Find moment generating function for given random variable

Let $X_1$ and $X_2$ be independent randıman variables without exponential distribution with parameter $\lambda$. Let Y be random variable $Y=X_1+X_2$.

If the the random variable Z has the following m.g.f.

$M_Z(t)= M_Y(M_Y’(t))$

where $M_Y’(t)=dM_Y(t)/dt$

Find E(Z) by using $M_z(t)$

My solution

I find $M_Y(t)=\frac{\lambda^2}{(\lambda - t)^2}$

And $M_Y’(t)=\frac{2\lambda^2}{(\lambda -t)^3}$

Then $M_z(t)=M_Y(\frac{\lambda^2}{(\lambda - t)^2})=E(e^{\frac{\lambda^2}{(\lambda - t)}^2y}) $

but from here, I cannot find mgf of Z.

Solution 1:

Using the chain rule: $$M_Z'(t)=\partial_t(M_Y(M_Y'(t)))=M''_Y(t)M_Y'(M_Y'(t))$$ so that $$E[Z]=M'_Z(0)=M''_Y(0)M_Y'(M_Y'(0))=\frac{6\lambda^2}{(\lambda-0)^4}\frac{2\lambda^2}{(\lambda-\frac{2\lambda^2}{(\lambda-0)^3})^3}=\frac{12\lambda^3}{(\lambda^2-2)^3}$$