I have the following question:
Let $X$ be a random variable, prove that for any real number $a$,
$$\mathrm{Var}(X) \le E[(X-a)^2].$$
The question gives a hint by saying:
Write $\mathrm{Var}(X) = E[(X-a-(E[X]-a)^2)$.
So I thought about just expanding the LHS:
\begin{align*}
E[(X-a)^2]
&= E[X^2 – 2aX – a^2] \\
&=E[X^2] – 2aE[X] – a^2
\end{align*}
Since $a$ is a constant $E[a^2]= a^2$.
I did not see anything obvious so I did the same thing to the RHS:
\begin{align*}
\mathrm{Var}(X)
&= E[(X – a – (E[X] – a))^2] \\
&= E[X] – E[a] -E[(E[X])^2 – aE[X] +a^2] \\
&= E[X] – a – (E[X])^2 – aE[X]+a^2
\end{align*}
But now I am stuck because I feel like a did a ton of algebra, and don't know how to proceed. Any help on to get unstuck (or if I misinterpreted something) would be greatly appreciated.
Best Answer
We know that $\operatorname{Var}(X) = E[X^2] - E^2[X]$. Let's see how can we massage $E[(X - a)^2]$:
\begin{eqnarray} E[(X - a)^2] &=& \\ &=& E[X^2 - 2 a E[X] + a^2] \\ &=& a^2 - 2aE[X] + E[X^2] \\ &=& (a - E[X])^2 + E[X^2] - E^2[X] \\ &=& (a - E[X])^2 + \operatorname{Var}(X) \end{eqnarray}
This concludes the proof :)