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Having trouble with this question.

Suppose $X$ is a random variable with probability function: $f_X(x)=k/x^2$

I need to use a "Basel Problem" to find k and prove that the expected value $E(X)$ does not exist.

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Is $f$ density function and is there some bound on $x$, like $x\ge 1$? – Berci Nov 1 '12 at 23:23
I think you cannot set simply a $k$ such that you will get a density. There is a problem as Berci indicated around $x$ close to zero. – Seyhmus Güngören Nov 1 '12 at 23:27
Presumably $x \in \mathbb{N}$ and $f_X$ is a pmf. – Niels Diepeveen Nov 1 '12 at 23:54
up vote 1 down vote accepted

Use that $\displaystyle\int_{\Bbb R}f_X(x)dx =1$ and that for any (measurable) function $g$, we have $$E(g(X))=\int_{\Bbb R} g(x)\cdot f(x)dx$$ In particular, $E(X)=\displaystyle\int^\infty_{1}x\cdot \frac k{x^2} dx = \infty$ (assuming, $X\ge 1$ always).

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If we are really supposed to use the "Basel Problem," then our probability is defined only on the positive integers, and $\Pr(X=n)=\dfrac{k}{n^2}$,

The sum of the probabilities over th sample space must be $1$. By Euler's solution to the Basel Problem, we have $$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots=\sum_1^\infty \frac{1}{n^2} =\frac{\pi^2}{6}.$$ It follows that we must have $k=\dfrac{6}{\pi^2}$.

The expectation of $X$ is then $$\sum_1^\infty n\frac{k}{n^2}=\frac{6}{\pi^2}\sum_1^\infty \frac{1}{n}.$$ But by the divergence of the harmonic series, the expectation does not exist, or, if one prefers, is infinite.

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