Derivation of the Rayleigh Distribution from 2D Normal Variables
Published July 25, 2025, 7:19 p.m. by james
Step 1: Start from 2D Normal Distribution
Let $X \sim \mathcal{N}(0, \sigma^2)$ and $Y \sim \mathcal{N}(0, \sigma^2)$ be independent normal random variables. Define:
$$R = \sqrt{X^2 + Y^2}$$
We want the probability density function (PDF) of $R$.
Step 2: Joint PDF of $X$ and $Y$
The joint PDF is:
$$f_{X,Y}(x,y) = \frac{1}{2\pi \sigma^2} \exp\left( -\frac{x^2 + y^2}{2\sigma^2} \right)$$
Step 3: Convert to Polar Coordinates
Use the transformation:
$x = r\cos\theta,\quad y = r\sin\theta,\quad r \geq 0,\ \theta \in [0, 2\pi)$
The Jacobian determinant is:
$$\left| \frac{\partial(x,y)}{\partial(r,\theta)} \right| = r$$
So the joint PDF in polar coordinates becomes:
$$f_{R,\Theta}(r,\theta) = \frac{1}{2\pi \sigma^2} \exp\left( -\frac{r^2}{2\sigma^2} \right) \cdot r$$
Step 4: Marginal PDF of $R$
Integrate over $\theta$ to get the marginal PDF of $R$:
\[ f_R(r) = \int_0^{2\pi} f_{R,\Theta}(r,\theta)\, d\theta = \frac{r}{2\pi \sigma^2} \exp\left( -\frac{r^2}{2\sigma^2} \right) \cdot \int_0^{2\pi} d\theta \]
\[ f_R(r) = \frac{r}{\sigma^2} \exp\left( -\frac{r^2}{2\sigma^2} \right),\quad r \geq 0 \]
Conclusion
The resulting distribution is the Rayleigh distribution:
\[ \boxed{f_R(r) = \frac{r}{\sigma^2} \exp\left( -\frac{r^2}{2\sigma^2} \right),\quad r \geq 0} \]
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