How to prove that $Var(X+Y)$=$Var(X)$+$Var(Y)$ if $X$ and $Y$ are independent? (Var indicates variance)

Question

What we know is $SE(X)=\sqrt{E[(X-E(X))^2]}$ ($E$ means expected value) and $SE$ is standard error. Also, we know that $Var(X+Y)=(SE(X+Y))^2$ and $SE(X+Y)=\sqrt{[SE(X)]^2+[SE(Y)]^2}$ when X and Y are independent of each other. One of my instructor provided the following proof:

$$Var(X+Y)=E[(X+Y)^2]-[E(X+Y)]^2=...$$

Here I am very confused with the first step, where does the formula $Var(X)= E[(X)^2]-[E(X)]^2 $ come from? Since we know that $SE(X)=\sqrt{E[(X-E(X))^2]}$ and $Var(X)=(SE(X))^2$, shouldn't $Var(X)$ be $E[(X-E(X))^2]$ and thus $Var(X+Y)$ be $E[(X+Y-E(X+Y))^2]$?

Answer 1

First,

\begin{align} Var(X) = E[(X-E[X])^2] &= E[X^2 - 2 X E[X] + E[X]^2]\\ &= E[X^2] - 2 E[X]^2 + E[X]^2\\ &= E[X^2]-E[X]^2. \end{align}

This is an extremely important equality. Computations of variance often use the last line instead of the original form.

So, to answer your question, the following are all equal: $$Var(X+Y) = E[(X+Y-E[X+Y])^2] = E[(X+Y)^2]-E[X+Y]^2.$$

Answer 2

\begin{align}Var(X)&=\mathbb{E}((X-\mathbb{E}(X))^2)\\&=\mathbb{E}(X^2-2X\mathbb{E}(X)+\mathbb{E}(X)^2)\\&= \mathbb{E}(X^2)-2\mathbb{E}(X\mathbb{E}(X))+\mathbb{E}(X)^2\\&= \mathbb{E}(X^2)-2\mathbb{E}(X)\mathbb{E}(X)+\mathbb{E}(X)^2\\&= \mathbb{E}(X^2)-2\mathbb{E}(X)^2+\mathbb{E}(X)^2\\&= \mathbb{E}(X^2)-\mathbb{E}(X)^2 \end{align}