Poisson Processes with Gamma Arrivals I think the title is the best description I can give of my problem (but I'm not 100% sure - the problem set-up has me very confused).
So, given a sequence of i.i.d. Gamma RV having parameters 3, $\lambda$, where for t $\ge$0, $$P(T_n \le t) = \int_0^t \frac{\lambda(\lambda x)^2e^{-\lambda x}}{2}dx.$$
Then, I'm told that {$ {N(t); t \ge 0} $} can be defined as $$ N(t) = \sum_{n=1}^\infty 1(S_n \le t),      t\ge0$$ {$1(S_n \le t)$} being the indicator function
Where for each integer $n \ge 1$, $S_n = T_1 + ... + T_n.$
I am asked to derive an expression for $P(N(t) = k)$, for each integer $k \ge0$.
A hint given is: Use the fact that each $T_n$ can be interpreted as the sum of three i.i.d. exponential random variables, having rate $\lambda$.
My answer follows this idea: This is essentially asking me to evaluate a Poisson counting process (I think), and because the inter arrival times $T_n$ are Gamma (3,$\lambda$), it should simplify to a counting process with 3 i.i.d. arrival processes each with rate $\lambda$ - This also means it should be an Erlang process, but that is irrelevant.
So, using the base representation for a Poisson Process, $$P(N(t) = k) = e^{-\lambda t} \frac{(\lambda t)^k}{k!}$$ I think the rate ($\lambda$) for my new process should be $\frac{\lambda}{3}$, giving me a final result of: $$P(N(t) = k) = e^{\frac{-\lambda t}{3}} \frac{(\frac{\lambda t}{3})^k}{k!}$$
Does this make sense to anyone else? The only reason I am concerned is this seemed far too simple for the course that I am currently taking.
 A: The basic idea stems from the relationship $$\Pr[S_k \le t] = \Pr[N(t) \ge k].$$  This is because $S_k$ is the random time of the $k^{\rm th}$ event, and $N(t)$ is the number of events that have occurred up to time $t$.  Intuitively, we can see that the probability that the $k^{\rm th}$ event occurs before time $t$ is equivalent to the probability that at least $k$ events occur by time $t$.
Next, note that because $S_k = T_1 + T_2 + \cdots + T_k$; that is to say, the time until the $k^{\rm th}$ event is the sum of the interarrival times of each event up to the $k^{\rm th}$ event, and because the sum of IID gamma variables is itself gamma, we have $$S_k \sim \operatorname{Gamma}(3k,\lambda), \quad f_{S_k}(x) = \frac{\lambda (\lambda x)^{3k-1} e^{-\lambda x}}{\Gamma(3k)}.$$ Consequently, $$\begin{align*} \Pr[N(t) = k] &= \Pr[N(t) \ge k] - \Pr[N(t) \ge k+1] \\ &= \Pr[S_k \le t] - \Pr[S_{k+1} \le t] \\ &= \int_{x=0}^t \frac{\lambda (\lambda x)^{3k-1} e^{-\lambda x}}{\Gamma(3k)} - \int_{x=0}^t \frac{\lambda (\lambda x)^{3k+2} e^{-\lambda x}}{\Gamma(3k+3)} \, dx.\end{align*}$$  We now employ integration by parts on the second integral with the choice $$u = (\lambda x)^{3k+2}, \quad du = (3k+2)\lambda (\lambda x)^{3k+1} \, dx, \\ dv = \lambda e^{-\lambda x} \, dx, \quad v = - e^{-\lambda x},$$ to obtain $$\begin{align*} \Pr[S_{k+1} \le t] &= \left[-e^{-\lambda x} \frac{(\lambda x)^{3k+2}}{(3k+2)!} \right]_{x=0}^t + \int_{x=0}^t \frac{\lambda(\lambda x)^{3k+1}e^{-\lambda x}}{\Gamma(3k+2)} \, dx \\ &= -e^{-\lambda t} \frac{(\lambda t)^{3k+2}}{(3k+2)!} + \int_{x=0}^t \frac{\lambda(\lambda x)^{3k+1}e^{-\lambda x}}{\Gamma(3k+2)} \, dx .\end{align*}$$  And we can notice that if we repeat the process twice more on the remaining integral, we would get $$\Pr[S_{k+1} \le t] = - e^{-\lambda t} \left( \frac{(\lambda t)^{3k+2}}{(3k+2)!} + \frac{(\lambda t)^{3k+1}}{(3k+1)!} + \frac{(\lambda t)^{3k}}{(3k)!}\right) + \Pr[S_k \le t].$$  This proves $$\Pr[N(t) = k] = e^{-\lambda t} \left( \frac{(\lambda t)^{3k+2}}{(3k+2)!} + \frac{(\lambda t)^{3k+1}}{(3k+1)!} + \frac{(\lambda t)^{3k}}{(3k)!}\right),$$ which happens to correspond to the sum of Poisson probabilities for a process that has rate $\lambda$, for a number of events being $3k$, $3k+1$, or $3k+2$.
