# Derive the conditional probability $P\left(X \in A \middle| \, Y \in B \right)$

I am trying to derive the following form of conditional probability: $P\left(X \in A \middle| \, Y \in B \right)$.

Let $g$ be a complex r.v. normally distributed as $\mathcal{CN}(0,1-\sigma^2)$, i.e. real and imaginary parts are independent and $\mathcal{N}(0,(1-\sigma^2)/2)$ distributed. Let $h$ be a complex r.v. normally distributed as $\mathcal{CN}(0,\sigma^2)$.

Question: How to derive $P\left(a_1\le|g+h|^2 <a_2 \middle| \ b_1\le|g|^2 <b_2 \right)$ ?
Here all the $a_i$ and $b_i$ are positive. Also, we have $b_2 \le a_1$ or $b_1 \ge a_2$.

Some remarks: $\small\bullet$ For a fixed $g$, $\frac{2}{\sigma^2}|g+h|^2 \sim\chi_2^2\left(\frac{2}{\sigma^2}|g|^2\right)$, which is a non-central chi-squared distribution. $\small\bullet$ We know that $|g|^2 \sim \frac{1-\sigma^2}{2} \chi_2^2$. $\small\bullet$ $g$ and $h$ are decorrelated. We can assume that they are independent if this can simplify the problem.

• If we are to make both sides of | non-trivial, i.e., $a_1 \lt a_2$ and $b_1 \lt b_2$, then your constants appear to be messed up. $a_2 \le b_1$. So $a_2 \lt b_2$. Meanwhile, $b_2 \le a_1$. So $b_2 \lt a_2$. This is a contradiction. – Mark L. Stone Feb 12 '16 at 18:04
• I missed the "or". Question edited.. Thx – din Feb 12 '16 at 18:16
• Are $g$ and $h$ independent? – Dilip Sarwate Feb 13 '16 at 2:12
• @DilipSarwate $g$ and $h$ are decorrelated. We can assume that they are independent if this can simplify the problem. I edited the question.. Do you have any idea on how to derive the probability ? – din Feb 13 '16 at 14:04