I am currently preparing for an upcoming Discrete choice models exam. I have come across an exercise where I need to calculate the conditional expected value.
Given:
$X$~$Pareto (a, b)$. Find $E(X | X > 11)$ when $(a, b) = (\frac{11}{2}, \frac{5}{2})$.
What I have so far:
$PDF = \frac{ab^a}{u^{a+1}}, \hspace{2mm} P(X > 11) = (\frac{a}{11})^b$
Also, $\hspace{2mm} E(X) = \frac{ab}{a-1}$
Problem: I do not quite understand what is the process for calculating a conditional expected value.
On
The calculation of the probability $\Pr[X > 11]$ is not consistent with your definition of the PDF. If you define $$f_X(x) = \frac{ab^a}{x^{a+1}}, \quad x \ge b,$$ then $$\Pr[X > x] = \int_{t = x}^\infty f_X(t) \, dt = \int_{t = x}^\infty \frac{ab^a}{t^{a+1}} \, dt = ab^a \left[ -\frac{1}{at^a} \right]_{t=x}^\infty = ab^a \left( 0 + \frac{1}{ax^a} \right) = \left(\frac{b}{x}\right)^a,$$ and not $(a/x)^b$ as you stated. So you need to make sure that you are not mixing up the parameters.
That said, we can compute the conditional expectation as follows:
$$\operatorname{E}[X \mid X > x] = \frac{\int_{t=x}^\infty t f_X(t) \, dt}{\Pr[X > x]} = \left(\frac{x}{b}\right)^a \int_{t=x}^\infty \frac{ab^a}{t^a} \, dt.$$ The remainder of the computation is left to you as an exercise; then setting $x = 11$, $a = 5/2$, $b = 11/2$ gives the desired numerical answer.
Let $X\sim\mathsf{Pareto}(a,b)$ and let $c>b$.
Then for $x>b$ we have $P(X>x)=\left(\frac{b}{x}\right)^a$ so for $x>c$ we find:$$P(X>x|X>c)=\frac{P(X>x)}{P(X>c)}=\left(\frac{c}{x}\right)^a$$showing that the conditional distribution is again Pareto but this time with parameters $a$ and $c$.
So if $a>1$ then we conclude that:$$\mathbb E[X|X>c]=\frac{ac}{a-1}$$
Also you can go for $$\mathbb E[X|X>c]$$ directly according to the formula:$$\mathbb E[X|X>c]=\frac{\mathbb EX1_{X>c}}{P(X>c)}$$ Here the numerator is $\frac{ac}{a-1}\left(\frac{b}{c}\right)^a$ and the denominator is $\left(\frac{b}{c}\right)^a$.