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?

  • 1
    $\begingroup$ There's a typo in your density function. That should be $\lambda^\alpha$ (not $\lambda^n$) in the numerator. $\endgroup$ May 4, 2023 at 12:26
  • $\begingroup$ Thanks, I'll change that $\endgroup$
    – 29703461
    May 4, 2023 at 12:28
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    $\begingroup$ 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. $\endgroup$
    – Yves
    May 4, 2023 at 12:48

1 Answer 1


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.

  • $\begingroup$ So both methods lead to a 95% confidence interval. Is one method prefered when both are possible? $\endgroup$
    – 29703461
    May 6, 2023 at 13:38
  • $\begingroup$ The method in the answer avoids the additional approximation, so I suppose it would be preferred. $\endgroup$ May 16, 2023 at 13:19
  • $\begingroup$ I have come across other posts relating to this topic and there exists extensive literature suggesting in most cases the inverse of the observed information matrix gives a better approximation to the true covariance matrix of the estimators, and this is therefore the preferred method in most applications. Efron and Hinkley (1978) published a paper on this. $\endgroup$
    – 29703461
    May 17, 2023 at 14:18

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