Expectation expression What is the difference between $E[X^2]$ and $E^2[X]$?
Consider $X$ is the number of heads when a fair coin is tossed twice.
Please explain clearly in general and w.r.t this example also.
 A: 
What is the difference between $E[X^2]$ and $E^2[X]$?

Answer: 


*

*The variance of $X$.

A: $E(X^2)=\sum_{i=1}^{n} x_{i}^{2} P(x_i)$ and $(E(X))^2= \left(\sum_{i=1}^{n} x_{i} P(x_i)\right)^2$ when the support of X is $x_1, x_2, ..., x_n$. On the other hand, the first one is the Mean Square but the second is Square of Mean.
For tossing a fair coin twice, $E(X^2)=3/2$ but $\left(E(X)\right)^2= (1)^2=1$. 
A: Consider the random variable given by $X_0 = X - \def\exp{\mathbb{E}}\exp[X]$.  Let me call this random variable the deviation of $X$ (from its expectation value).
The key observation is that the expectation of the deviation of $X$ is always $0$, no matter what $X$ is:
$$\exp[X_0] = \exp[X] - \exp[\exp[X]] = \exp[X] - \exp[X] = 0.$$
In contrast, the square of the deviation of $X$, namely $X_0^2$, is always non-negative, and therefore $\exp[X_0^2] > 0$.1  I'll call this random variable $X_0^2$ the squared deviation of $X$.
It's not difficult to derive an expression for $\exp[X_0^2]$ in terms of $X$.  First, expand $X_0^2$:
$$X_0^2 = (X - \exp[X])^2 = X^2 - 2 X\cdot\exp[X] + \exp[X]^2\;.$$
Now, take expectations on both sides of $=$, and apply the linearity of expectation to the RHS:
$$
\begin{array}{rcl}
\exp[X_0^2] & = & \exp[X^2] - 2\exp[X]\cdot\exp[X] + \exp[\exp[X]^2] \\
            & = & \exp[X^2] - 2\exp[X]^2 + \exp[X]^2 \\
            & = & \exp[X^2] - \exp[X]^2
\end{array}
$$
(In going from the first to the second equality, I used the fact that $\exp[\exp[X]^2] = \exp[X]^2$, since $\exp[X]^2$ is a constant for any fixed random variable $X$.)
Note that the expression on the last line of the derivation above is precisely the difference $\exp[X^2] - \exp[X]^2$.  IOW, the derivation shows that this difference is nothing more than the expectation of the squared deviation of $X$.
This realization makes it particularly easy to see what the value of this difference would be when $X$ is the number of heads out of two tosses of a fair coin.  In this case, the deviation $X_0$ can be either $-1, 0,$ or $1$, and therefore, the squared deviation $X_0^2$ can be either $0$ or $1$.  Only the latter case contributes to the expectation of $X_0^2$.  It corresponds to the event where both flips turn out heads or both turn out tails.  This event has probability $\frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.  Therefore, for this $X$, the expectation of the squared deviation is $1 \cdot \frac{1}{2} = \frac{1}{2}$.
1  To be more precise, $\exp[X_0^2] \geq 0$, but if $\mathbb{P}[X \neq \exp[X]] > 1$, then $\exp[X_0^2] > 0$.

