_Z \equiv \sqrt{\sigma_X^2 + \sigma_Y^2}$.  In terms of the cdf of the standard normal, this is $$ \Phi\left(\frac{\beta - \mu_Z}{\sigma_Z}\right) - \Phi\left(\frac{\alpha - \mu_Z}{\sigma_Z}\right)$$
The numerator of (1) can be rewritten as $$ \int_{\mathbb{R}^2} v\,\phi\left(v\right) \phi\left(\frac{y - \mu_Y}{\sigma_Y}\right) \mathbb{1}_{\left[\alpha-\mu_X,\beta-\mu_X\right)}\left(\sigma_X v+y\right)\,dv\,dy$$
$$ \int_\mathbb{R} v\,\phi\left(v\right) \int_{\alpha-}^{\beta-x} \phi\left(\frac{y - \mu_Y}{\sigma_Y}\right)  \,dy\,dx $$
$$ \int_\mathbb{R} v\,\phi\left(v\right) \left(\Phi\left(\frac{\beta - \left(\mu_X + \sigma_X v + \mu_Y\right)}{\sigma_Y}\right) - \Phi\left(\frac{\alpha - \left(\mu_X + \sigma_X v + \mu_Y\right)}{\sigma_Y}\right)\right) dx \tag{*}$$
Here I used the result
$$ \int_\mathbb{R} v\,\phi\left(v\right) \Phi\left(c_0 - c_1 v\right) dv = -\frac{c_1 e^{-\frac{c_0^2}{2 c_1^2+2}}}{\sqrt{2 \pi } \sqrt{c_1^2+1}} = -\frac{c_1}{\sqrt{c_1^2+1}} \phi\left(\frac{c_0}{\sqrt{c_1^2+1}}\right), \quad c_1 > 0 $$
to evaluate (*)
$$ (*) = \frac{\sigma_X}{\sqrt{\sigma_X^2 + \sigma_Y^2}} \left(\phi\left(\alpha' \right) - \phi\left(\beta' \right)\right),$$
where $\alpha' = \frac{\alpha - \left(\mu_X + \mu_Y\right)}{\sigma_Z} $ and $\beta' = \frac{\beta - \left(\mu_X + \mu_Y\right)}{\sigma_Z} $
So the final answer is
$$ E\left[X \mid \alpha < X + Y < \beta\right] = \mu_X + \sigma_X \frac{\sigma_X}{\sqrt{\sigma_X^2 + \sigma_Y^2}}\left(\phi\left(\alpha'\right) - \phi\left(\beta'\right)\right)\left(\Phi\left(\beta'\right) - \Phi\left(\alpha'\right)\right)^{-1}$$
By inspection, this is the expectation of a truncated normal variable with mean $\mu_X$ and standard deviation $\frac{\sigma_X}{\sqrt{\sigma_X^2 + \sigma_Y^2}} \sigma_X$, where the two sided truncation occurs at $\left(1 - \frac{\sigma_X^2}{\sigma_Z^2}\right) \mu_X + \frac{\sigma_X^2}{\sigma_Z^2} \left(\alpha - \mu_Y\right)$ and  $\left(1 - \frac{\sigma_X^2}{\sigma_Z^2}\right) \mu_X + \frac{\sigma_X^2}{\sigma_Z^2} \left(\beta - \mu_Y\right)$.

As a final note, the case of correlated normal variables $X$ and $Y$ can be accounted for by rotating $X$ and $Y$ so that the transformed variables $X'$ and $Y'$ are independent, and then taking a linear combination of the expectations of $X'$ and $Y'$.
A: The conditional expectation $E[X_1\mid \alpha<X_1+X_2<\beta]$ should be a function of $X_2$. So this the answer I propose :
$$E[X_1\mid \alpha<X_1+X_2<\beta] = E[X_1\mid \alpha-X_2 < X_1 < \beta-X_2] = \mu_{X_1} + \frac{\phi(a)-\phi(b)}{\Phi(b)-\Phi(a)}\sigma_{X_1}\,,$$ where $a = \frac{\alpha-X_2-\mu_{X_1}}{\sigma_{X_1}}$ and $b = \frac{\beta-X_2-\mu_{X_1}}{\sigma_{X_1}}$.
