Sum of two independent Gaussian r.v.s. given that one is negative Let $X \sim \mathcal N(0, \sigma_x^2)$ and $Y \sim \mathcal N(0, \sigma_y^2)$ be two independent r.v.s.
Is there a simple solution for $p(X+Y = a ~|~ X \leq 0)$?
I'm aware that without the conditioning on $X \leq 0$ we have that $Z = X+Y \sim \mathcal N(0, \sigma_x^2 + \sigma_y^2)$, but for me it seems that conditioning on $X \leq 0$ makes the problem considerably harder.
Here's what I have so far:
\begin{equation}
    \begin{split}
    p(X+Y=a ~|~ X \leq 0) &= \frac{p(X+Y=a, X \leq 0)}{P(X \leq 0)} \\
                          &= 2\int_{-\infty}^0 p(X+Y=a, X=b) db \\
                          &= 2 \int_{-\infty}^0 p(Y=a-b) ~ p(X=b) db \\
                          & \quad\quad\quad\quad \vdots \quad \text{(following the proof for the non-conditioned case)}\\
                          &= 2 ~ \frac{1}{\sqrt{2 \pi (\sigma_x^2 + \sigma_y^2)}} \exp \left(- \frac{a^2}{2 (\sigma_x^2 + \sigma_y^2)} \right) \int_{-\infty}^0 \frac{1}{\sqrt{2 \pi \frac{\sigma_x^2 \sigma_y^2}{\sigma_x^2 + \sigma_y^2}}} \exp \left(- \frac{\left(b - \frac{\sigma_x^2 a}{\sigma_x^2 + \sigma_y^2}\right)^2}{2 \frac{\sigma_x^2 \sigma_y^2}{\sigma_x^2 + \sigma_y^2}} \right) db \\
                          &= 2 ~ p(X+Y=a) \int_{-\infty}^0 \frac{1}{\sqrt{2 \pi \frac{\sigma_x^2 \sigma_y^2}{\sigma_x^2 + \sigma_y^2}}} \exp \left(- \frac{\left(b - \frac{\sigma_x^2 a}{\sigma_x^2 + \sigma_y^2}\right)^2}{2 \frac{\sigma_x^2 \sigma_y^2}{\sigma_x^2 + \sigma_y^2}} \right) db \\
                          &= 2 ~ p(X+Y=a)~p \left(W \geq \frac{\sigma_x a}{\sigma_y \sqrt{\sigma_x^2 + \sigma_y^2}} \right), \quad\quad W \sim \mathcal N(0,1)\\
    \end{split}
\end{equation}
I'm trying to simplify this further such that its entropy can be 'neatly' computed.

In your third equality line, the integrand is zero as these are absolutely continuous random variables. Dont you get an answer of zero just from that?
