Q:

Calculating expected value of a pareto distribution

Suppose that you have a Pareto product distribution function defined by:
$$
f(x;k;\theta)= \begin{cases}
\frac{k\theta^k}{x^{k+1}}  & x \ge\theta \\
0 & x \lt \theta
\end{cases}
$$
How would one go about deriving the expression used to calculate the expected value $E[X]$?
I understand that in general the expected value can be calculated for any generic product distribution function $f(x)$ by the following integral:
$$E[X]=\int^\infty_{-\infty} x f(x)dx$$
Does the presence of multiple parameters change this at all? I'm not sure how to set up this integral properly using the above for $f(x; k; \theta)$.

A:

No, it makes no difference: your expectation will just be a function of $\theta$ and $k$, viz.
$$ E[X] = \int_{-\infty}^{\infty} x f(x;k;\theta) \, dx = \int_{\theta}^{\infty} \frac{k\theta^k}{x^k} \, dx = \frac{k\theta}{k-1}, $$
provided $k>1$. (In other words, the other two arguments to $f$ are just parameters: $x$ is the thing in the probability space you're integrating over.)

