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<a<b $, 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 $E(X^2)=\int_\theta^\infty{x^2\frac{k\theta^k}{x^{k+1}}}dx=k\theta^k\int_\theta^\infty{x^{-k+1}dx}=\frac{-k\theta^2}{-k+2}$

  • $\begingroup$ Please see the wiki page of the Pareto distribution which likely has all the answers to your questions. $\endgroup$ – StubbornAtom Nov 9 '18 at 17:53
  • $\begingroup$ 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? $\endgroup$ – D. Wei Nov 9 '18 at 18:23
  • $\begingroup$ Yes, but you have to show your work. $\endgroup$ – StubbornAtom Nov 9 '18 at 18:24
  • $\begingroup$ By the way, your answers are correct now if you mention the conditions under which the moments exist. $\endgroup$ – 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}

You should be able to proceed now.

  • $\begingroup$ 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! $\endgroup$ – D. Wei Nov 10 '18 at 3:12
  • $\begingroup$ @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 $k>2$. $\endgroup$ – StubbornAtom Nov 10 '18 at 6:12
  • $\begingroup$ Yes, thank you for pointing that out, that was just a typo though, thanks to your help I have solved this problem :D $\endgroup$ – D. Wei Nov 12 '18 at 16:54

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