Asymptotic Maxwell MLE distribution

Question

Consider i.i.d. random samples $X_1,...,X_n$ from the Maxwell Density:

$$ f_\theta(x)=\sqrt{\frac{2}{\pi}}\dfrac{x^2}{\theta^3}e^{-\frac{x^2}{2\theta^2}}I_{(0,\infty)}(x) $$

with $\theta > 0$. Derive the Maximum likelihood estimator (MLE) for $\theta$ and find its asymptotic distribution.

So far, I can reason that

$$ \hat{\theta}_{MLE}=\sqrt{\frac{1}{3n}\sum_{i=1}^{n}x_i^2} $$

but I am having trouble deriving the asymptotic distribution. I can reason that if I set $Y=X^2$, then

$$ f_\theta(y)=\sqrt{\dfrac{y}{2\pi}}\dfrac{1}{\theta^3}e^{-\frac{y}{2\theta^2}}I_{(0,\infty)}(y) $$

but I don't recognize this distribution, and I certainly don't know what the distribution of its sum or sample mean look like.

Two thoughts I've had are 1. using the central limit theorem on the sample mean of $Y$, but this seems unlikely, as $Y^2$ is nonnegative and 2. Using the fact that for all distributions, $\hat{\theta}_{MLE}\approx N(\theta,\text{CRLB}_\theta)$, but that seems too easy.

Thanks!

Answer 1

Hint: The delta-method applied to $\sum\limits_{k=1}^nx_k^2$ yields

$$\sqrt{n}(\hat\theta_n-\theta)\to N(0,\tfrac23).$$

To prove this, one starts with the CLT expansion $$\sum\limits_{k=1}^nx_k^2=nE_\theta(x_1^2)+\sqrt{n}\sigma_\theta(x_1)Z_n,$$ where $Z_n\to N(0,1)$ when $n\to\infty$, with $E_\theta(x_1^2)=3\theta^2$ and $\sigma^2_\theta(x_1)=$ $_____$, thus...

Answer 2

Under suitable regularity conditions, the MLE is asymptotically normal. That is,

$$ \sqrt{n}\left(\hat{\theta}_{mle}{}-{}\theta\right)\to\mathcal{N}\left(0,\dfrac{1}{I(\theta)}\right)\,, $$

where, in the present case, $I(\theta){}={}-\mathbb{E}\left[\dfrac{\partial^2}{\partial\theta^2}\log\left(\dfrac{1}{\theta^3}\sqrt{\dfrac{2}{\pi}}x^2e^{-x^2/2\theta^2}\right)\right]{}={}\dfrac{3!}{\theta^2}$.

So, $$ \sqrt{n}\left(\hat{\theta}_{mle}{}-{}\theta\right)\to\mathcal{N}\left(0,\dfrac{\theta^2}{3!}\right)\,. $$

Edit: As @Did hints, the delta-method can be used to verify this result, so that

$$ \sqrt{n}\left(\hat{\theta}_{mle}{}-{}\theta\right)\approx\dfrac{\sqrt{n}}{2\theta}\left(\hat{\theta}_{mle}^2-\theta^2\right)\to\mathcal{N}\left(0, \dfrac{\theta^2}{3!}\right)\,, $$ where $\,\,\hat{\theta}_{mle}{}={}\sqrt{\dfrac{1}{3n}\sum\limits_{i=1}^{n}X_i}\,\,$ and $\mathbb{V}ar\left(\hat{\theta}^2_{mle}\right){}={}\dfrac{2}{3n}\theta^4$.

Answer 3

There is no need to use asymptotic approximations here, since we can get the exact distribution of the MLE. The distribution you didn't recognise is $Y = X^2 \sim \text{Gamma}(\text{Shape} =\tfrac{3}{2}, \text{Scale} = 2 \theta^2)$, which has density function:

$$\begin{equation} \begin{aligned} f_Y(y) &= \frac{1}{\Gamma(\tfrac{3}{2}) (2 \theta^2)^{3/2}} \cdot y^{3/2 - 1} \exp \Big( - \frac{y}{2 \theta^2} \Big) \\[6pt] &= \frac{\sqrt{y}}{\tfrac{1}{2} \sqrt{\pi} \text{ } 2^{3/2} \theta^3} \cdot \exp \Big( - \frac{y}{2 \theta^2} \Big) \\[6pt] &= \sqrt{\frac{y}{2 \pi}} \frac{1}{\theta^3} \cdot \exp \Big( - \frac{y}{2 \theta^2} \Big) \quad \quad \text{for } y > 0. \\[6pt] \end{aligned} \end{equation}$$

From here you have $\sum_{i=1}^n X_i^2 \sim \text{Gamma}(\text{Shape} = \tfrac{3}{2} n, \text{Scale} = 2 \theta^2)$, which means that the MLE has a scaled Nakagami distribution:

$$\hat{\theta}_{\text{MLE}} = \frac{1}{\sqrt{3n}} \cdot \sqrt{\sum_{i=1}^n X_i^2} \sim \frac{1}{\sqrt{3n}} \cdot \text{Nakagami}(\text{Shape} = \tfrac{3}{2} n, \text{Spread} = 3 n \theta^2).$$

The exact moments of the estimator are:

$$\mathbb{E}(\hat{\theta}_{\text{MLE}}) = \Pi (n) \cdot \theta \quad \quad \mathbb{V}(\hat{\theta}_{\text{MLE}}) = ( 1 - \Pi(n)^2 ) \cdot \theta^2,$$

where:

$$\Pi (n) \equiv \frac{\Gamma(\tfrac{3}{2} n + \tfrac{1}{2})}{\Gamma(\tfrac{3}{2} n) \sqrt{\tfrac{3}{2}n}}.$$

As $n \rightarrow \infty$ we have $\Pi (n) \rightarrow 1$ which gives the limiting values $\mathbb{E}(\hat{\theta}_{\text{MLE}}) \rightarrow \theta$ and $\mathbb{V}(\hat{\theta}_{\text{MLE}}) \rightarrow 0.$ Hence, the estimator is a consistent estimator.