Rayleigh distribution with unequal variances

Question

Suppose we have two independent, uncorrelated random variables $X\sim N\left(0,a^2\right)$ and $Y\sim N\left(0,b^2\right)$ (i.e. $X$ and $Y$ are Normally distributed with mean 0 and standard deviations $a$ and $b$ respectively.)

How do I find the probability that $\sqrt{X^{2}+Y^{2}}\le r$, where r is a positive real number?

I know that if the two normals had the same standard deviations, then I would use the Rayleigh CDF to answer my question. But when the standard deviations are unequal, I can't seem to find much on how to obtain this probability.

Answer 1

In this case the random variable $X^2+Y^2$ is a weighted sum of two chi-squared random variables each with one degree-of-freedom. There is no closed form distribution for the probability of interest in the general case where $a \neq b$. To facilitate our analysis, let $S \equiv Y^2/b^2 \sim \text{ChiSq}(1)$. We can then write the probability of interest as:

$$\begin{align} \mathbb{P}(\sqrt{X^2+Y^2} \leqslant r) &= \mathbb{P}(X^2+Y^2 \leqslant r^2) \\[16pt] &= \mathbb{P}(X^2+ b^2 S \leqslant r^2) \\[8pt] &= \int \limits_0^\infty \mathbb{P}(X^2+b^2 S \leqslant r^2|S=s) \cdot \text{ChiSq}(s|1) \ ds \\[6pt] &= \int \limits_0^\infty \mathbb{P}(X^2+b^2 s \leqslant r^2) \cdot \text{ChiSq}(s|1) \ ds \\[6pt] &= \int \limits_0^\infty \mathbb{P} \bigg( \Big( \frac{X}{a} \Big)^2 \leqslant \frac{r^2-b^2 s}{a^2} \bigg) \cdot \text{ChiSq}(s|1) \ ds \\[6pt] &= \frac{1}{\sqrt{2} \pi} \int \limits_0^{r^2/b^2} \gamma \Big( \frac{1}{2}, \frac{r^2-b^2 s}{2 a^2} \Big) \cdot \frac{1}{\sqrt{s}} \cdot \exp \Big( -\frac{s}{2} \Big) \ ds, \\[6pt] \end{align}$$

where $\gamma$ is the lower incomplete gamma function. You can evaluate this integral numerically to obtain the probability of interest, but it has no closed form in general (so far as I'm aware).