1.Find the mean and variance of a uniform distribution [0,1] by MGF.
2.Y=a+bX1+cX2+d find the MGF of X1 and X2 ?
Please help
2. Makes no sense to me. Perhaps you have not posted the whole question?
1. First calculate $\displaystyle m(t) = E(e^{tX}) = \int_0^1 e^{tx} \cdot 1 dx$. Then remember that $\displaystyle E(X) = \left. \frac{dm}{dt} \right|_{t = 0}$ and $\displaystyle E(X^2) = \left. \frac{d^2 m}{dt^2} \right|_{t = 0}$.
are X1 and X2 uniform (0,1)?
$\displaystyle E(e^{Yt})=E(e^{t(a+bX1+cX2+d)})$
Now using indep (which I assume you have)
$\displaystyle =e^{(a+d)t}E(e^{t(bX1+cX2)}) =e^{(a+d)t}E(e^{btX1+ctX2)})=e^{(a+d)t}E(e^{btX1}) E(e^{ctX2})$.
which is...
$\displaystyle e^{(a+d)t}MGF_{X1}(bt)MGF_{X2}(ct)$.
And I don't get the a and d, why two constants?
You're expected to know and use the following two results:
1. If $\displaystyle U = aX + b$ then $\displaystyle M_U (t) = e^{bt} M_X (at)$.
2. If $\displaystyle U = X + Y$ and X and Y are independent random variables then $\displaystyle M_U (t) = M_X (t) \cdot M_Y (t)$.