Question

If the following is a probability distribution function: `f(x)=k(1/x^2)`, what is k and what is the variance?

Answer 1

The Trick is to define the proper domain of the density.
`f_k(x)=k/(x^2) AA x>k`, `f_k(x)=0` otherwise. With that useful restriction you can determine, that every `k>0` works.

Explanation:

Now the Integration is a piece of cake and the density fulfills the two properties for a density function:

1. `f(x)>0 AA x`
2. `int_k^oof(x)dx=1`

`int_k^ook/x^2dx=[-k/x|_k^oo]=lim_(x->oo)(-1/x)-(-k/k)=1`.

The expected value `E(X)=oo`, hence also the variance `V(X)=oo`.