6
$\begingroup$

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.

  • 0
    Due to formatting issues, i could not post here but i have a neat derivation on my blog on this [link](http://mwanziamath.blogspot.com/search/label/mean%20of%20pareto%20derived "Just click")2015-03-14

5 Answers 5

8

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}. $

  • 0
    @Manasa : In view of your questions in comments under this answer and another posted answer, I have added more to my answer. See above.2012-11-24
5

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 $E(X)=1+\dfrac{1}{a-1}=\dfrac{a}{a-1}$.

  • 0
    This is a standard alternate formula (mentioned in the question) for computing the expectation of a positive random variable. The [Wikipedia article](https://en.wikipedia.org/wiki/Expected_value) on expectation mentions it near the end.2016-01-21
1

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.

0

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.

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

0

$z = \frac {|{P}g_i|^2\alpha}{2-\alpha} h_i e^{j \theta_{i}}$ 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.....