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.
$\endgroup$ 25 Answers
$\begingroup$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 $$ \int_1^\infty xf(x)\,dx. $$
Later addendum in response to comments:
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 $$ \int_1^\infty xf(x)\,dx = \int_1^\infty x\,ax^{-a-1}\,dx = a\int_1^\infty x^{-a}\,dx $$ $$ =a\left[\frac{x^{-a+1}}{-a+1}\right]_1^\infty = 0 - a\left(\frac{1}{-a+1}\right) = \frac{a}{a-1}. $$
$\endgroup$ 8 $\begingroup$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 $E(X)=1+\dfrac{1}{a-1}=\dfrac{a}{a-1}$.
$\endgroup$ 6 $\begingroup$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.
$\endgroup$ 0 $\begingroup$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.
$\endgroup$ $\begingroup$$$z = \frac {|{P}g_i|^2\alpha}{2-\alpha} h_i e^{j \theta_{i}}$$ where $\theta$, $h$, $g$ and $\alpha$ are random variables. Is it possible to find expected value using integration formula? OR anything like Taylor series.i just want to know how to start this.....
$\endgroup$
