For $E (X - EX)^2$ to exist, do we need $EX$ to exist and be finite? For $E (X - EX)^2$  to exist (may be infinite), 


*

*according to $E (X - EX)^2 = E X^2 - (EX)^2$,  I think a necessary
and sufficient condition is $EX$ exists and is finite, because  $ E
X^2 \geq (E X)^2$ by Jensen's inequality and $\infty - \infty$ is
not defined.

*But as long as $EX$ exists, $(X - EX)^2$ is a nonnegative measurable
function and $E (X - EX)^2$ should exists. So it seems a necessary
and sufficient condition is $EX$ exists.


Why do the above two not agree with each other? Do I miss something? Thanks.

Note that Glen_b uses a different definition for variance (see comment), which might exists when the traditional definition doesn't. 
In the definition he uses, if $X$ has a Pareto distribution with $α=1$, then  $EX = \infty$ and  $Var X = \infty$, from https://stats.stackexchange.com/a/91515/1005.
 A: I'm going to answer a simpler but related question.  You can work out the details by noting that your expression $E(X - E(X))^2$ expands to a expression in $E(X^2)$ and $E(X)$.
Yes, it is necessary for $E[X]$ to exist for $E[X^2]$ to exist.
This is because if
$$
\int x^m f(x) dx
$$ 
is undefined (un-integrable), then so is ($n > m$)
$$
\int x^n f(x) dx
$$ 
It can be meaningful to say that these integrals are $\infty$, with the usual caveats about talking about $\infty$ in the extended real numbers.  But you're not going to get any meaningful information out of the higher order integrals, once you start hitting $\infty$.
On the other hand, the existence of $E[X]$ is not a sufficient condition for $E[X^2]$ to exist, though I am having a hard time thinking of an example at the moment.  (I suspect that the square root of a Cauchy random variable might do it.)
A: $E (X - EX)^2 = EX^2 - 2 EX EX + (EX)^2 = E X^2 - (EX)^2$ needs $EX$ to be finite besides its existence. Otherwise $\infty - \infty$ in the last equality.
But I still confirmation for the second part: is it true that $Var X := E (X - EX)^2$ exists iff $EX$ exists?
A: seems to me like one way this could make sense is by defining the limiting process - how each term gets to "infinity". for example suppose I define a sequence of random variables $X_n\sim Uniform(n,n+1)$. then we have $Var(X_n)=\frac{1}{12}$ and $E(X_n)=n+\frac{1}{2}$. Then as $n\to\infty$ the expectation diverges to infinity but the variance does not. you could define countless other examples like this. This example has finite central moments of all orders and infinite raw moments of all orders.
perhaps the other question is whether you consider how the parameters of the limiting process work, and whether or not they're linked. I think something like this is how you could generally define a variance...
$$\lim_{a,b,c,d\to\infty}\int_{-a}^{b}x^2 f(x)dx - \left[\int_{-c}^{d}x f(x)dx\right]^2$$
it may be that you need to the limits $a,b$ to relate somehow to the limits $c,d$ when the mean is infinite.
