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. $\endgroup$ Commented May 4, 2023 at 12:26
  • $\begingroup$ Thanks, I'll change that $\endgroup$
    – 29703461
    Commented May 4, 2023 at 12:28
  • 1
    $\begingroup$ 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
    Commented 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
    Commented May 6, 2023 at 13:38
  • $\begingroup$ The method in the answer avoids the additional approximation, so I suppose it would be preferred. $\endgroup$ Commented 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
    Commented May 17, 2023 at 14:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.