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

May 16, 2016

The Trick is to define the proper domain of the density.
${f}_{k} \left(x\right) = \frac{k}{{x}^{2}} \forall x > k$, ${f}_{k} \left(x\right) = 0$ 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. $f \left(x\right) > 0 \forall x$
2. ${\int}_{k}^{\infty} f \left(x\right) \mathrm{dx} = 1$

${\int}_{k}^{\infty} \frac{k}{x} ^ 2 \mathrm{dx} = \left[- \frac{k}{x} {|}_{k}^{\infty}\right] = {\lim}_{x \to \infty} \left(- \frac{1}{x}\right) - \left(- \frac{k}{k}\right) = 1$.

The expected value $E \left(X\right) = \infty$, hence also the variance $V \left(X\right) = \infty$.