Probability of Independent Normal Random Variables

Question

There are two Random Variables X and Y. Both of them are Normally distributed with 0 mean and variance a and b respectively. X and Y are independent of each other.

What is the probability of

What should be the easiest way to do this?

Answer 1

The joint probability distribution of $X$ and $Y$ is $\frac{1}{2\pi \sqrt{ab}}e^{\frac{-1}{2} \left(\frac{x^2}{a^2} + \frac{y^2}{b^2} \right)}$.

Hence, we are looking for the value of the following integral.

\begin{align*} \frac{1}{2 \pi \sqrt{ab}} \iint_{x+y>0, y>0}e^{\frac{-1}{2} \left(\frac{x^2}{a^2} + \frac{y^2}{b^2} \right)} dxdy \end{align*} We make the substitution $s = \frac{x}{a}$ and $t = \frac{y}{b}$, and this becomes

\begin{align*} \frac{\sqrt{ab}}{2 \pi} \iint_{as+bt>0, t>0}e^{\frac{-1}{2} \left(s^2 + t^2 \right)} dsdt \end{align*}

Now, we want to switch to polar coordinates, but we have to be careful about the domain over which we integrate. You might want to draw a picture to find that it is the sector $0 < \theta < \pi + \arctan\left( \frac{-a}{b} \right)$. This gives the following expression.

\begin{align*} \frac{\sqrt{ab}}{2 \pi} \int_{0}^{\pi + \arctan\left( \frac{-a}{b} \right)}\int_0^{\infty} e^{\frac{-r^2}{2}} rdrd\theta =& \frac{\sqrt{ab}\left( \pi + \arctan \left( \frac{-a}{b}\right) \right)}{2\pi} \int_0^{\infty}e^{\frac{-r^2}{2}}d\left( \frac{r^2}{2} \right) \\ =& \frac{\sqrt{ab}\left( \pi + \arctan \left( \frac{-a}{b}\right) \right)}{2\pi} \end{align*} which is your solution.

Answer 2

The easiest way to do this is transforming the integral in the $(x,y)$ space to an integral using polar coordinates.

First, we take care of the variances by transforming $X$ and $Y$ into standard normals, $S$ and $T$ (both with variance equal to 1): let $S=X/\sqrt{a}$ and $T=Y/\sqrt(b)$. The desired probability is rewritten as: \begin{align*} \Pr\left(\{X+Y>0\}\cap \{Y>0\}\right) = \Pr\left(\{\sqrt{a}S+\sqrt{b}T>0\}\cap \{T>0\}\right). \end{align*}

Next, we write the probability as an integral of the joint probability distribution of two standard normals, $S$ and $T$: \begin{align*} \Pr\left(\{\sqrt{a}S+\sqrt{b}T>0\}\cap \{T>0\}\right) = \frac{1}{2 \pi } \iint_{s\sqrt{a}+t\sqrt{b}>0, t>0}e^{-\frac{1}{2} \left(s^2 + t^2\right)} dsdt. \end{align*}

Then, we convert $(s,t)$ coordinates into polar coordinates using $r^2 = s^2 + t^2$, $\tan(\theta)= t/s$. The two things to take care are: (i) $dxdy$ changes to $rdrd\theta$, and (ii) we need to convert the area we are integrating to polar coordinates.

The area is a cone from the origin (that's why using polar coordinates is going to be useful), so we are integrating over rays from the origin of any (positive) length, that is over $r\in[0,\infty]$. We now need the angles of the rays on the edge of the cone: (i) the x-axis (between the quadrants I and IV), which is a ray of angle $\theta=0$, and (ii) the line $t=-\sqrt{\frac{a}{b}}s$ (in quadrant II, where $t>0$ and $s<0$). The angle of this ray is $\theta = \pi +\arctan(-\sqrt{a/b}) = \pi -\arctan(\sqrt{a/b})$.

The resulting integral is the following: \begin{align*} \frac{1}{2 \pi} \int_{0}^{\pi - \arctan\left(\sqrt{ \frac{a}{b}} \right)}\int_0^{\infty} e^{-\frac{r^2}{2}} rdrd\theta. \end{align*} This integral is easy to evaluate. We can integrate with respect to $\theta$ to get: \begin{align*} \frac{1}{2 \pi} \left(\pi - \arctan\left(\sqrt{ \frac{a}{b}} \right)\right)\int_0^{\infty} e^{-\frac{r^2}{2}} rdr. \end{align*}

For the remaining integral, if we make the change of variable $z=r^2$ with $dz = 2 r dr$, we realize that the integral is equal to one, and we are done!

\begin{align*} \Pr\left(\{X+Y>0\}\cap \{Y>0\}\right) &= \frac{\left( \pi - \arctan \left( \sqrt{\frac{a}{b}}\right) \right)}{2\pi}\\ &= \frac{1}{2}- \frac{1}{2\pi}\arctan \left(- \sqrt{\frac{a}{b}}\right). \end{align*}