# Consistency of MLE of $\alpha$ when pdf is $f(x;\alpha,\beta)=\frac{\alpha x^{\alpha-1}}{\beta^\alpha}1_{0<x<\beta}$

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

$$f(x;\alpha,\beta)=\frac{\alpha x^{\alpha-1}}{\beta^\alpha}1_{00$$

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?

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

## 2 Answers

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 $$\hat\alpha$$.

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 $$\hat{\alpha}$$.