3
$\begingroup$

This question already has an answer here:

If $a$ and $b$ are both independent, identically distributed normal random variables, what is the distribution of $\phi = \arctan{(-\frac{b}{a})}$? It is uniform, across $0$ to $2\pi$, but I am trying to see why. $a \sim \mathcal{N}(0,\sigma^2)$ and $b$ is identically distributed and independent to $a$.

For example, setting $X = \arctan{(-\frac{b}{a})}$, my strategy is to first calculate the cumulative distribution function (CDF), and then compare the CDF to known CDFs.

$$F_X(x) = P(X\leq x) = P\left(\arctan{\left(-\frac{b}{a}\right)}\leq x\right) = P\left(-\frac{b}{a}\leq\tan(x)\right)$$

Now, $a$ could be either positive or negative, since $a$ is a normal random variable with mean $0$ and non-zero variance. Then, $$\begin{aligned}P\left(-\frac{b}{a}\leq\tan(x)\right) & = P\left(-\frac{b}{a}\leq\tan(x)|a>0\right)P(a>0)\\ & +P\left(-\frac{b}{a}\leq\tan(x)|a\leq 0\right) P(a\leq 0) \end{aligned}$$

Each of the terms $P\left(-\frac{b}{a}\leq\tan(x)|a\leq 0\right) $ and $P\left(-\frac{b}{a}\leq\tan(x)|a > 0\right)$ have terms depending on the value of $x$. I am drawing a plot of $a$ on the x-axis and $b$ on the y-axis and trying to use polar coordinates to calculate the integral using the joint probability function of $a$ and $b$, which is simply the product of the two distributions. Can I have a hint on how to proceed?

If I ignore the value of $a$ entirely (which is wrong) and ignore one of the probability terms (which is also wrong) and ignore the value of $x$, I am getting a result very similar to $\frac{x}{2\pi}$ (which is desirable).

Thanks.

$\endgroup$

marked as duplicate by whuber Jun 18 '13 at 17:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ It is (essentially) the probability integral transform of a Cauchy distributed random variable. $\endgroup$ – cardinal Feb 14 '13 at 3:03
  • 4
    $\begingroup$ As an alternative to the above, if you like geometry, think about this geometrically and use spherical symmetry of iid normals. $\endgroup$ – cardinal Feb 14 '13 at 3:18
5
$\begingroup$

This answer is basically an expansion on the hint in cardinal's second comment.

Perhaps you should consider that $-b$ is a zero-mean normal random variable that has the same variance as $b$ and is also independent of $a$, and so we may as well find the distribution of $\arctan(b/a)$ which saves carrying around a negative sign. Then, one way of doing this problem with a "standard" approach is to find the joint distribution of $\sqrt{a^2+b^2}$ and $\arctan(b/a)$ which just looks at the polar coordinates of the point $(a,b)$ in rectangular coordinates. Plugging and chugging via the Jacobians etc. used in finding the joint distribution of $(g(X,Y), h(X,Y))$ from the joint distribution of $(X,Y)$, we readily get that $\sqrt{a^2+b^2}$ has a Rayleigh distribution and $\arctan(b/a)$ has a uniform distribution on $[0,2\pi)$ and that these random variables are independent. (This is just the Box-Muller transform method worked backwards).

But to really use cardinal's hint, note that for $0 \leq \theta < 2\pi$, the event $\{\arctan(b/a) \leq \theta\}$ occurs only if and only if $(a,b)$ lies in a sector of the plane between the $a$ axis and the line at angle $\theta$ to the $a$ axis, and since the joint distribution of $(a,b)$ has circular symmetry, $$P\{\arctan(b/a) \leq \theta\} = \frac{\theta}{2\pi}$$ which proves the result needed without any use of explicit integration or Jacobians etc.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.