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Given is a a random sample $X_1,\ldots,X_n$ from a distribution with a pdf $f(x;\theta) = 2\theta^{-1} x^{(2/\theta)-1}$ for $0<x<1$, zero otherwise. We know that maximum likelihood estimator is $\hat{\theta} = -\dfrac 2 n \sum_i \log X_i$. Determine $c$ such that $c\left(\dfrac{n}{-2\sum_i \log X_i} - \dfrac{1}{\theta}\right) \xrightarrow{\text{d}} Z \sim N(0,1)$, i.e. converges in distribution.

So in the correction sheet it says the answer is $\dfrac{1}{\sqrt{CRLB}}$ where CRLB stands for the Cramer Rao Lower Bound. I don't really understand this. I have found in my book that for large $n$ $\hat{\theta}_n \sim N(\theta, CRLB)$, but how can we normalise this result to a $N(0,1)$?

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  • $\begingroup$ If $X_n \to X$ in distribution, $a X_n + b \to a X + b$ in distribution. This is a consequence of slutsky's theorem, though easy enough to prove from first principles. Also, the asymptotics of the MLE are usually $\sqrt{n} (\hat{\theta}_{MLE} - \theta) \to N(0, \sigma_{\theta}^2)$ where $\sigma_\theta^2$ is the asymptotic variance, which is 1/ fisher information. So, check the statement in your book more carefully You'd expect a $\sqrt{n}$ normalizing factor for Gaussianity after the $c$ in the problem. $\endgroup$
    – Batman
    Jun 8, 2017 at 23:35

1 Answer 1

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$X_1, X_2, \ldots, X_n$ is a sample from distribution with pdf $f(x; \theta) = 2\theta^{-1}x^{(2/\theta)-1}$ for $x \in (0,1)$ and zero otherwise.

Now, the distribution of $Y=-2\log X$ is derived as follows.

$$X=e^{-\frac{Y}{2}}$$

$$\frac{dX}{dY} = -\frac{1}{2}e^{-\frac{Y}{2}}$$

The pdf of $Y$ is given by,

$$g(y;\theta) = f(e^{-\frac{Y}{2}};\theta)\left | \frac{dX}{dY} \right | = \frac{1}{\theta}e^{-\frac{y}{\theta}}$$

Therefore, $Y_1, Y_2, \ldots, Y_n$ is transformed sample with $Y_i = -2\log X_i$ with $EY_i = \theta$ and $VarY_i = \theta^2$.

Now, one can use central limits theorem to get,

$$\sqrt{n}\frac{(\overline{Y_n} - \theta)}{\theta} \rightarrow \mathcal{N}(0,1)$$

Let $h(z) = \frac{1}{z}$. Using the first order taylor approximation to estimate the limiting distribution of $h(\overline{Y_n})$ (which is $\frac{1}{\hat{\theta}}$ as per your notation), we get,

$$h(\overline{Y_n}) \approx h(\theta) + h^{'}(\theta)(\overline{Y_n}-\theta)$$

$$\sqrt{n}\frac{(h(\overline{Y_n}) - h(\theta))}{|h^{'}(\theta)|\theta} \approx \sqrt{n}\frac{(\overline{Y_n} - \theta)}{\theta} \rightarrow \mathcal{N}(0,1)$$

$$\sqrt{n}\theta(h(\overline{Y_n}) - h(\theta)) = \sqrt{n}\theta(\frac{1}{\hat{\theta}} - \frac{1}{\theta})\rightarrow \mathcal{N}(0,1)$$

Note that I used $\frac{|h^{'}(\theta)|\theta}{\sqrt{n}}$ in the denominator because it is the standard deviation of $h(\overline{Y_n})$ which is the square root of the variance $\frac{h^{'}(\theta)^2\theta^2}{n}$. That's why || came up.

So the $c$ you are looking for is $\sqrt{n}\theta.$

Edit 1. If $X \sim \mathcal{N}(\mu,\sigma^2)$ then $\frac{X-\mu}{\sigma} \sim \mathcal{N}(0,1)$

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