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

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?

• Couldn't you avoid going down the route of invoking any exponential family results and just directly show that $T$ is unbiased for $\theta^2$ and has variance proportional to $n^{-1}$? May 29, 2019 at 23:43
• Could you provide a full answer? Because this is the only approach I am able to handle.
– user242554
May 30, 2019 at 0:06
• I don't follow. You don't know how to calculate $E(T)$ directly from the density and the definition of $T$? May 30, 2019 at 0:08
• Hmm, I got it. I'll try to do it. If it doesn't work, I will edit it asking for help.
– user242554
May 30, 2019 at 0:09
• Just in case ... I mean using en.wikipedia.org/wiki/Law_of_the_unconscious_statistician ... naturally there are other approaches as well. May 30, 2019 at 0:11

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 $$\theta^2$$.