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I found on the book Casella & Berger: "Statistical Inference" the following theorem (2.3.4):

If $X$ is a random variable with finite variance, then for any constants $a$ and $b$ $$\operatorname{Var}(aX+b)=a^2\operatorname{Var}X$$

This makes me wonder the following problem:

Give an example of a random variable $X$ (with infinite variance) and real numbers $a,b$ such that $$\operatorname{Var}(aX+b)\ne a^2\operatorname{Var}X$$

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    $\begingroup$ What happens if $a=0$? Since $\operatorname{var}(X)=\infty$, do you want to interpret $a^2\operatorname{var}(X) = 0^2\cdot\infty$ as $\infty$ or an arbitrary (not necessarily zero) positive finite real number? $\endgroup$
    – Dilip Sarwate
    Commented Oct 4, 2014 at 20:45
  • $\begingroup$ I have learned from measure theory that $0\cdot\infty=0$. $\endgroup$
    – juniorstatisticstudent
    Commented Oct 4, 2014 at 20:47
  • $\begingroup$ Measure theory does not teach that. The "$\infty$" in measure theory is a special value attained by a measure, subject to strict and limited rules of application, but in this context it would be a way of asserting that an integral diverges, which is not the same thing at all. $\endgroup$
    – whuber ♦
    Commented Oct 5, 2014 at 1:41

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Even though $\text{Var}(X)$ is infinite, $aX$ when $a=0$ is not like $0\cdot \infty$; it's perfectly well defined.

As a result, $\text{Var}(aX+b)$ is perfectly well defined -- in that case, it's not indeterminate.

On the other hand $a^2\cdot\text{Var}(X)$ when $a=0$ is of the form $0\cdot\infty$.

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