Rayleigh as exponential family - compute $\mathbb{E}(Y)$ and Var$(Y)$ where Y is a sum of independent squared Rayleigh's distribution

Question

Prove that X of Rayleigh distribution with pdf $f(x, \sigma) = \frac{x}{\sigma^2}e^{-\frac{x^2}{2\sigma^2}}\mathbb{1}_{(0, \infty)}(x)$ comes from the exponential family and then compute $\mathbb{E}(Y)$ and Var$(Y)$ where $Y = \sum_{k=1}^nX^2_k$ (all $X_k$ are independent).

I did the first part: $$ f(x, \sigma) = \frac{x}{\sigma^2}e^{-\frac{x^2}{2\sigma^2}}\mathbb{1}_{(0, \infty)}(x) = xe^{-\frac{1}{2\sigma^2}x^2 - 2\ln(\sigma)}\mathbb{1}_{(0, \infty)}(x) $$ so $$ h(x) = x; \eta_1(\theta) = -\frac{1}{2\sigma^2}; T_1(x)=x^2; B(\theta)=2ln(\sigma) $$

Why is the fact that Rayleigh's comes from an exponential family useful? Are there any equations to quickly solve the second part of the task? I appreciate any help.

Answer 1

The Rayleigh distribution is a single parameter exponential family if we can write it in the form

$$ f(x: \sigma) = h(x) \exp\left( \eta(\sigma) T(x) - A(\sigma) \right)$$

Here we have $$f(x: \sigma) = x \mathbb{1}_{[0,\infty)} (x) \exp \left( \frac{-1}{2\sigma^2} x^2 - 2 \log \sigma \right)$$

so it is indeed an exponential family, with $h(x) = x \mathbb{1}_{[0,\infty)} (x), \ T(x) = x^2, \ \eta(\sigma) = \frac{-1}{2\sigma^2}$ and $A(\sigma) = 2 \log \sigma.$

The moment generating function of a member of an exponential family has a particularly nice form:

$$ M_T(t) = \mathbb{E}[\exp(t T(x))] = \exp\left( A(\eta + t) - A(\eta) \right) $$

Here, since $\eta = \frac{-1}{2\sigma^2}$ and $A(\sigma) = 2 \log \sigma,$ we can write $A(\eta) = - \log( -2 \eta).$

This gives $M_{T}(t) = \frac{\eta}{\eta + t},$ and differentiating this gives

$$ \frac{d}{dt} M_{T}(t) = \frac{-\eta}{(\eta + t)^2} \ , \ \frac{d^2}{dt^2} M_{T}(t) = \frac{2\eta}{(\eta + t)^3}.$$

Plugging in $t=0$ into these (and remembering $T = X^2$) we have

$$ \mathbb{E}[X^2] = \frac{-1}{\eta} = 2\sigma^2 \ , \ \mathbb{E}[X^4] = \frac{2}{\eta^2} = 8 \sigma^4$$

From these we deduce $\text{Var}(X^2) = \mathbb{E}[X^4] - \mathbb{E}[X^2]^2 = 4 \sigma^4$ and therefore

$$ \mathbb{E}[Y] = 2n \sigma^2 \ , \ \text{var}(Y) = 4n \sigma^4. $$

Answer 2

Let $E$ have exponential distribution with parameter $\lambda=\frac{1}{2\sigma^{2}}$ and let $Z=\sqrt{E}.$

Then for $x>0$ we find: $$P\left(Z>x\right)=P\left(E>x^{2}\right)=e^{-\lambda x^{2}}=e^{-\frac{x^{2}}{2\sigma^{2}}}$$ showing that $Z$ has Rayleigh distribution with parameter $\sigma$.

So if $X_{1},X_{2},\dots$ are iid with Rayleigh distribution and parameter $\sigma$ then we can write $X_{k}=\sqrt{E_{k}}$ where the $E_{k}$ are iid with exponential distribution with parameter $\lambda=\frac{1}{2\sigma^{2}}$ .

Then $Y=\sum_{k=1}^{n}X_{k}^{2}=\sum_{k=1}^{n}E_{k}$ and has Erlang distribution with parameters $n$ and $\lambda=\frac{1}{2\sigma^{2}}$.

This with: $$\mathbb EY=\frac{n}{\lambda}=2n\sigma^{2}$$ and: $$\mathsf{Var}Y=\frac{n}{\lambda^{2}}=4n\sigma^{4}$$