Rayleigh Distribution

The distribution with Probability Function

$$P(r) = {re^{-r^2/2s^2}\over s^2}$$ (1)

for $r\in [0,\infty)$. The Moments about 0 are given by

$$\mu_m' \equiv \int_0^\infty r^m P(r)\,dr = s^{-2} \int_0^\infty r^{m+1}e^{-r^2/2s^2}\,dr$$

$$= s^{-2} I_{m+1}\left({1\over 2s^2}\right),$$ (2)

where $I(x)$ is a Gaussian Integral. The first few of these are

$$I_1(a^{-1}) = {\textstyle{1\over 2}}a$$ (3)

$$I_2(a^{-1}) = {\textstyle{1\over 4}}a\sqrt{a\pi}$$ (4)

$$I_3(a^{-1}) = {\textstyle{1\over 2}}a^2$$ (5)

$$I_4(a^{-1}) = {\textstyle{3\over 8}} a^2\sqrt{a\pi}$$ (6)

$$I_5(a^{-1}) = a^3,$$ (7)

so

$$\mu_0' = s^{-2} {\textstyle{1\over 2}}(2s^2)=1$$ (8)

$$\mu_1' = s^{-2} {\textstyle{1\over 4}}(2s^2)\sqrt{2s^2\pi} = {\textstyle{1\over 2}}s\sqrt{2\pi}=s\sqrt{\pi\over 2}$$ (9)

$$\mu_2' = s^{-2} {\textstyle{1\over 2}}(2s^2)^2=2s^2$$ (10)

$$\mu_3' = s^{-2} {\textstyle{3\over 8}}(2s^2)^2\sqrt{2s^2\pi} = {\textstyle{3\over 2}}s^3\sqrt{2\pi} =3s^3\sqrt{\pi\over 2}$$ (11)

$$\mu_4' = s^{-2} (2s^2)^3=8s^4.$$ (12)

The Moments about the Mean are

$$\mu_2 = \mu_2'-(\mu_1')^2={4-\pi\over 2} s^2$$ (13)

$$\mu_3 = \mu_3'-3\mu_2'\mu_1'+2(\mu_1')^3=\sqrt{\pi\over 2}\,(\pi-3)s^3$$ (14)

$$\mu_4 = \mu_4'-4\mu_3'\mu_1'+6\mu_2'(\mu_1')^2-3(\mu-1')^4$$

$$= {32-3\pi^2\over 4} s^4,$$ (15)

so the Mean, Variance, Skewness, and Kurtosis are

$$\mu = \mu_1'=s\sqrt{\pi\over 2}$$ (16)

$$\sigma^2 = \mu_2={4-\pi\over 2} s^2$$ (17)

$$\gamma_1 = {\mu_3\over\sigma^3} = {2(\pi-3)\sqrt{\pi}\over (4-\pi)^{3/2}}$$ (18)

$$\gamma_2 = {\mu_4\over\sigma^4}-3 = {2(-3\pi^2+12\pi-8)\over(\pi-4)^2}.$$ (19)