I have a problem the answer for which is intuitively obvious but is more difficult than anticipated to actually show.
Claim: If X is a random variable such that the expectation of X does not exist, then the variance of X does not exist either.
This seems true on an intuitive level, as the variance is a measure taken relative to the mean. My issue is in showing that, for a discrete random variable X with E[X] and E[X^2] both infinity, V[X] does not exist. In this case, using the definition V[X] = E[X^2]-E[X]^2, I have a sum which diverges to infinity take (the square of) another sum which diverges to infinity, which I don't really know how to treat.
Any tips would be appreciated.
Please see if the below is useful
\begin{align*} \int\limits_{x=-\infty}^\infty x^2 f_X(x) dx &= \int\limits_{x=-\infty}^0 x^2 f_X(x) dx + \int\limits_{x=0}^1 x^2 f_X(x) dx+\int\limits_{x=1}^\infty x^2 f_X(x) dx\\ &\geq \int\limits_{x=-\infty}^0 x^2 f_X(x) dx + \int\limits_{x=0}^1 0 f_X(x) dx+\int\limits_{x=1}^\infty x^2 f_X(x) dx\\ &\geq \int\limits_{x=-\infty}^0 x f_X(x) dx + \int\limits_{x=0}^1 0 f_X(x) dx+\int\limits_{x=1}^\infty x f_X(x) dx\\ &= \int\limits_{x=-\infty}^\infty x f_X(x) dx + \int\limits_{x=0}^1 (-x) f_X(x) \\ &\geq \int\limits_{x=-\infty}^\infty x f_X(x) dx - \int\limits_{x=0}^1 1 f_X(x) \\ &\geq \int\limits_{x=-\infty}^\infty x f_X(x) dx - \int\limits_{x=-\infty}^\infty f_X(x) \\ &= \left[\int\limits_{x=-\infty}^\infty x f_X(x) dx\right]-1\\ &= \mathbb E [X]-1 \end{align*} Therefore $\mathbb E [X^2] \geq \mathbb E [X]-1 = \infty$.
Let's call $\mathbb E[X] = \mu$. Then, $Var(X)=\mathbb E[(X-\mu)^2]$. \begin{align*} \lim\limits_{\mu\to\infty}Var [X] &=\lim\limits_{\mu\to\infty}\int\limits_{x=-\infty}^\infty (x-\mu)^2 f_X(x) dx\\ &=\lim\limits_{\mu\to\infty}\int\limits_{x=-\infty}^\infty [x^2 + \mu(\mu-2x)] f_X(x) dx\\ &=\int\limits_{x=-\infty}^\infty x^2f_X(x) dx +\lim\limits_{\mu\to\infty} \int\limits_{x=-\infty}^\infty\mu(\mu-2x) f_X(x) dx\\ &=\int\limits_{x=-\infty}^\infty x^2f_X(x) dx +\lim\limits_{\mu\to\infty} \mu\int\limits_{x=-\infty}^\infty(\mu-2x) f_X(x) dx\\ &\geq\int\limits_{x=-\infty}^\infty x^2f_X(x) dx +\lim\limits_{\mu\to\infty} \mu\int\limits_{x=-\infty}^\infty f_X(x) dx\\ &=\int\limits_{x=-\infty}^\infty x^2f_X(x) dx +\lim\limits_{\mu\to\infty} \mu\cdot 1\\ &=\mathbb E [X^2] +\lim\limits_{\mu\to\infty} \mu\cdot 1\\ &=\infty \end{align*}