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 ?

2. Originally Posted by Jason2009
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 ?

2. Makes no sense to me. Perhaps you have not posted the whole question?

1. First calculate $m(t) = E(e^{tX}) = \int_0^1 e^{tx} \cdot 1 dx$. Then remember that $E(X) = \left. \frac{dm}{dt} \right|_{t = 0}$ and $E(X^2) = \left. \frac{d^2 m}{dt^2} \right|_{t = 0}$.

3. are X1 and X2 uniform (0,1)?

$E(e^{Yt})=E(e^{t(a+bX1+cX2+d)})$

Now using indep (which I assume you have)

$=e^{(a+d)t}E(e^{t(bX1+cX2)}) =e^{(a+d)t}E(e^{btX1+ctX2)})=e^{(a+d)t}E(e^{btX1}) E(e^{ctX2})$.

which is...

$e^{(a+d)t}MGF_{X1}(bt)MGF_{X2}(ct)$.

And I don't get the a and d, why two constants?

4. No X1 and X2 are not uniform distribution .
The 2nd question is :X1 and X2 are independent variables and Y= a+bX1+cX2+d Find the MGF of Y in terms of MGF of X1 and X2.

1. If $U = aX + b$ then $M_U (t) = e^{bt} M_X (at)$.
2. If $U = X + Y$ and X and Y are independent random variables then $M_U (t) = M_X (t) \cdot M_Y (t)$.