Let X and Y be discrete random variables and a; b are real numbers.
Prove E(aX + bY ) = aE(X)+ bE(Y ) and Var(aX + bY ) = a^2Var(X)+
b^2Var(Y )+ 2abCov(X; Y ).
Both these seem easy but I can't proof why E(X+Y)=E(X)+E(Y)
Look at the definition of expectation for discrete RVs:
$\displaystyle E(f(U))=\sum f(u_i) p(u_i)$
So (assuming suitable conditions allowing us to change orders of summation etc hold):
$\displaystyle E(X+Y)=\sum_{i,j} (x_i+y_j) p(x_i,y_j) = \sum_i \sum_j x_i p(x_i,y_j) + \sum_j \sum_i y_j p(x_i,y_j)$ $\displaystyle =\sum_i x_i p(x_i) +\sum_j y_j p(y_j)$
(here we are writing the marginal distribution $\displaystyle p(x_i)=\sum_j p(x_i,y_j)$ and similar for the other case)
etc
CB
$\displaystyle Var(X+Y)=E([(X+Y)-(\overline{X}+\overline{Y})]^2 $ $\displaystyle =E([(X-\overline{X})+(Y-\overline{Y})]^2)$ $\displaystyle =E((X-\overline{X})^2 + 2(X-\overline{X})(Y-\overline{Y})+(Y-\overline{Y})^2)$
Now use what has already been proven (that the expectation of a sum is the sum of the expectations) to get:
$\displaystyle Var(X+Y)=$ $\displaystyle =Var(X)+Var(Y) + 2 CoVar(X,Y)$
CB