I am having trouble with the following statement (from one of the answers from SOA exam, don't worry about the question as it has no bearing on the question I am asking).
The marginal distribution of X has probability $1/5 + a$ at $0$, $2a + b$ at $1$, and $1/5 + a$ at $2$. Due to symmetry, the mean is $1$ and so the variance is $(0-1)^2 (1/ 5+a ) +(1- 0)^2 (1/ 5 +a) = 2/5+2a$
I do not recognize this expression for variance from anywhere that I remember. The only one I am familiar with is $${E(X^2)-{[E(X)]}^2}$$
Can somebody please explain to me this other expression that is being used here, $(0-1)^2 (1/ 5+a ) +(1- 0)^2 (1/ 5 +a)$, or even just provide a link?
\begin{align} E[(X-E[X])^2] &= E[X^2-2XE[X]+(E[X])^2] \\ &=E[X^2]-2E[X]E[X]+(E[X])^2 \\ &=E[X^2]-2(E[X])^2+(E[X])^2 \\ &=E[X^2]-(E[X])^2 \\ \end{align}
They use an alternative expression for the variance which is just $E[((X-E[X])^2] $.
We have three values, $x_1, x_2, x_3$ where $x_2$ is the mean and $(x_3-x_2)^2=(x_1-x_2)^2$ in this question.