Question

I understand the formula for Variance (x) = E(x-myu)^2 = Sigma(x-myu)^2 * P(X=x). However, I don't understand how the above also equals E(x^2)-myu^2. Please help.

Answer 1

Variance \(V(X)\) is defined as
\[V(X)=E\bigg[(X-\mu)^2\bigg]\]
Expand the binomial:
\[E\bigg[X^2-2X\mu+\mu^2\bigg]\]
The expected value function is linear, which means you can write the expected value of a linear combination as a linear combination of expected values, i.e. the above is equivalent to
\[E\bigg[X^2\bigg]-E\bigg[2X\mu\bigg]+E\bigg[\mu^2\bigg]\]
Now, because \(E\) is a linear operator, you can also pull out constants, like so:
\[E\bigg[X^2\bigg]-2\mu E\bigg[X\bigg]+\mu^2E\bigg[1\bigg]\]
The expected value of a constant, like 1, is equal to that constant. Also, the expected value of a random variable is the mean of that random variable:
\[E\bigg[X^2\bigg]-2\mu \mu+\mu^2\]
Simplifying, you get the desired result, often written as
\[E\bigg[X^2\bigg]-\bigg[E(X)\bigg]^2\]