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Let $X: \Omega \rightarrow \Bbb R$ be a random variable. Prove that $$\min_{a \in \Bbb R} E((X - a)^2) = Var(X)$$

First, we note that $E((X - a)^2) = E(X^2 - 2aX + a^2) = E(X^2) - 2aE(X) + a^2$.

In order to minimize the function, we have to take the first derivate, which is $2E(X) - 2a$. Now, we have to solve

$$2E(X) - 2a = 0$$

This is obviously only true for $a = E(X)$, so there is no other possible root. In order to show that this is indeed the minimum, we calculate the second derivative, which is $2 > 0$, hence, this is a minimal turning point trivially.

Is that the correct solution?

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    Looks good to me! – 2017-01-02
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    It seems fine. You have a little typo writing $+2a$ instead of $-2a$. – 2017-01-02
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    Corrected it, thanks! – 2017-01-02

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Another simple method: Note $E(X-a)^2=E(X-E(X))^2+(E(X)-a)^2=Var(X)+(EX-a)^2\geq Var(X)$

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    Hi can you please clarify why this is true? I am trying to expand it and am not seeing why it is so clear – 2018-11-13