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I have a sample of size $n$ from the following distribution:

$$f(x;\alpha,\beta)=\frac{\alpha x^{\alpha-1}}{\beta^\alpha}1_{0<x<\beta}\quad,\,\alpha>0$$

I found that the MLEs are

$$\hat{\beta}=x_{(n)}$$

and $$\hat{\alpha}=\frac{n}{n\log(x_{(n)})-\sum\limits_{i=1}^n \log(x_i)}$$

Now, since $\hat{\beta}$ is an order statistic, it's pretty straightforward to find its pdf and to show that it's consistent using its expectation and its variance. However, I'm having trouble showing the same for the $\hat{\alpha}$. Does $\hat{\alpha}$ following a common distribution that I can't see? Or is there another trick to use?

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    $\begingroup$ take a single $X$ and transform to $Y = \log{X}$ and see what you get $\endgroup$ – Cyan Nov 19 '19 at 15:46
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    $\begingroup$ If you need any more of a hint, notice that when $X$ has this distribution, $X/\beta$ has a Beta$(\alpha,1)$ distribution. $\endgroup$ – whuber Nov 19 '19 at 16:27
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Regarding the distribution of $\hat\alpha$:

The MLE of $\alpha$ can be rewritten as $$\hat\alpha=\frac{n}{\sum\limits_{i=1}^n \ln \left(\frac{X_{(n)}}{X_i}\right)}$$

Now you can use the change of variables suggested by @whuber in comments, or equivalently note that whenever $X$ has the pdf in your question, $Y=\frac{\beta}{X}$ has a Pareto distribution with pdf $$f_Y(y)=\frac{\alpha}{y^{\alpha+1}}1_{y>1}$$

As $X_i=\frac{\beta}{Y_i}$, you have $X_{(n)}=\frac{\beta}{Y_{(1)}}$. Therefore $$\sum\limits_{i=1}^n \ln \left(\frac{X_{(n)}}{X_i}\right)=\sum\limits_{i=1}^n \ln \left(\frac{{Y_i}}{Y_{(1)}}\right)\,,$$

which has a Gamma distribution courtesy of this result.

So one can say $\hat\alpha$ has an Inverse-Gamma type of distribution. Once you figure out the exact parameters of the Gamma distribution, you can find the mean and variance of $\hat\alpha$.

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Re-arranging your equation for $\hat{\alpha}$ gives the useful form:

$$\frac{1}{\hat{\alpha}} = \log(x_{(n)}) - \frac{1}{n} \sum_{i=1}^n \log(x_i).$$

This form is useful for showing the convergence of the MLE. From the laws of large numbers you are clearly going to get $x_{(n)} \rightarrow \beta$ and $\frac{1}{n} \sum_{i=1}^n \log(x_i) \rightarrow \mathbb{E}[\log(X)]$, which gives you:

$$\frac{1}{\hat{\alpha}} \rightarrow \log(\beta) - \mathbb{E}[\log(X)].$$

All you need to do now is to find the expected value of $\log(X)$ (which is a simple exercise) and then you will get a convergence result for $\hat{\alpha}$.

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