Understanding $E[x^2]$ in a variance question

Question

I know variance is equal to $V[x]=E[x^2] - (E[x])^2 $, but how do you expand $E[x^2]$ for some x if your given the necessary information...

What I mean for example is if you suppose $x = ys+(1-y)r$ then would $E[x^2] = E[(ys+(1-y)r)^2] = E[(ys+(1-y)r) * (ys+(1-y)r)]$?

I'm trying to solve a problem dealing with variance, but feel like I maybe just don't understand how to properly use $E[x^2]$.

Answer 1

That's right. If $x=ys+(1-y)r$ then $E(x^2)=E((ys+(1-y)r)^2)$. You can expand this and use properties like: if $x$ is a random variable and $a,b\in \mathbb R$ then $E(ax+b)=aE(x)+b$.

Answer 2

If $x$ is your random variable, then $x^2$ is just a 'transformation' of that random variable.

Remember that the expectation operator $\text{E}[ \cdot]$, when applied to a random variable $x$, just gives you back the 'weighted average' of your support's values i.e. $\text{E}[ x] = \sum_{i \in S}\mathbb{P}(x = i)\times x$ (where the weights are each of the $\mathbb{P}(x = i)$'s). Here $S$ is just the support of the random variable. If $x\sim \text{binary}(p)$ then $S = \{0, 1\}$ for example.

When we ask what $\text{E}[ x^2]$ is, we literally mean "what is the 'weighted average' when we square the random variable $x$". Since the value of $x^2$ depends on the value of $x$, $\text{E}[ x^2] = \sum_{i = 0}^{S}\mathbb{P}(x = i)\times x^2$.