# "Flattening" a 2D Normal Distribution

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?

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 $$F_R(r; \sigma) = \dfrac1{2 \pi \sigma^2} \int_{\rho=0}^{r} \int_{\theta=0}^{2 \pi} \exp \left( - \dfrac{\rho^2}{2 \sigma^2} \right) \rho d \rho d \theta\\ =\dfrac1{\sigma^2} \int_{\rho=0}^{r} \rho \exp \left(- \dfrac{\rho^2}{2 \sigma^2} \right) d \rho \\ = -\left. \exp \left( - \dfrac{\rho^2}{2 \sigma^2}\right) \right \rvert_{\rho=0}^{\rho=r} \\ = 1 - \exp \left( - \dfrac{r^2}{2 \sigma^2} \right)$$ $$f_R(r) = \dfrac{d F_R(r)}{dr} = \dfrac{r}{\sigma^2} \exp \left( - \dfrac{r^2}{2 \sigma^2}\right)$$
• 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 - yes: the actual standard deviation of a Rayleigh distribution with scale parameter $\sigma$ is in fact $\sqrt{\frac{4-\pi}{2}}\sigma \approx 0.655 \sigma$ Oct 5, 2019 at 11:33