Double Integral to compute conditional expectation

Question

Let $V=X+Y$ where $X$ and $Y$ are independent random variables, both normally distributed with $\mu=0$ and $\sigma=1$. I am interested in computing $E[X \mid V\geq c]$ where $c$ is an arbitrary constant. This is $$ E[X \mid V>c] = \frac{1}{1-\Phi_v(c)} \int_c^\infty \int_{-\infty}^{\infty} x \cdot f_X(x) \cdot f_Y(v-x) \quad dx ~dv $$ In Mathematica, with $c=0.5$, I am coding

1/(1 - CDF[NormalDistribution[0, 1], 0.5])
  Integrate[
  x PDF[NormalDistribution[0, 1], x] PDF[NormalDistribution[0, 1], 
    v - x], {x, -Infinity, Infinity}, {v, 0.5, Infinity}]

but this seems a very difficult problem for Mathematica to solve, taking a long time and sometimes not solving at all. If I use NIntegrate it is done instantly instead. My question is whether this is to be expected, or whether there is a way to do it symbolically in a more efficient way.

Answer 1

Thanks to @AlexTourneau's slight modification to my approach exploiting the power of Mathematica:

Clear[x, y, c];
Expectation[x \[Conditioned] x + y > c, 
  {x \[Distributed] NormalDistribution[],
  y \[Distributed] NormalDistribution[]}]

$$\frac{e^{-\frac{c^2}{4}}}{\sqrt{\pi } \text{erfc}\left(\frac{c}{2}\right)}$$