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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.

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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).2012-01-19
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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!2012-01-19
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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.2012-01-19

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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 $\mathrm{Var}(X)$.