show that E(Y ) = rE(X_1)

Question

For $Y= X_1 + X_2 + ... + X_r$, with the X's being independent and identically distributed random variables, I'm asked to prove that $E(Y) = rE(X_1)$

All I could do is show $(M_X(t))^r = (E( e^{tx}))^r = E(e^{tx_1}) E(e^{tx_2}) ... E(e^{tx_r}) = E( e^{tx_1} e^{tx_2} ... e^{tx_r}) = E(e^{ty}) = M_Y(t) $

Can anyone help/guide me? Thanks

Answer 1

If you are going to use moment generating functions (not all distributions have them, but assuming $X$ does) then you can say

$$M'_Y(t) = \frac{d}{dt} (M_X(t))^r = r (M_X(t))^{r-1} M'_X(t)$$

and since $M_X(0)=1$ and $M'_X(0)= E[X]$ you can say $$E[Y] = M'_Y(0) = r M'_X(0)=rE[X]$$