# moment generating function of a summation

Suppose the time between independent incoming calls to a call center is well modeled by an exponential distribution having $\lambda = 0.025$. Let $T_i$ be the time between the $i$th and the $(i-1)$st call, and define $S_k = \sum^k_{i=1} T_i$. What are:

(a) The moment generating function of $T_4$?

(b) The moment generating function of $S_6$?

(c) $E[S_6]$?

I did (a) by solving $E(e^{tX}) = \int^\infty_00.025e^{tx}e^{-0.025x}dx$ and got that the mgf of $T_4 = \frac{0.025}{0.025-t}$, defined for $t<0.025$.

I'm not quite sure how to solve (b), but I think that since all $T_i$ are independent, then the mgf of $S_6$ should be the product of all the individual mgf of $T_1,T_2,...,T_6$. Since the mgf of all $T_i$ should be the same, then should the mgf of $S_6 = {(\frac{0.025}{0.025-t})}^6$?

And for (c), I should just find the first moment of $S_6$ right?

-

If $X\sim \exp(\lambda)$ for $i=1,\ldots k$, then \begin{align} M_X(t)&=E[e^{tX}]=\int_0^\infty e^{tx}\lambda e^{-\lambda x}\,\mathrm dx=\lambda\int_0^\infty e^{x(t-\lambda)}\,\mathrm dx\\ &= \begin{cases} \infty,\quad &\text{for }t\geq \lambda,\\ \lambda(t-\lambda)^{-1},&\text{for }t<\lambda, \end{cases} \end{align} so you're right about (a). Since $T_1,\ldots, T_6$ are independent, we have $$M_{S_6}(t)=E\left[e^{tS_6}\right]=E\left[e^{tT_1}\cdots e^{tT_6}\right]=E\left[e^{tT_1}\right]\cdots E\left[e^{tT_6}\right]=M_{T_1}(t)\cdots M_{T_6}(t),$$ where we have used that also $e^{tT_1},\ldots, e^{tT_6}$ are independent. So your answer is correct here as well.
And yes, (c) just asks for the first moment of $S_6$, but note that this can be calculated via its moment-generating function, i.e. $$E\left[S_6\right]=M_{S_6}'(0).$$