An approximate confidence interval for the $\alpha$ parameter of a Pareto Type II distribution when $\lambda$ is known

The Pareto Type II distribution, also known as the Lomax distribution, has the following density, $$f(x|\alpha,\lambda)=\frac{\alpha\lambda^{\alpha}}{(\lambda+x)^{\alpha+1}}, \qquad x>0,\ \alpha>1,\ \lambda>0$$ with $$\lambda$$ known. I'm trying to find an approximate confidence interval for $$\alpha$$. For context this is an old exam question so students would have access to statistical tables, so I'm guessing that the confidence interval will involve either Gaussian or Student T distributed random variable.

Work so far:

So assume we observe a sample $$x_1,\dots,x_n$$ from the aforementioned distribution. I found the MLE for $$\alpha$$ to be $$\hat\alpha=\frac{n}{\sum_{i=1}^n\log\big(\frac{x_i}{\lambda}+1\big)}$$

The following result holds for the MLE, $$\sqrt{\mathbb{E}[\mathcal{I}(\hat\alpha)]}(\hat\alpha-\alpha)\sim \mathcal{N}(0,1)$$

Now $$\sqrt{\mathbb{E}[\mathcal{I}(\hat\alpha)]}=\frac{\sqrt{n}}{\hat \alpha}$$ so the CI should look something like this,

$$\hat\alpha \pm q\frac{\hat\alpha}{\sqrt{n}}$$

where $$q$$ is a quantile from the standard Gaussian distribution that corresponds to a specific significance level.

Is this correct?

• There's a typo in your density function. That should be $\lambda^\alpha$ (not $\lambda^n$) in the numerator. Commented May 4, 2023 at 12:26
• Thanks, I'll change that Commented May 4, 2023 at 12:28
• Seems correct. However mind that for this distribution (a.k.a Lomax) the distribution of the estimate differs from the normal for moderate $n$ (say $n < 100$) and even a profile likelihood interval has a misleading coverage rate. Checking this is a very good exercise.
– Yves
Commented May 4, 2023 at 12:48

Your MLE is correct. The asymptotic result we want to use here is that the MLE, $$\hat\alpha$$, converges in distribution to $$\mathrm{N}(\alpha, \mathcal{I}(\alpha)^{-1})$$ as $$n \rightarrow \infty$$. The Fisher information is $$\mathcal{I}(\alpha)=\frac{n}{\alpha^2}$$.
At this point, we can approximate $$\mathcal{I}(\alpha)$$ by $$\mathcal{I}(\hat\alpha)$$ to obtain the CI you give in your answer, but notice that the quantity $$\sqrt{\mathcal{I}(\alpha)}(\hat\alpha-\alpha) = \frac{\sqrt{n}}{\alpha}(\hat\alpha-\alpha)=\sqrt{n}\left(\frac{\hat\alpha}{\alpha}-1\right)$$ is approximately standard normal for large $$n$$.
It follows that we can construct a CI of the form $$\left[ \left(1+\frac{q}{\sqrt{n}} \right)^{-1} \hat{\alpha}, \left(1-\frac{q}{\sqrt{n}} \right)^{-1} \hat{\alpha} \right]$$ where $$q$$ is a quantile from the standard normal.