Is there an easy way to see that $E(X^2) \geq E^2(X)$? I'm trying to memorize the Steiner translation theorem/König–Huygens formula.
The English name seems to be "Algebraic formula for the variance"
$$Var(X) = E[(X-E(X))^2] = E(X^2) -E^2(X)\;\;\;[1]$$
I assume that since $Var(X) \geq 0$, $E(X^2) \geq E^2(X)$ holds. Correct?
There is a not too complicated proof for [1] on the wikipedia page linked that I understand. But it takes some time to reproduce it. However, I can perfectly remember the outcome of $E(X^2), -, E^2(X)$ just not the order. So (since the difference has to be $\geq 0$) is there an easy way to see which of $E(X^2)$ and $E^2(X)$ is bigger?
Best shot so far from @wiskundeliefhebber:
Remembering that one is at least as big as the other.
Then with $P(X=1) = P(X=-1) = 0.5$ follows $E(X)=0$ and $E(X^2)=1$
Ergo $E^2(X) \leq E(X^2)$
footnote: $E^2(X)$ means $(E(X))^2$
 A: If you need help with remembering: just take an $X$ with $E[X] = 0$, then it is clear that $E[X^2] \ge E[X]^2$
A: Which is greater than which? Use the expression you already have! 
Remembering how expected value is defined (and/or that the expected value of a positive quantity has to be nonnegative), write
$$0 \leq \int_{X} (X-E[X]) d\mu = E[(X-E[X])^2] = E[X^2]-E[X]^2$$
to get $E[X^2]\geq E[X]^2$.
However, this doesn't solve your problem as to remembering which is greater or not. Well, in this case, just memorize it. It's simple. The alternative is to rederive the above. That is, write
$$E[(X-E[X])^2] = E[X^2-2 X E[X] + E[X]^2] = E[X^2] -2 E[X] E[X] + E[X]^2 = E[X^2] - E[X]^2.$$
You don't have to worry about getting the order of $(X-E[X])$ vs $(E[X]-X)$ right. After squaring, they both give you the same result.
What's an even quicker way of remembering? Well, you get an expression $A-B$ but when you look at $(a-b)^2$ you have $(a+b)^2=a^2+b^2+\mbox{stuff}$. This means that the subtraction comes from the cross term. This means that the term subtracted has to be a factor of the cross term, i.e., $2E[X E[X]] = 2 E[X] E[X] = 2 E[X]^2$. The 2 of course goes away from the algebra. Hope that helps.
A: Suppose there are only 2 values, $X\in \{x_{1},x_{2}\}$,
$\displaystyle E^{2}(X)=\frac{(x_{1}+x_{2})^2}{4}...(1)$
$\displaystyle E(X^{2})=\frac{x_{1}^{2}+x_{2}^{2}}{2}...(2)$
$\displaystyle (2)-(1)=\frac{2x_{1}^{2}+2x_{2}^{2}-x_{1}^{2}-2x_{1}x_{2}-x_{2}^{2}}{4}$
$\displaystyle =\frac{(x_{1}-x_{2})^{2}}{4}\ge0$,
hence $E(X^{2})\ge E^{2}(X)$.
A: One way to see the thing very very easily is to think of it as a special case of 
Cauchy-Schwarz inequality,

where Yis are all Pis.
Then by the given Cauchy-Schwarz Inequality, the result follows. 
