The derivation of the formula for variance for a Pareto Distribution

I need help with the step-by-step process of how to derive the formula for variance. 
First I need to see how to calculate E[X^2] and then how to get variance using Vax[X] = E[X^2] - (E[X])^2.

I'm pretty sure I'm doing something wrong with the integration of x^2*ab/(b+x)^a+1 and the algebra that follows. 

I've been trying this for 2 days now and I cannot seem to get it.

I've posted a picture of my attempt so you can see what I'm expecting. 


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Erdos Erdos
  • Erdos Erdos

    Leave a comment if you need any clarifications.

  • How did you know to multiply and divide by(a-k) lambda^(a-k)????

  • Erdos Erdos

    We would like the function inside the integral to be a Pareto distribution, since we know it integrates to 1. So you try to create a Pareto distribution algebraic manipulations.

  • Shouldnt the end of be negative?? for when we integrate and then plug in infinity, that part goes to 0 as infinity is in the denominator but then it's minus a minus one?? To be more specific what is the step between a*lambda^a/(a-k)/((a-k)lambda^(a-k) multiplied by the integration of (a-k)lambda*^(a-k)/x^(a-k+1) dx and the last step. Because I get (a*lambda^a)/(a-k)lambda^(a-k) multiplied by 0 - 1

  • OOOO are you using the fact that the density integrates to 1 therefore in that last step for E[X^k] a is (a-k)? then since it is the integral of the density it's just equal to 1!!

  • Erdos Erdos

    Yes, I revised my solution and added some more details.

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