What is $Var(X+X)$ for a random variable X?

Question

Given that $Var(X)=5$ and $E(X)=20$, how can one compute $Var(X+X)$?

I got that $Var(X+X)=2\cdot Var(X) + 2\cdot (E(XX)-E(X)+E(X))$.

Then $Var(X+X)=2(5)+2\cdot (E(XX)-20+20)$.

Then $Var(X+X)=10+2\cdot E(XX)$.

But how can I compute $E(XX)$? I know that in general, $E(XY)=E(X)\cdot E(Y)$ if $X$ and $Y$ are independent, but I don't think $X$ is independent with itself so I don't see how to get around that.

Edit

Graph of $X$ (Blue) and $X+X$ (Red) below

It appears that $Var(X+X)=4\cdot Var(X)$ based on testing and the comment from Mark. I still don't understand how that's derived (I was confused by both of the answers unfortunately), though, and now I'm confused on how the distribution of $X+X$ quadruples in variance (not sure if I'm allowed to ask a "follow-up question" or what).

I can see it visually from the graph but it doesn't make sense to me - why would the distribution of $X+X$ not be the same as $X$, just shifted by $E(X)$? How does adding $X$ to $X$ change the shape of the distribution so drastically? I don't understand it intuitively.

Answer 1

It depends on what you mean by $X+X$. If you mean one observation added to itself, i.e. the same value doubled, then this is $2X$ and $$Var(2X)=4Var(X)$$

On the other hand, if you mean two independent observations $X_1$ and $X_2$ added together, the the variance is different:

$$Var(X_1+X_2)=Var(X_1)+Var(X_2)=2Var(X)$$

Answer 2

Use the definition of variance. For any random variable $Y$ we have that

$$Var(Y) = E((Y - E(Y))^2)$$

Now let $Y = X + X$, rewrite and solve for $E(X + X)$.

Answer 3

$\mathrm{Var}(X+X) = \mathrm{Var}(2X) = 2^2\mathrm{Var}(X)$. You can easily show: $\mathrm{Var}(aX)=a^2\mathrm{Var}X$.

If you want to compute $\mathrm E(XX)$ you are looking for the second moment. For this you need to calculate $$ \int_\Omega X^2\,\mathrm d\mathbb P$$