Prove: $\mathbb{V}(X)=\mathbb{E}[(X-\mathbb{E}[X])^2]=\mathbb{E}[X^2]-\mathbb{E}[X]^2$

Question

I want to prove that:

$$\mathbb{V}(X)=\mathbb{E}[(X-\mathbb{E}[X])^2]=\mathbb{E}[X^2]-\mathbb{E}[X]^2$$

Here's my attempt:

\begin{equation} \label{eq1} \begin{split} \mathbb{V}(X) & =\mathbb{E}[(X-\mathbb{E}[X])^2] \\ & = \mathbb{E}[(X-\mathbb{E}[X])(X-\mathbb{E}[X])] \\ & = \mathbb{E}[X^2-X\mathbb{E}[X]-X\mathbb{E}[X]+\mathbb{E}[X]^2] \\ & = \mathbb{E}[X^2] - \mathbb{E}[2X \mathbb{E}[X]] + \mathbb{E}[\mathbb{E}[X]^2] \end{split} \end{equation}

From this step onwards, I got a bit confused. I assume that the expected value of an expected value of a random variable is the same as the expected value of a random variable, i.e. $\mathbb{E}[\mathbb{E}[X]] = \mathbb{E}[X]$. So I'd get:

\begin{equation} \label{eq2} \begin{split} \mathbb{V}(X) & =\mathbb{E}[(X-\mathbb{E}[X])^2] \\ & = \mathbb{E}[(X-\mathbb{E}[X])(X-\mathbb{E}[X])] \\ & = \mathbb{E}[X^2-X\mathbb{E}[X]-X\mathbb{E}[X]+\mathbb{E}[X]^2] \\ & = \mathbb{E}[X^2] - \mathbb{E}[2X \mathbb{E}[X]] + \mathbb{E}[\mathbb{E}[X]^2] \\ & = \mathbb{E}[X^2] - 2\mathbb{E}[X \mathbb{E}[X]] + \mathbb{E}[X]^2 \end{split} \end{equation}

But now how do I deal with the middle term? How can I simplify $2\mathbb{E}[X \mathbb{E}[X]]$?

Answer 1

Since $\mathbb{E}[X]$ is a constant - call it $\mu$ - you have $\mathbb{E}[\mu X]=\mu \mathbb{E}[X]=\mu^2=\mathbb{E}[X]^2$.

Answer 2

Note that ,$$\underbrace{\mathbb{E}[X]}_{is -a-constant}$$ $$2\mathbb{E}[X \mathbb{E}[X]]\\=2\mathbb{E}[X \underbrace{\mathbb{E}[X]}_{\leftarrow}]\\=2\mathbb{E}[X]\mathbb{E}[X]$$

as ahint: if you take $\mathbb{E}[X]=\mu$ from the first line...it can shows easier