# Pareto distribution moments

I have 2 questions where I have to use the Pareto distribution.

1. A family of pdf’s that has been used to approximate the distribution of income, city population size, and size of firms is the Pareto family. The family has two parameters, $$k$$ and $$y$$, both $$> 0$$, and the pdf is: $$f(x;k,\theta)=\frac{k\cdot\theta^k}{x^{k+1}} \ for \ x\geq\theta$$ and is $$0$$ otherwise.

a) Sketch the graph of $$f(x;k,\theta)$$

b) Verify the the total area under the graph is $$1$$.

c) For $$\theta, obtain an expression for the probability $$P(a\leq{X}\leq{b}).$$

For this question, I have integrated and proved part b) because all $$k$$ and $$\theta$$ cancel. I have also got an expression for part c) by just changing the bounds to $$a$$ and $$b$$. However, I have no idea how to graph the 3 variables, any thoughts?

1. Let X have the Pareto pdf introduced in Exercise 1.

a) If $$k>1$$, compute $$E(X)$$

b) What can you say about $$E(X)$$ if $$k=1?$$

c) If $$k>2$$, show that $$V(X)=k\theta^2(k-1)^{-2}(k-2)^{-1}$$.

d) If $$k=2$$, what can you say about $$V(C)$$?

e) What conditions on $$k$$ are necessary to make $$E(X^n)$$finite?

I got $$E(X)=\int_\theta^\infty{x\frac{k\theta^k}{x^{k+1}}}dx=k\theta^k\int_\theta^\infty{x^{-k}}dx=-\frac{k\theta}{-k+1}$$. And I got $$E(X^2)=\int_\theta^\infty{x^2\frac{k\theta^k}{x^{k+1}}}dx=k\theta^k\int_\theta^\infty{x^{-k+1}dx}=\frac{-k\theta^2}{-k+2}$$

• Please see the wiki page of the Pareto distribution which likely has all the answers to your questions. – StubbornAtom Nov 9 '18 at 17:53
• Right, I got the graph, but now my E(X) and V(X) for question 2 aren't matching what they give V(X) to be, any help? – D. Wei Nov 9 '18 at 18:23
• Yes, but you have to show your work. – StubbornAtom Nov 9 '18 at 18:24
• By the way, your answers are correct now if you mention the conditions under which the moments exist. – StubbornAtom Nov 9 '18 at 18:59

In light of the fact that $$\frac{1}{x^\alpha}\to 0\text{ as }x\to\infty\quad\text{ provided } \alpha>0$$
, we have the $$r$$-th order raw moment of $$X$$ about $$0$$ :
\begin{align} E(X^r)&=\int_{\theta}^\infty \frac{x^r\,k\theta^k}{x^{k+1}}\,dx \\&=k\theta^k\int_{\theta}^\infty x^{r-k-1}\,dx \\&=k\theta^k\lim_{A\to\infty}\left[\frac{x^{-(k-r)}}{-(k-r)}\right]_{\theta}^A \\&=\frac{k\theta^r}{k-r}\qquad,\text{ if }k>r \end{align}
• Right so I get $V(X)=\frac{k\theta^2}{k-2}-(\frac{k\theta}{k-1})^2=k\theta^2(\frac{1}{k-2}-\frac{k}{(k-1)^2})=k\theta^2(\frac{k^2-2k+1-k^2-2k}{(k-2)(k-1)^2})$. Thanks! – D. Wei Nov 10 '18 at 3:12
• @D.Wei The second $2k$ in the numerator has a $+$ in front, so variance is just $\frac{k\theta^2}{(k-2)(k-1)^2}$ for $k>2$. – StubbornAtom Nov 10 '18 at 6:12