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.
 A: 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
$$
\int_1^\infty xf(x)\,dx = \int_1^\infty x\,ax^{-a-1}\,dx = a\int_1^\infty x^{-a}\,dx
$$
$$
=a\left[\frac{x^{-a+1}}{-a+1}\right]_1^\infty = 0 - a\left(\frac{1}{-a+1}\right) = \frac{a}{a-1}.        
$$
A: 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 $E(X)=1+\dfrac{1}{a-1}=\dfrac{a}{a-1}$.
A: 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. 
A: 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.
A: $$z = \frac {|{P}g_i|^2\alpha}{2-\alpha} h_i e^{j \theta_{i}}$$
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.....
