If X and Y are two random variables, then show that E(X+Y) =E(X) +E(Y)?

Answer 1

E(x+y) = (x+y)*Prob(x+y)

= (E(x)+E(y))Prob(x+y) <--- x = E(x) and y = E(y)

= E(x) * Prob(x+y) + E(y)*Prob(X+y)

= x*Prob(x)*Prob(x+y) + y*Prob(y)*Prob(x+y)

= Prob(x+y) [ x*Prob(x) + y*Prob(y) ]

= Prob(x+y) [ E(x) + E(y)]

= 1 * [ E(x) + E(y) ] <--- probability of all events is 1

= E(x) + E(y)

Answer 2

This is only true if X and Y have the same distribution function, f(x)

E(X+Y) = ∑(x+y)f(x) = ∑xf(x) + ∑yf(x) = E(X) + E(Y)