Let $X\sim\text{Rayleigh}(\theta^{2})$. Prove that $T_{n}$ is consistent, given that $T_{n}(\textbf{X}) = \frac{1}{2n}\sum_{i=1}^{n}x^{2}_{i}$

Question

Let $X\sim\text{Rayleigh}(\theta^{2})$. Prove that $T_{n}$ is consistent, given that $$T_{n}(\textbf{X}) = \frac{1}{2n}\sum_{i=1}^{n}x^{2}_{i}$$

MY ATTEMPT

To begin with, let us notice that \begin{align*} p(x|\theta) = \frac{x}{\theta^{2}}\exp\left\{-\frac{x^{2}}{2\theta^{2}}\right\} \end{align*}

which can be rewritten as in the canonical form as \begin{align*} p(x,\eta) = x\exp\left\{\eta x^{2} + \ln(-2\eta)\right\} \end{align*}

where $\eta = -1/2\theta^{2}$. Consequently, $A(\eta) = -\ln(-2\eta)$, from which we obtain that \begin{align*} \textbf{E}(X^{2}) = A^{\prime}(\eta) = -\frac{1}{\eta} = 2\theta^{2}\quad\text{and}\quad \textbf{Var}(X^{2}) = A^{\prime\prime}(\eta) = \frac{1}{\eta^{2}} = 4\theta^{4} \end{align*}

Based on this, we may assert that \begin{align*} \textbf{E}(T_{n}(\textbf{X})) = \frac{1}{2n}\textbf{E}\left(\sum_{i=1}^{n}x^{2}_{i}\right) = \frac{1}{2n}\times 2n\theta^{2} = \theta^{2} \end{align*}

Analogously, we have \begin{align*} \textbf{Var}(T_{n}(\textbf{X})) = \frac{1}{4n^{2}}\textbf{Var}\left(\sum_{i=1}^{n}x^{2}_{i}\right) = \frac{1}{4n^{2}}\times 4n\theta^{4} = \frac{\theta^{4}}{n}\xrightarrow{n\rightarrow\infty} 0 \end{align*}

from whence we conclude that $T_{n}$ is consistent, as previously stated.

My question is: is there another approach to this problem?

Answer 1

Invoking the exponential family here is not very helpful. What you can do is find the mean and variance of your estimator directly as suggested in comments. Alternatively, you can use the following:

For i.i.d $X_1,X_2,\ldots,X_n$ with $E(X_1^2)<\infty$, by weak law of large numbers we have

$$\frac{1}{n}\sum_{i=1}^n X_i^2\stackrel{P}\longrightarrow E(X_1^2)$$

This of course implies $$\frac{1}{2n}\sum_{i=1}^n X_i^2\stackrel{P}\longrightarrow \frac{1}{2}\times E(X_1^2)$$

For your question, this shows $\frac{1}{2n}\sum\limits_{i=1}^n X_i^2$ is a consistent estimator of $\theta^2$.