Adding two distributions with same moment generating function

[trojansc82]

Homework Statement

I wanted to know what the result would be if you added two distributions with the same moment generating function.

For example, what would the result be of:

M_x(t) + M_y(t) if M_x(t) = (1/3 + 2/3e^t) and M_y(t) = (1/3 + 2/3e^t)

Homework Equations

The Attempt at a Solution

Would the solution be (1/3 + 2/3e^t)²?

 

[LCKurtz]

Homework Statement

I wanted to know what the result would be if you added two distributions with the same moment generating function.

For example, what would the result be of:

M_x(t) + M_y(t) if M_x(t) = (1/3 + 2/3e^t) and M_y(t) = (1/3 + 2/3e^t)

Homework Equations

The Attempt at a Solution

Would the solution be (1/3 + 2/3e^t)²?

No. Denoting the mgf of X by M_X(t) we have the relationship
M_{a+bX}(t) = E(e^{t(a+bX)}) = E(e^{at}e^{btX})=e^{at}E(e^{btX})<br /> =e^{at}M_X(bt)

You are asking about the mgf of X + X = 2X. So using the above with a = 0 and b = 2 gives M_{2X}(t) = M_X(2t).

 

[Ray Vickson]

Homework Statement

I wanted to know what the result would be if you added two distributions with the same moment generating function.

For example, what would the result be of:

M_x(t) + M_y(t) if M_x(t) = (1/3 + 2/3e^t) and M_y(t) = (1/3 + 2/3e^t)

Homework Equations

The Attempt at a Solution

Would the solution be (1/3 + 2/3e^t)²?

There is a difference between adding distributions (or mgf's) and adding random variables. Which do you mean? If you want to find the MGF of X+Y, and assuming X,Y are independent, you get (1/3 + 2/3*exp(t))^2. If you want to find the MGF of the sum of the distributions of X and Y, you get 2*(1/3 + 2/3*exp(t)), but this thing does not really have much meaning.

RGV