How do I calculate the probability with joint normal pdf

Question

i have a question who is very difficult for me. Its very much hard for find the solution:

$X,Y$ are random variables, and $R^2=(X-a)^2+(Y-b)^2$ with $a,b$ constants

$X,Y$ have distribution normal $N~(0,\sigma)$

How do I calculate the probability $P=[R\leq r]$?

could you help me? Plz.

Answer 1

Possible approach. Let $Z$ be the distance of the point $(X, Y)$ from the origin; then

$$ P(Z \leq z) = 1-e^{-\frac{z^2}{2}} = 1-e^{-\frac{X^2+Y^2}{2}} $$

At this point, I don't see any easier way than to integrate this expression over the disc $(X-a)^2+(Y-b)^2 \leq r^2$. Is this something you're supposed to be able to do symbolically?

Answer 2

Let X'=X-a and Y'=Y-b. The joint density function is $\frac{1}{2\pi \sigma^2}exp-((x'+a)^2+(y'+b)^2)$. Transform to polar coordiates wth$R^2=x'^2+y'^2$ and intrgrate over $0\le R \le r$ and full circle in angle.