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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?

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

1 Answer 1

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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 $\theta^2$.

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