Ashley P.

asked • 07/19/22

Statistics - Pareto Distribution

I'm taking an Mathematical Stat course at university and I came up with this question.


Question can be found via(the question involves hard to type notations, so I thought of uploading it and sharing: drive(dot)proton(dot)me/urls/FVHX8Y08RR#jb7eaDEeff3H


As for this question(the one I've marked in red colour), we need to find a parameter estimation for alpha.

I have completed part (i) and my immediate next approach for solving part (ii) was to find E(x) and E(x^2) and then find estimator for alpha.


However, here I'm facing trouble with finding E(x^2) using integration. I'm getting the answer for integration related to E(x^2) as follows: E(x^2) = [ alpha*((beta)^alpha)/(2-alpha) ]*[((infinity)^(2-alpha)) - ((beta)^(2-alpha))


Obviously alpha cannot be equal to 2 in this case given above.

Also, from part 1, we know E(x) will only hold when alpha>1.


Your help on finding a parameter estimator for alpha is highly appreciated!

1 Expert Answer

By:

Steven S. answered • 07/19/22

Tutor
5.0 (466)

Experienced Tutor Specializing in Probability and Statistics

Ashley P.

Thank you very much for pointing that out! However, in case Beta is not know how do we evaluate this integral to estimate Alpha & Beta?
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07/20/22

Steven S.

Since you have two unknown parameters, you will need another equation. The most straightforward way is to find the expectation of X^2 in terms of alpha and beta. That way, you'll have two equations (the first is given in part (i)). As you noted before, calculating the second moment, E[X^2], is tricky. The approach to use is to utilize the fact that the original distribution integrates to 1 for x > 0. So, the integral of the pareto distribution from 0 to infinity is equal to 1. You would then try find E[X^2] using integration and manipulate the integral such that it looks like integral of the pareto distribution from 0 to infinity (with appropriate variable substitution). Once you have your two equations, you can try solving for alpha and beta in terms of the two moments. Solving this algebraically will be very difficult, so what people would end up doing is use a computer to solve the system of equations.
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07/20/22

Ashley P.

Thank you for the great explanation, now that makes sense!
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07/20/22

Roger R.

tutor
The improper integral for E[X^2] is convergent only if α >2. In this case, your "infinity" term becomes zero, and the second moment becomes E[X^2] = α(β^2)/(α -2).
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07/20/22

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