# On variance of a random variable

I am relatively very new to probability distribution and after understanding the concept of Expected value of a discrete random variable,I am trying to understand the variance of the same here is an extract from my book:

$Var(X) = E[(X - E(X))^2] = \sum_{\text{all }X} (X - E(X))^2 P_x(X) = E(X^2) - (E(X))^2$

What my question is that I don't understand this simplification.Any pointers in this regard will be highly appreciated.

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To expand on the first sentence of Matt's answer: the third bit of the formula of your post (the one with a sum over all $X$) has nothing to do here. If really the authors of your book wrote this, they should be blamed. –  Did Jan 27 '11 at 13:46
Exactly I gave me some woes in understanding but Mat's answers clears down everything in a very lucid away :) –  Quixotic Jan 27 '11 at 13:58

Forget the middle bit. What you need to know is that $E(X)$ is a linear function, this means that $E(X + Y) = E(X) + E(Y)$ and $E(aX) = aE(X)$ where $a$ is a constant.

Per definition, $Var(X) = E((X - E(X)^2))$.

Multiplying out the argument of $E$ gives you

$$E((X - E(X)^2)) = E(X^2 - 2XE(X) +E^2(X)) =$$ $$E(X^2) - E(2XE(X)) + E(E^2(X)) = E(X^2) - 2 E(X)E(X) +E^2(X)$$

where the last equality holds because $E(X)$ is a constant, so $E(E(X)) = E(X)$.

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