I have a problem with two competing solutions giving different answers. I know the right one, but I want to understand why the other solution does not work.
(Paraphrase from an ACTEX SOA Exam P Handbook problem)
Random variables $J$, $K$, and $L$ have moment generating functions $M_J(t)=(1-2t)^{-3}$, $M_K(t)=(1-2t)^{-2.5}$, and $M_L(t)=(1-2t)^{-4.5}$. Set $X=J+K+L$. Compute $\mathbb{E}[X^3]$.
My first approach was as follows. The question is what the error in this solution attempt is.
According to the ACTEX study guide, moment generating functions are linear for weighted averages (that is, if $S=\displaystyle\sum a_i S_i$, where $\displaystyle\sum a_i=1$, $a_i\geq0$, then the moment generating function is the corresponding linear combination of the moment generating functions of the component random variables $M_S(t)=\displaystyle\sum a_iM_{S_i}(t)$).
Since $X/3=\frac{1}{3}J+\frac{1}{3}K+\frac{1}{3}L$ is a weighted average, I applied the above fact to $X/3$. Thus $$M_{X/3}(t)=\frac{1}{3}((1-2t)^{-3}+(1-2t)^{-2.5}+(1-2t)^{-4.5}).$$ I then computed $\frac{1}{27}\mathbb{E}[X^3]=\mathbb{E}[(X/3)^3]={M_{X/3}}'''(0)=694$, which then gives $\mathbb{E}[X]=18738$. This is the wrong answer, and I want to understand the error in my above reasoning.
To save answerers the trouble, I will give the correct (and easier!) approach. In fact the moment generating function turns sums of independent random variables into products as follows: if $S=\displaystyle\sum c_jS_j$ (not a weighted average), then $M_S(t)=\displaystyle\prod M_{S_i}(c_it)$. Thus $$M_X(t)=(1-2t)^{-3}\cdot(1-2t)^{-2.5}\cdot(1-2t)^{-4.5}=(1-2t)^{-10}.$$
From this we can compute the correct answer of $\mathbb{E}[X^3]={M_X}'''(0)=10560$. Thanks!