Expected value of a strange PMF

383 Views Asked by At

Say we have a PMF defined as $p_X(x)=\frac{6}{\pi^2}\frac{1}{x^2} $.

Our expected value $E[X]=\sum_{x=1}^\infty x\frac{6}{\pi^2}\frac{1}{x^2}$.

Can we merely simplify this to be $E[X]=\sum_{x=1}^\infty \frac{6}{\pi^2x}$? How else could we define $E[X]$?

1

There are 1 best solutions below

0
On BEST ANSWER

The result $\sum_{x=1}^{\infty} \frac{1}{x^2} = \frac{\pi^2}{6}$ is due to the well known Basel problem solved by Euler in his 20's. The PMF in question is the scaled version of this famous infinite series.

As indicated by the comment by @angryavian, this is the discrete analogue of the Cauchy distribution.

\begin{align} E[X] & =\sum_{x=1}^\infty x\frac{6}{\pi^2}\frac{1}{x^2} \\ & = \frac{6}{\pi^2} \sum_{x=1}^\infty \frac{1}{x} \end{align}

which is a constant times the divergent harmonic series and as such the $EX$ does not exist.