Probability Distribution of Random Sums of Exponential RVs

[spitz]

Homework Statement

Z=X_1+\ldots+X_N, where:

X_i\sim_{iid}\,\text{Exponential}(\lambda)

N\sim\,\text{Geometric}_1(p)

For all i,\,N and X_i are independent.

Find the probability distribution of Z

Homework Equations

G_N(t)=\frac{(1-p)t}{1-pt}
M_X(t)=\frac{\lambda}{\lambda-t}

The Attempt at a Solution

M_Z(z)=G_N(M_X(z))=\frac{(1-p)\left(\frac{\lambda}{\lambda-z}\right)}{1-p\left(\frac{ \lambda}{\lambda-z}\right)}
Z\sim\,\text{Geometric}_1\left(p \frac{ \lambda}{\lambda-z}\right)

Is that even correct? Should I be looking for E[Z] and V[Z] ?

 

[Ray Vickson]

Homework Statement

Z=X_1+\ldots+X_N, where:

X_i\sim_{iid}\,\text{Exponential}(\lambda)

N\sim\,\text{Geometric}_1(p)

For all i,\,N and X_i are independent.

Find the probability distribution of Z

Homework Equations

G_N(t)=\frac{(1-p)t}{1-pt}
M_X(t)=\frac{\lambda}{\lambda-t}

The Attempt at a Solution

M_Z(z)=G_N(M_X(z))=\frac{(1-p)\left(\frac{\lambda}{\lambda-z}\right)}{1-p\left(\frac{ \lambda}{\lambda-z}\right)}
Z\sim\,\text{Geometric}_1\left(p \frac{ \lambda}{\lambda-z}\right)

Is that even correct? Should I be looking for E[Z] and V[Z] ?

Z is a continuous random variable, so does not have a discrete generating function M_Z(z). You should be looking at its MGF M_Y(u) = E e^{u Y}, or its Laplace transform L_Y(s) = E e^{-s Y}. You almost had it right, but you switched the roles of the two types of transforms.

Another, perhaps more direct approach is to get the density f_Y(t) of Y from
f_Y(t) dt = \sum_{n=1}^{\infty} P\{N=n\} P\{ Y \in (t,t+dt) | N=n \},
and noting that given {N=n}, Y has an n-Erlang distribution.

RGV

 

[spitz]

I'm still confused... this is what I was doing:
M_Z(z)=E(e^{zZ})=E[E(e^{zZ}|N)]=E[(Ee^{zX_1})(Ee^{zX_2})\ldots(Ee^{zX_N})]
=E[(Ee^{zX})^N]=E[(M_X(z))^N]=G_N(M_X(z))
Where am I going wrong? Should I be doing this:
M_Z(s)=M_X(G_N(s))