Let $X$ and $Y$ be independent random variables each with normal density $N(0, σ^2)$. Find $P(X^2+Y^2\leq 1)$. Use polar coordinates.

Question

Let $X$ and $Y$ be independent random variables each with normal density $N(0, σ^2)$. Find $P(X^2+Y^2\leq 1)$. Use polar coordinates.

Attempt

Since $X$ and $Y$ are independent, and because $μ=0$, then we know that $$f_{X,Y}(x,y)=f_X(x)f_Y(y)=\frac{1}{2\pi σ^2}\exp \left(\frac{-(x^2+y^2)}{2σ^2}\right).$$ Using polar coordinates, we have $$P(X^2+Y^2\leq 1)=\frac{1}{2\pi}\int_{0}^{1}\int_{0}^{2\pi}\, \exp \left( \frac{-r^2}{2\sigma ^2}\right)\, r \, dr\, d\theta$$

Answer 1

$ \displaystyle P(X^2+Y^2\leq 1) = \frac{1}{2\pi \color {blue }{\sigma^2}}\int_{0}^{1}\int_{0}^{2\pi}\, \exp \left( \frac{-r^2}{2\sigma ^2}\right)\, r \, dr\, d\theta$

$ \displaystyle = \frac{1}{ \sigma^2}\int_{0}^{1} \, \exp \left( \frac{-r^2}{2\sigma ^2}\right)\, r \, dr$

Now substituting $ \displaystyle t = - \frac{r^2}{2 \sigma^2}, dt = - \frac{r}{\sigma^2} dr$

$P(X^2+Y^2\leq 1) \displaystyle = \int_{- 1 / (2\sigma^2)}^0 \, e^t \, dt = 1 - \frac{1}{e^{1/(2\sigma^2)}}$