# Turning Nearest Neighbour Distribution of Poisson Scatter Theorem to Rayleigh Distribution By Multiplying Constant

Consider a Poisson random scatter of points in a plane with mean intensity $$\phi$$ per unit area. Let R be the distance from 0 to closest point of the scatter.

Show that $$\sqrt{2\phi\pi}$$R has the Rayleigh distribution.

I get $$F_R(r) = 1 - e^{-\phi \pi r^2}$$.This is the CDF of an exponential distribution with parameter $$\lambda = \phi\pi r$$.

I get that $$2\phi\pi \times E[R] = \frac{2}{r}$$ which in turn means that new distribution will have CDF $$F_{r\phi\pi R} = 1 - e^{-\frac{1}{2}r^2}$$, due to the expectation of an exponential distribution W with rate $$\lambda$$ being $$E[W] = \frac{1}{\lambda}$$. This is indeed the CDF of a Rayleigh distribution.

However, I don't get how taking the square root of $$2\phi\pi$$ gives us the correct final distribution? What am I missing here?

Okay, I just realized I needed a different approach. This is a common problem shown for undergrads studying probability theory, so instead of deleting the question I will just answer it myself.

The key is to consider the distribution $$R_1 = \sqrt{2 \phi \pi}R$$, such that $$P(R_1 \leq r_1) = P( \sqrt{2 \phi \pi} R \leq r_1) = P(R \leq \frac{r_1}{ \sqrt{2 \phi \pi}} ) = 1 - e^{-\phi \pi (\frac{r_1}{ \sqrt{ 2 \phi \pi}})^2} = 1 - e^{-\frac{1}{2} r_1^2}$$

Which is exactly the CDF of the Rayleigh distribution!