Let $X$ ~ Exponential with $\lambda =1$ and $Y$ ~ $N(0,1)$ If I want the moment generating function for $Z=X+Y$, assuming independence, isn't $M_{X+Y}(t)=M_{X}(t)M_{Y}(t)$? In other words, I would just multiply the two generating functions to get the new one.
I have a question about the expected value of the combined function, that can be calculated from the combined mgf, I assume by $M'(0)$. How else can you calculate the variance of a sum of random variables?
Thanks!
 A: For the first question, yes, under independence, the MGF of a sum is simply the product of the MGFs. You can see this by the following argument.
$$
M_Z(t)=E[e^{t(X+Y)}] = E[e^{tX}e^{tY}] = E[e^{tX}]E[e^{tY}] = M_X(t)M_Y(t),
$$
where independence let us split the expectation into a product of two expectations.
For the second part, the variance of a sum, you can actually just directly apply the formulas and expand:
$$
\text{Var}[Z] = E[Z^2] - E[Z]^2
$$
Let's calculate those two parts individually.
$$
E[Z^2] = E\left[(X+Y)^2\right] = 
E\left[X^2+2XY+Y^2\right] = 
E\left[X^2\right]+2E\left[XY\right]+E\left[Y^2\right]
$$
$$
= E\left[X^2\right]+2E\left[X\right]E\left[Y\right]+E\left[Y^2\right]
$$
Note that we used independence to say $E\left[XY\right]=E\left[X\right]E\left[Y\right]$ above. Now, the second part
$$
E\left[Z\right]^2 = E\left[X+Y\right]^2 = (E\left[X\right]+E\left[Y\right])^2 = 
E[X]^2 + 2E[X]E[Y] + E[Y]^2
$$
Putting that all together:
$$
\text{Var}[Z] = E\left[X^2\right]+2E\left[X\right]E\left[Y\right]+E\left[Y^2\right] - E\left[X\right]^2 - 2E\left[X\right]E\left[Y\right] - E\left[Y\right]^2
$$
$$
 = (E\left[X^2\right]- E\left[X\right]^2)+(E\left[Y^2\right]   - E\left[Y\right]^2)
$$
$$
= \text{Var}[X]+\text{Var}[Y]
$$
No MGFs are needed here. We simply derived the well known rule that for independent variables, that the variance of the sum is the sum of the variances.
