# Moment generating function(2 variables for $Z = X+Y$)

$$Q)$$ Let the $$Z = X+Y$$ (Here the variables $$X$$ and $$Y$$ are independent.)

Find the Moment generating function , $$M_z(t)$$ for $$Z = X+Y$$

The $$f(t)$$ is a $$P.D.F$$ both of the $$X$$ and $$Y$$, defined like the below.

## $$f(t) = \begin{cases} 3e^{-3t} & \text{t \gt 0} \\ 0 & \text{o.w.} \end{cases}$$

By definition we can find the $$M_z(t)$$ for $$0

## $$M_z(t) = E(e^{(x+y)t}) = \int\int 9e^{({(t-3)x})({(t-3)y})}dxdy$$ = $$1 \over (1- {t \over3})^2$$

But I tried the different ways by finding the $$f_z(z)$$ (Surely $$f(x,y) = 9e^{-3(x+y)}$$)

Let the C.D.F, $$F(z) = P(X+Y \leq Z) = \int_{0}^{z}\int_{0}^{z-x} f(x.y)dxdy$$

Then P.D.F $$f_z(z) = {d \over dz} F(z) = 9ze^{-3z}$$ for $$z>0$$ in my calculation.

Hence $$M_z(t) = \int_{0}^{\infty} e^{tz}f_z(z) dz$$. But this value is not the $$1 \over (1- {t \over3})^2$$

What the point did I mistake? Any help would be appreciated.

First, remark that $$M_Z(t)=\mathbb E[e^{tX+tY}]\underset{(1)}{=}\mathbb E[e^{tX}][e^{tY}]\underset{(2)}{=}\mathbb E\left[e^{tX}\right]^2,$$ where $$(1)$$ comes comes from the independence and $$(2)$$ from the fact that $$X$$ and $$Y$$ are identically distributed.
Now $$M_X(t)=\int_{\mathbb R}e^{tx}f_X(x)\,\mathrm d x,$$ (and not what you wrote), i.e. $$M_X(t)=3\int_{0}^\infty e^{tx}e^{-3x}\,\mathrm d x=3\int_0^\infty e^{(t-3)x}\,\mathrm d x=\frac{3}{t-3}.$$ Therefore $$M_Z(t)=\frac{9}{(t-3)^2},$$ as wished.