Mathematical Expectation. E[E[x]]=E[x]

Question

Is it true that $ E[E[X]]=E[X]$? I can't find this property. If it isn't true, then why $E[(X −E[X])^2]=E[X^2]−E[X]^2$?

Answer 1

Yes, $E[E[X]] = E[X]$. This is because $E[X]$ is just a number, it's not random in any way. So when we ask for $E[E[X]]$, i.e., our best guess for the number $E[X]$, well since $E[X]$ is just a constant number which is not random, we know its value, so our best guess for it should be it, i.e., $E[E[X]] = E[X]$.

To calculate $E[ (X - E[X])^{2}]$, we first multiply the inside and get:

$$(X - E[X])^{2} = X^{2} - 2XE[X] + E[X]^{2}.$$ Now, taking the expectation of both sides gives:

\begin{split} E[(X - E[X])^{2}] &= E[X^{2} - 2XE[X] + E[X]^{2}] \\ &= E[X^{2}] - \underbrace{E[2XE[X]]}_{2E[X]*E[X]} + \underbrace{E[E[X]^2]}_{E[X]^2} \end{split}

where we used $E[2XE[X]] = 2E[X]*E[X]$ since $2E[X]$ is just a constant number and we know $E[cX] = cE[X]$ for any constant number $c$.

So, the above last line equals $E[X^{2}] - 2E[X]^{2} + E[X]^{2}$ which simplifies to $E[X^{2}] - E[X]^{2}$.