$P(X^2+Y^2<1)$ of two independent n(0,1) random variables

Question

Suppose that X and Y are independent n(0,1) random variables.

a) Find $P(X^2+Y^2<1)$

Attempt:

a) Let $U = X^2 + Y^2$, $V = Y$.

Then $X = \sqrt{V^2 -U}$, $Y = V$.

$J = \left| \begin{array}{ccc} \frac{-1}{\sqrt{V^2-U}} & \frac{V}{V^2-U} \\ 0 & 1\\ \end{array} \right| $

Then the joint distribution of $f_{u,v}(u,v)$ is:

$$f_{u,v}(u,v)= \frac{1}{2\pi}e^{\frac{-\sqrt{v^2-u}}{2}}e^{\frac{-u^2}{2}}\frac{1}{\sqrt{v^2-u}}$$

Then $P(X^2 +Y^2 <1)$ is:

$$\int_0^\infty \int_0^{v^2-u} \frac{1}{2\pi}e^{\frac{-\sqrt{v^2-u}}{2}}e^{\frac{-u^2}{2}}\frac{1}{\sqrt{v^2-u}}dudv$$

However, at this point I simply do not know how any tricks to complete this integration.

Answer 1

Since $X$ and $Y$ are independent, the joint density of $X$ and $Y$ is the product of the individual densities, so it is $$\frac{1}{2\pi}e^{-(x^2+y^2)/2}$$ on the entire $x$-$y$ plane. We want to find $$\int_D \frac{x^2+y^2}{2\pi}e^{-(x^2+y^2)/2}\,dy\,dx,$$ where $D$ is the unit disk. Switch to polar coordinates. We want to find $$\int_{r=0}^{1}\int_{\theta=0}^{2\pi} \frac{r^3}{2\pi}e^{-r^2/2}\,d\theta\,dr.$$ The integration with respect to $\theta$ gets rid of the $2\pi$ in the denominator. So we want $$\int_0^1 r^3 e^{-r^2/2}\,dr.$$ This can be done by integration by parts. To make things more familiar, you may wish to first make the substitution $t=r^2/2$. Our integral becomes $$\int_0^{1/2} 2te^{-t}\,dt.$$

Answer 2

Although the following method is more convoluted than necessary, I think it provides a fun way to see the convolution method for computing the density of a sum of independent random variables in action. It also requires a few fun integration tricks.

First we will find the distribution for $X^2$ (and, by symmetry, the distribution of $Y^2$). We consider the cdf $F_{X^2}(t)$, and we see that $$F_{X^2}(t) = P(X^2 \leq t) = P(|X| \leq \sqrt{t}) = P(X \in [-\sqrt{t},\sqrt{t}] = P(X \leq \sqrt{t}) - P(X < \sqrt{t}) $$ Continuing this logic, and noting that $X$ has density $f_X = \frac{1}{\sqrt{2 \pi}} e^{-x^2/2}$, we see that $$ F_{X^2}(t) = F_X(\sqrt{t}) - F_X(-\sqrt{t}) = \displaystyle\int_{-\infty}^{\sqrt{t}} \frac{1}{\sqrt{2 \pi}} e^{-x^2/2} dx - \displaystyle\int_{-\infty} ^{-\sqrt{t}} \frac{1}{\sqrt{2 \pi}} e^{-x^2/2} dx$$ We can then find the density $f_{X^2}$ by differentiating our equation for $F_{X^2}(t)$ with respect to $t$, which gives us $$f_{X^2}(t) = \frac{d}{dt}\left(F_{X^2}(t) \right) = \frac{1}{\sqrt{2 \pi}} e^{- (\sqrt{t})^2 / 2} \left( \frac{1}{2 \sqrt{t}} \right) - \frac{1}{\sqrt{2 \pi}} e^{- (-\sqrt{t})^2 / 2} \left(- \frac{1}{2 \sqrt{t}} \right) $$ and this simplifies to tell us that $$f_{X^2}(t) = f_{Y^2}(t) = \frac{1}{\sqrt{t} \sqrt{2 \pi}} e^{- t/2} $$

Because we know that $X$ and $Y$ are independent, it follows that $X^2$ and $Y^2$ are independent as well. Therefore we can apply the convolution method to find the probability density for $Z = X^2 + Y^2$. We can write the probability distribution as the convolution $$f_Z(z) = \displaystyle\int_{0}^z f_{X^2}(z-y) f_{Y^2}(y) dy $$ where the lower-limit of the integral is 0 because $Y^2$ is a non-negative random variable, and the upper limit is $z$ because it cannot be the case that both $Y^2 = y > z$ and $X^2 < z - y$, as $X^2 $ is a non-negative random variable. Plugging our value for the densities $f_{X^2}(z-y) = \frac{1}{\sqrt{z-y} \sqrt{2 \pi}} e^{-(z-y)/2} $ and $f_{Y^2} = \frac{1}{\sqrt{y} \sqrt{2 \pi}} e^{-y/2}$, we have that $$f_Z(z) = \displaystyle\int_0^z \left( \frac{1}{\sqrt{z-y} \sqrt{2 \pi}} e^{-(z-y)/2} \right) \left( \frac{1}{\sqrt{y} \sqrt{2 \pi}} e^{-y/2} \right) dy = \frac{1}{2 \pi} e^{-z/2} \displaystyle\int_0^z \displaystyle\frac{dy}{\sqrt{zy - y^2}}$$

Now, what we have left to do is to compute the integral on the right. Completing the square under the square root sign we have $$\displaystyle\int_0^z \displaystyle\frac{dy}{\sqrt{zy - y^2}} = \displaystyle\int_0^z \displaystyle\frac{dy}{\sqrt{\frac{z^2}{4} - (\frac{z^2}{4} - zy + y^2) }} = \displaystyle\int_0^z \displaystyle\frac{dy}{\sqrt{\frac{z^2}{4} - (y - \frac{z}{2})^2 }} $$ Making the u-substitution $u = y - \frac{z}{2}$ (with $du = dy$ and with transformed limits of integration $y = 0 \mapsto u = -z/2$ and $y = z \mapsto z/2$), we obtain the integral $$ \displaystyle\int_0^z \displaystyle\frac{dy}{\sqrt{zy - y^2}} = \displaystyle\int_{-z/2}^{z/2} \displaystyle\frac{du}{\sqrt{\frac{z^2}{4} - u^2}} = \frac{2}{z} \displaystyle\int_{-z/2}^{z/2} \displaystyle\frac{du}{\sqrt{1 - (\frac{2u}{z})^2}} = \frac{2}{z} \left( \left(\frac{1}{\frac{2}{z}} \right) \sin^{-1}(\frac{2u}{z}) \bigg|_{-z/2}^{z/2} \right) $$ This simplifies to $$ \displaystyle\int_0^z \displaystyle\frac{dy}{\sqrt{zy - y^2}} = \sin^{-1}(1) - \sin^{-1}(-1) = \frac{\pi}{2} - (- \frac{\pi}{2}) = \pi $$ and therefore we have that $$f_{Z}(z) = \frac{1}{2 \pi} e^{-z/2} \left( \pi \right) \Longrightarrow \boxed{f_{Z}(z) = \frac{1}{2} e^{-z/2}} $$

And we can compute that $$P(X^2 + Y^2 <1 ) = P(Z < 1) = \displaystyle\int_0^1 f_{Z}(z)dz = \displaystyle\int_0^1 \frac{1}{2} e^{-z/2} dz = - e^{-z/2} \bigg|_0^1 $$ and we obtain our final answer $$ \boxed{P(X^2 + Y^2 < 1) = 1 - e^{-1/2}} $$

Answer 3

Box-Muller transform suggests that if $X, Y$ are two independent standard normal random variables, then if we let $X = R cos\Theta, Y = R sin\Theta, R \in (0, \infty), \Theta \in [0, 2\pi)$ (polar coordinate), then this bivariate transform is 1-to-1 and onto. Further, $R, \Theta$ are also independent. And $\frac{1}{2}R^2 \sim Exponential(1), \Theta \sim Uniform(0, 2\pi)$, i.e., $\frac{1}{2}R^2$ follows an Exponential distribution with $\lambda = 1$ and $\Theta$ follows a Uniform distribution on $(0, 2\pi)$.

Hence, $P(X^2 + Y^2 < 1) = P(R^2cos^2\Theta + R^2sin^2\Theta < 1) = P(R^2 < 1) = P(\frac{1}{2}R^2 < \frac{1}{2})$. And since $\frac{1}{2}R^2 \sim Exponential(1)$, we can denote $Z = \frac{1}{2}R^2$. Using the pdf of an $Exponential(1)$ distribution, $P(\frac{1}{2}R^2 < \frac{1}{2}) = P(Z < \frac{1}{2}) = \int_{-\infty}^{\frac{1}{2}} e^{-z} \, dz = 1 - e^{-\frac{1}{2}}$.