Derive the expected value for a Pareto distribution?

X is a random value that is Pareto distributed with parameter $a>0$, if $\Pr(X>x)=x^{-a}$ for all $x≥1$.

Show that $EX=a/(a-1)$ if $a>1$ and $E(X)=∞$ if $0< a \le1$.

I can derive the latter using the fact that the expected value is the integral between $0$ and $\infty$ of $\Pr(X>x)$ but I'm not sure how to go about showing the first case (i.e. when $a>1$)?

Any help would be appreciated.

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Also asked (with identical wording) on stats.SE – Dilip Sarwate Nov 24 '12 at 18:52
Due to formatting issues, i could not post here but i have a neat derivation on my blog on this link – Josephat Mwanzia Mar 14 '15 at 6:40

The density is $$f(x) = \frac{d}{dx} F(x) = \frac{d}{dx} \Pr(X\le x) = \frac{d}{dx} (1-\Pr(X> x)).$$ The expected value is $$\int_1^\infty xf(x)\,dx.$$

In the posted question, we are told that for $x\ge 1$ we have $\Pr(X>x) = x^{-a}$. It follows that for $x=1$, $\Pr(X>x)=1^{-a}=1$, so this random variable is always $\ge 1$.

Above I wrote $\dfrac{d}{dx}(1-\Pr(X>x))$. Now we can see that that is equal to $$\frac{d}{dx}(1-x^{-a}) = -(-ax^{-a-1}) = ax^{-a-1}.$$ Therefore this is the density on the interval $(1,\infty)$, and the density is $0$ everywhere else. Thus the expected value is $$\int_1^\infty xf(x)\,dx = \int_1^\infty x\,ax^{-a-1}\,dx = a\int_1^\infty x^{-a}\,dx$$ $$=a\left[\frac{x^{-a+1}}{-a+1}\right]_1^\infty = 0 - a\left(\frac{1}{-a+1}\right) = \frac{a}{a-1}.$$

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How would I find f(x) from the given information above? Would it just be the same as P(X>x)? – Mathlete Nov 24 '12 at 17:53
You know what $\Pr(X>x)$ is. Put that in the appropriate place in $\dfrac{d}{dx}(1-\Pr(X>x))$, which is the last expression on the first "displayed" line above. – Michael Hardy Nov 24 '12 at 17:54
Sorry, I didn't read that line carefully enough! – Mathlete Nov 24 '12 at 17:58
Is that integral not just used when working with a discrete random variable? – Mathlete Nov 24 '12 at 18:02
Quite the opposite: It's used only for continuous random variables. – Michael Hardy Nov 24 '12 at 18:08

This is about the convergence of mean.You can generalized it for moments of Pareto Distribution. Note that $$E|X|^r=\int_1^\infty |x|^r ax^{a-1}~dx=a.\int_1^\infty \frac{1}{x^{a-r+1}}~dx$$which converges iff $a-r+1>1$ iff $r<a$.

For $E(X)$ we have $r=1$. Hence we get the result.

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We evaluate the integral $$\int_0^\infty \Pr(X\gt x)\,dx$$ of the post.

Note that if $0\le x\lt 1$, then $\Pr(X\gt x)=1$. And if $x\ge 1$, then $\Pr(X\gt x)=x^{-a}$. Since $\Pr(X\gt x)$ is given by two different formulas, it is natural to break up the integral at $x=1$.

The integral of $\Pr(X\gt x)$, from $0$ to $1$, is $1$.

By a standard integral calculation, $$\int_1^\infty x^{-a}\,dx=\frac{1}{a-1}.$$ So $E(X)=1+\dfrac{1}{a-1}=\dfrac{a}{a-1}$.

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I think I understand what you're saying - should that not be "a" in your inequalities though? – Mathlete Nov 24 '12 at 18:06
Actually, ignore my last comment. Could you explain how you know what the values of P(X>x) are when 0<=x<1 and x=>1? – Mathlete Nov 24 '12 at 18:11
When $x\le1$ then $\Pr(X\ge x)$ is necessarily $1$, since this random variables is always $\ge1$. When $x>1$, then $\Pr(X>x)$ is given in the posted question. – Michael Hardy Nov 24 '12 at 18:52
@Manasa: If you are at a point $x\le 1$, the whole mass is to the right of you. – André Nicolas Nov 24 '12 at 18:59
How do you know to integrate from $0$ to $\infty$ instead of from $-\infty$ to $\infty$? – Jeremy Jan 21 at 17:45

The cumulative density function is $F(x)=P(x \leq X)=1-P(x>X)=1-x^{-a}.$ The derivative of $F(x)$ is density function, so $F'(x)=f(x)$. Then mean is given by standard formula: $$EX=\int_1^\infty x\cdot f(x)dx=\int_1^\infty x \cdot ax^{-a-1}dx.$$ Sometimes when $F$ does not have a derivative, then you can write $$EX=\int_\mathbb{R}xdF(x),$$ which is more general formula. This is as well useful when you have to do partial integration.

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