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Let $x$ be a Pareto distribution with a known scale parameter $m>0$, i.e. $x\sim f(x|a)=\frac{am^a}{x^{a+1}}, x>a, a>0$

$\mathrm{E}\left[X\right]=\frac{am}{a-1}$

Using method of moments estimator for the shape parameter, $\frac{\hat{a}m}{\hat{a}-1}=\sum_{i=1}^{n}{\frac{x_i}{n}}, \hat{a}=\frac{\sum{x_i}}{\sum{x_i}-mn}$

How does one calculate the bias of the estimator? What's $\mathrm{E}\left[\hat{a}\right]$? Is there a known distribution for sum of Pareto variables?

If there's no closed form expression for the expectation, is the bias always one way?

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  • $\begingroup$ This seems rather like a textbook-style exercise. Is this for some class? $\endgroup$ – Glen_b -Reinstate Monica Dec 16 '15 at 7:48
  • $\begingroup$ @Glen_b no, I wish. It's for modeling losses associated with business activities, where the threshold is determined by business judgment $\endgroup$ – C.J. Jackson Dec 16 '15 at 16:24
  • $\begingroup$ I've made some additional comments with that in mind. $\endgroup$ – Glen_b -Reinstate Monica Dec 16 '15 at 22:37
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The distribution of a sum of Pareto variates is not especially simple, but has been done. [1] [2]

Without loss of generality, we can take $m=1$; we can simply divide through by $m$ to work with $X^*=X/m$ and the lower limit for $X^*$ is then $1$. Since $m$ is then just a scale factor applied to the data we can translate any results back to the original data scale.

In this answer I haven't attempted to compute the exact bias from those published results. If it were me faced with this exercise I probably would focus first on using simulation to obtain a clear understanding how the bias relates to the $a$ parameter and the sample size (though I think we can say something about how it should work as a function of sample size).

However, we can do the last part easily enough.

$\bar{x}$ will be unbiased for $E(X) = \frac{a}{a-1}$, but we have

$$\hat{a}=\frac{\bar{x}}{\bar{x}-1}=\frac{1}{{1-\frac{1}{\bar{x}}}}$$

Taking $Y=\bar{X}$, we can show that $\varphi(Y)=\frac{1}{{1-\frac{1}{Y}}}$ is convex.

From Jensen's inequality, we can then show that the estimator is biased and in which direction. Jensen's inequality says that for $\varphi$ convex:

$$\varphi \left(\mathbb {E} [Y]\right)\leq \mathbb {E} \left[\varphi (Y)\right]$$

(The conditions under which equality will hold don't apply here; the inequality will be strict.)

This shows that the estimator will be too high on average.

Here's some results of a simulation with $a=2$

enter image description here

(10000 samples each with n=100, so here we have 10000 $\hat{a}$ values. The blue line is the mean estimate from those 10000 samples.)

This simulation doesn't prove anything, but it shows that we don't contradict the derived result; looks like there was no error in concluding the bias was upward.

Such simulations allow us to see how the bias changes with $a$ -- by doing the same thing across a variety of $a$ values -- and with sample size (again, by using different $n$).


If this is not for a class exercise, I'm very curious why you wouldn't use maximum likelihood in this case:

  • it's very simple - for $m=1$ it's the reciprocal of the mean of the logs; if $m$ is not 1, you subtract $log(m)$ from the mean of the logs before taking reciprocals.

  • For this parameterization, it's not unbiased either, but it makes better use of the data, and it will have lower variance (and lower bias, by the look of some simulations).

  • The bias is also easy to compute!


[1] Blum, M. (1970),
"On the Sums of Independently Distributed Pareto Variates"
SIAM Journal on Applied Mathematics, 19:1 (Jul.), pp. 191-198

[2] Ramsay, Colin M. (2008)
"The Distribution of Sums of I.I.D. Pareto Random Variables with Arbitrary Shape Parameter"
Communications in Statistics - Theory and Methods, 37:14, pp 2177-2184

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  • $\begingroup$ Thanks for the help. I have estimated both MLE in addition to MoM with software. It just that in simulation MoM seem to produce much lower tail losses and I'm trying to figure out why theoretically. If bias on shape parameter is upward, that means tails are thinner, so it makes sense along with what you wrote here. $\endgroup$ – C.J. Jackson Dec 16 '15 at 23:02
  • $\begingroup$ That would be a good question to post, but briefly, if MoM is more positively biased you'll tend to get less of an upper tail on average (since larger $\hat{a}$ values will be more common). The larger variance of the MoM would be expected to complicate that picture a bit (since it will lead to both more of the larger and smaller $\hat{a}$ values, and the interplay between the way they together impact the distribution of $\hat{a}$ and the way that $\hat{a}$ impacts the tail will determine the overall effect). $\endgroup$ – Glen_b -Reinstate Monica Dec 16 '15 at 23:08
  • $\begingroup$ The quantity I'm concerned with is covering some high cumulative percentile of losses. The upward bias on $\hat{a}$ would drive it down. But would the higher variance of $\hat{a}$ in this case drive up the cumulative losses? $\endgroup$ – C.J. Jackson Dec 16 '15 at 23:18
  • $\begingroup$ I'd expect it to, yes, since I expect the extreme upper tail will be more impacted by a slightly lower estimate of the parameter than it will be by a slightly higher one. I can't tell for certain until I actually work it through (calculating $\frac{d^2S(x)}{da^2}$ might be sufficient) or simulate but I'd bet on yes. $\endgroup$ – Glen_b -Reinstate Monica Dec 17 '15 at 0:05

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