A particularly challenging integral

**Dave2718** · June 21st 2010, 09:08 PM

Hello, Actex recently posted a really good challenge problem on their facebook page, and I've been trying to figure it out thus far, which eventually turned into deriving the variance of Pareto Distribution. The PDF is

f(x)= (a(n^a))/((x+n)^(a+1)) for x>0, a>0, n>0.

So I've been able to calculate E[x], because I'm going to have to square it in the end anyways, and got

E[x] = n/(1-a)

However I'm having trouble computing the integral for E[x^2], the second movement of the distribution.

Could anyone at least give me some sort of hint on it? I know partial fraction decomp won't work because we have a power of a..I've been thinking maybe trig sub as well, however I'm not entirely sure.

If you aren't particularly up on probability, computing E[x^2] is the same as computing(for this problem) the integral from 0 to infinity of (x^2)f(x) dx

**Chris L T521** · June 21st 2010, 10:03 PM

Originally Posted by Dave2718

Hello, Actex recently posted a really good challenge problem on their facebook page, and I've been trying to figure it out thus far, which eventually turned into deriving the variance of Pareto Distribution. The PDF is

f(x)= (a(n^a))/((x+n)^(a+1)) for x>0, a>0, n>0.

So I've been able to calculate E[x], because I'm going to have to square it in the end anyways, and got

E[x] = n/(1-a)

However I'm having trouble computing the integral for E[x^2], the second movement of the distribution.

Could anyone at least give me some sort of hint on it? I know partial fraction decomp won't work because we have a power of a..I've been thinking maybe trig sub as well, however I'm not entirely sure.

If you aren't particularly up on probability, computing E[x^2] is the same as computing(for this problem) the integral from 0 to infinity of (x^2)f(x) dx

$\displaystyle E[X^2]=an^a\int_0^{\infty}\frac{x^2}{(x+n)^{a+1}}dx$

Let $u=x+n\implies x=u-n$ . Therefore, $du=dx$ .

Thus, the integral becomes $\displaystyle an^a\int_n^{\infty}\frac{(u-n)^2}{u^{a+1}}du=an^a\int_n^{\infty} u^{1-a}-2nu^{-a}+n^2u^{-(a+1)}du$

This gives us $an^a\left.\left[\tfrac{1}{2-a}u^{2-a}-\tfrac{2n}{1-a}u^{1-a}-\frac{n^2}{a}u^{-a}\right]\right|_n^{\infty}=a\left[\frac{n^2}{a-2}-\frac{2n^2}{a-1}+\frac{n^2}{a}\right]$

There *may* be a mistake somewhere, but I hope this gives you the general idea of how to do it.

**chisigma** · June 21st 2010, 10:06 PM

Is...

$\displaystyle \int_{0}^{\infty} \frac{x^{2}}{(x+n)^{a+1}} dx = |- \frac{x^{2}}{a\ (x+n)^{a}}|_{0}^{\infty} + \frac{2}{a} \int_{0}^{\infty} \frac{x\ dx}{(x+n)^{a}}$ (1)

If $a>2$ the first term of (1) converges and its value is 0 so that we have to compute...

$\displaystyle \int_{0}^{\infty} \frac{x\ dx}{(x+n)^{a}}= |- \frac{x}{(a-1)\ (x+n)^{a-1}}|_{0}^{\infty}+ \frac{1}{a-1} \int_{0}^{\infty} \frac{dx}{(x+n)^{a-1}}$ (2)

Also in this case the first term of (2) is 0 so that it remains to compute...

$\displaystyle \int_{0}^{\infty} \frac{dx}{(x+n)^{a-1}} = \frac{1}{(a-2)\ n^{a-2}}$ (3)

... so that the final result, if $a>2$ and no mistakes of me [

], is ...

$\displaystyle E[x^{2}] = \frac{n^{2}}{(a-1)\ (a-2)}$ (4)

Kind regards

$\chi$ $\sigma$

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