Compute $\mathbb{P}(1<X^2+Y^2<2)$ when $(X,Y)$ is i.i.d. standard normal

Question

Assume that $(X,Y)$ is i.i.d. standard normal. Compute $\mathbb{P}(1<X^2+Y^2<2)$.

So I've decided to use polar coordinates to solve and I've gotten to this point:

$$\iint_{1\lt X^2+Y^2\lt2} e^{\frac{-(x^2+y^2)}{2}}dxdy=\iint_{1\lt r^2\lt2}e^{\frac{-r^2}{2}}drd\theta$$

Where $1\lt r\lt\sqrt2$ and $0\lt \theta \lt 2\pi$.

How do I go about integrating dr? I know I need to use the $\Phi$ function but I'm not so sure how to go about doing that. Any help would be muchappreciated.

Answer 1

Your forgot the Jacobian factor $r$. If you include it, the integral becomes a lot easier to evaluate.

Answer 2

Since $X,Y\sim N(0,1)$ then $X^2,Y^2\sim \chi^2(1)\implies X^2+Y^2\sim \chi^2(2)$ Now use the$\chi^2$ distribution's CDF or PDF as required..

Answer 3

To elaborate on @probablyme 's comment, you can use the characteristic function to derive the "common fact".

We have that if $X \sim N(0,1)$, then the characteristic function (CF) of $X^2$ (for any $t \in \mathbb{R}$) is $$ \begin{align} \varphi_{X^2}(t)&=E \left(e^{itX^2} \right) \\ \\ &= \int_{-\infty}^{\infty} e^{itx^2} \frac{1}{\sqrt{2 \pi}} e^{\frac{-x^2}{2}} dx \\ \\ &= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{-x^2 \left(1-2it\right)2^{-1}} dx \\ \\ &=\frac{1}{\sqrt{2 \pi}} \frac{\sqrt{\pi}}{\sqrt{\left(1-2it\right)2^{-1}}} \, , \quad \mbox{see Gaussian integral on Wikipedia for this result} \\ \\ &= (1-2it)^{-1/2} \quad \mbox{which is the CF of a } \chi^2_{(1)} \end{align} $$

Let $Z$ be a r.v. distributed as $\chi^2_{(1)}$. By the Uniqueness Theorem, if $\varphi_{X^2}(t) = \varphi_{Z}(t)$, then $X^2 \sim Z$. So, $X^2 \sim \chi^2_{(1)}$. The same is valid for $Y^2$.

Now, since $X^2, Y^2 \stackrel{iid}{\sim} \chi^2_{(1)}$, we can calculate the CF of $X^2+Y^2$ for any $t \in \mathbb{R}$:

$$ \begin{align} \varphi_{X^2+Y^2}(t) &= E \left(e^{it(X^2+Y^2)}\right) \\ &= E \left(e^{itX^2}e^{itY^2}\right) \\ &= E \left(e^{itX^2} \right) E\left(e^{itY^2}\right) \, , \quad \mbox{by independence} \\ &= \varphi_{X^2}(t) \varphi_{Y^2}(t) \\ &= (1-2it)^{-1/2} (1-2it)^{-1/2} \\ &= (1-2it)^{-2/2} \, , \quad \mbox{which is the CF of a } \chi^2_{(2)} \end{align} $$

Therefore, $X^2+Y^2 \sim \chi^2_{(2)}$, and you can use its PDF to get your answer.

Also, we have that $\chi^2_{(1)}$ is a special case of Gamma ($Gamma(1/2,1/2)$, as stated by @probablyme). The same way, $\chi^2_{(2)}$ can be seen as $Gamma(2/2,1/2) = Gamma(1,1/2)$ or $Exponential(1/2)$ (another special case of Gamma).