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

1 Answer
PhilippG
May 16, 2016

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.

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