conditional expectation of power law distribution

Question

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 $1<b<2$

Tnx

Answer 1

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 $$ E(X;X\gt x)=\int_x^\infty tf_X(t)\,\mathrm dt. $$