Why is $E(nX)=nE(X)$?
I know that this is correct for expected value, but then I don't know how it's possible.$$ E\ [n X]=n \sum_i x_i \times P_{X}(X = n x_i) $$ but $ P_{X}(X = n x_i) $ is not equal to $ P_X(X = x{_i}) $ which would make the equation as : $$ n \sum x_i\times P_X(X=x_i)$$ that is equal to $ \ \ n \ E[X] .$
PS_edit: some one told me it's because probability indices are not linear, what does it have to do with my question exactly?
$\endgroup$ 92 Answers
$\begingroup$$$E(cX)=\sum_{x\in\Omega}cxP(X=x)=c\sum_{x\in\Omega}xP(X=x)=cE(x)$$
$\endgroup$ 6 $\begingroup$By definition,
$$E(g(X)) := \sum_j y_j \Pr(g(X)=y_j)$$
But in general, if $X$ is a discrete random variable and a function $g$ gives $g(X)$ also a discrete random variable, * then:
$$E(g(X)) = \sum_i g(x_i) \Pr(X=x_i)$$
Here $g(X) = cX$. You are correct that $\Pr(X=x)\ne \Pr(X=g(x))$ in general.
There is an analogous theorem for $X$ and $g(X)$ continuous RV's.
This theorem is known as the Law of the Unconscious Statistician. It's called so because statisticians sometimes use this law without even realizing they are invoking a theorem. Proofs can be found on this website.
Edit: here's a proof of the discrete case:
Proving the Law of the Unconscious Statistician
Second edit:
- according to Did, $g(X)$ is always discrete if $X$ is discrete, which sounds reasonable
