Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I would like to model the probability of a point being at a certain place on a 2D grid. The X coordinate of the point varies according to a normal distribution with mean $0$ and standard deviation $\sigma$. The Y coordinate varies according to a normal distribution with the same mean and same standard deviation.

I know that the probability of being a certain distance from $(0, 0)$ is the same in all directions. I would therefore like to "flatten" my 2 distributions into a single distribution where the random variable is the distance from $(0, 0)$. In other words, if I know the distributions of x and y, what is the distribution of $\sqrt{x^2+y^2}$?

Am I right in my intuition that this will also be a normal distribution? (Or, half of a normal distibution, since the distance cannot be below zero). If so, how do I calculate the standard deviation of this distribution? If not, what type of distribution would it be, and what are the pdf and cdf functions of it?

share|cite|improve this question
up vote 5 down vote accepted

The distribution you are after is called the Rayleigh distribution function. The pdf is given by $$f_R(r) = \dfrac{r}{\sigma^2} \exp \left(-\dfrac{r^2}{2 \sigma^2} \right) \text{ where } r \geq 0$$

$$F_R(r; \sigma) = \mathbb{P}(R \leq r) = \mathbb{P}(X^2 + Y^2 \leq r^2) = \int_{-r}^{r} \mathbb{P} \left(Y \in \left[ \sqrt{r^2-x^2},\sqrt{r^2-x^2} \right] \right) f_X(x) dx\\ = \int_{-r}^{r} \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} f_Y(y)f_X(x) dx = \dfrac1{2 \pi \sigma^2} \int_{-r}^{r} \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} \exp \left( -\left( \dfrac{y^2 + x^2}{2 \sigma^2} \right) \right) dy dx$$ Changing variable from $(x,y) \rightarrow (\rho,\theta)$, we get that $$F_R(r; \sigma) = \dfrac1{2 \pi \sigma^2} \int_{\rho=0}^{r} \int_{\theta=0}^{2 \pi} \exp \left( - \dfrac{\rho^2}{2 \sigma^2} \right) \rho d \rho d \theta\\ =\dfrac1{\sigma^2} \int_{\rho=0}^{r} \rho \exp \left(- \dfrac{\rho^2}{2 \sigma^2} \right) d \rho \\ = -\left. \exp \left( - \dfrac{\rho^2}{2 \sigma^2}\right) \right \rvert_{\rho=0}^{\rho=r} \\ = 1 - \exp \left( - \dfrac{r^2}{2 \sigma^2} \right)$$ $$f_R(r) = \dfrac{d F_R(r)}{dr} = \dfrac{r}{\sigma^2} \exp \left( - \dfrac{r^2}{2 \sigma^2}\right)$$

share|cite|improve this answer
... which can be regarded as a particular case (dimension=2) of the Chi distribution. – leonbloy Jun 3 '12 at 1:17
@leonbloy True. – user17762 Jun 3 '12 at 1:19
So, just to confirm, the $\sigma$ in your Rayleigh pdf function is the same variable as the $\sigma$ that represents the standard deviation of the initial normal distributions, right? – Ord Jun 3 '12 at 15:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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