<s>[INST] 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.
 [/INST] 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?
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