I'd be interested in solving problem 2.2 here. The OP posted three questions in a single thread so thought it could be useful to split the questions. Apologies if this is a duplicate thread.

Suppose that $X$ has a power law distribution $P(X > x) = a x^{-b}$ for $x > x_0 > 0$ and some $a > 0$, $b > 1$. Calculate the conditional expectation $ E(X|X>x)$ , $x> x_0$

Here's my attempt. This may be interesting in tackling the problem

$P(X>0) = \int_x^{\infty} a s^{-b} ds = - \frac{a}{1-b} x^{-b+1}$, using $b>1$

$P(X|X>x) = \frac{P(X \in ds, X>x)}{P(X>x)} = \frac{P(X\in ds)}{P(X>x)}$

$E(X|X>x) =- \frac{1-b}{a}x^{b-1} \int_x^{\infty} a s s^{-b} ds$

This is probably wrong because the problem only states that $b>1$ and the integral in the expression above leads to $ \frac{a}{2-b}s^{2-b} $ where $s$ needs to be evaluated as $s\rightarrow \infty$, where the expression does not get squeezed if $1<b<2$


  • $\begingroup$ What does exactly mean the notation "$P(X\in ds, X>x)$"? I have seen it already several times but I cannot find any reference in the literature to what it exactly means. Thanks. $\endgroup$ – RandomGuy Aug 17 '16 at 13:17

Let $b\gt1$, then $x_0=a^{1/b}$ and, for every $x$, $$ E(X\mid X\gt x)=\frac{b}{b-1}\,\max\{x,x_0\}. $$ To show this, it may help to use that the density $f_X$ of $X$ is such that, for every $x$, $$ f_X(x)=\frac{ab}{x^{b+1}}\,\mathbf 1_{x\geqslant x_0}, $$ and that $$ E(X;X\gt x)=\int_x^\infty tf_X(t)\,\mathrm dt. $$

  • $\begingroup$ Is this answer correct? For $b > 2$ I get a different result. I find also counter-intuitive that the result above does not depend on $x$. Thanks. $\endgroup$ – IcannotFixThis Feb 21 '14 at 15:33
  • $\begingroup$ It was not, in particular for the reason you mention. See revised version and thanks for asking. $\endgroup$ – Did Feb 21 '14 at 15:54
  • $\begingroup$ Got it, thanks. One last question: why is $x_0 = a^{1/b}$? $\endgroup$ – IcannotFixThis Feb 21 '14 at 18:01
  • $\begingroup$ Because $P(X\gt x_0)=1$. $\endgroup$ – Did Feb 21 '14 at 18:03
  • 1
    $\begingroup$ @SimonMarkett No, you are confusing E(X;A) with E(X|A) = E(X;A) / P(A), for A = [X>x]. $\endgroup$ – Did Mar 10 '15 at 16:30

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