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This is from page 25 of this book:

In general, it may be shown that $$\textit{Var}(X) = E(X^2) - [E(X)]^2$$

I can't remember ever seeing that "In general" elsewhere.

So if this identity only holds "in general" are there cases where $$\textit{Var}(X) \neq E(X^2) - [E(X)]^2$$

Even in cases where there is no second moment, I think the identity should hold. Because neither the variance nor the second moment exist. But maybe this is what they're talking about.

JasonMond
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    I think you're just reading 'in general' as in the common language use 'almost always.' the author though means 'holds w/o further assumptions', and you agree (existence of second moment an obvious precondition). – gnometorule Jan 19 '12 at 15:44
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    Yes, mathematicians use the expression "in general" to mean "always" or "in all cases", whereas in everyday informal language, "in general" sometimes just means "usually".

    This is a bit unfortunate, and I'll try to bear this in mind the next time I have reason to say "in general" in class!

     – idmercer Jan 19 '12 at 18:15
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    In general means not merely in the kind of cases we've been considering, but in all cases. To say that it holds in general means precisely that there are no cases where the identity fails to hold. – Michael Hardy Jan 19 '12 at 21:49

1 Answers1

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If $X$ has no second moment, then the formula does not make sense because $\mathrm{Var}(X)$ is not defined.

If $X$ has second moment, then the formula holds without any additional assumptions---it follows easily from the definition of $\mathrm{Var}(X)$.

Rasmus
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