# Cumulative Distribution Function of $S_{N_{t}}$ where $S_{N_{t}}$ is the time of the last arrival in $[0, t]$ I am confused on this problem. My professor gave this as the solution:

$$S_{N_{T}}$$ is the time of the last arrival in $$[0, t]$$. For $$0 < x \leq t, P(S_{N_{T}} \leq x) \sum_{k=0}^{\infty} P(S_{N_{T}} \leq x | N_{T}=k)P(N_{T}=k)$$

$$= \sum_{k=0}^{\infty} P(S_{N_{T}} \leq x | N_{T}=k) * \frac{e^{- \lambda t}*(\lambda t)^k}{k!}$$.

Let $$M=max(S_1, S_2, ..., S_k)$$ where $$S_i$$ is i.i.d. for $$i = 1,2,.., k$$ and $$S_i$$~ Uniform$$[0,t]$$.

So, $$P(S_{N_{T}} \leq x) = \sum_{k=0}^{\infty} P(M \leq x)\frac{e^{- \lambda t}*(\lambda t)^k}{k!} = \sum_{k=0}^{\infty} (\frac{x}{t})^k \frac{e^{- \lambda t}*(\lambda t)^k}{k!} = e^{- \lambda t} \sum_{k=0}^{\infty} \frac{(\lambda t)^k}{k!} = e^{- \lambda t}e^{- \lambda x} = e^{\lambda(x-t)}$$

If $$N_t = 0$$, then $$S_{N_{T}} = S_0 =0$$. This occurs with probability $$P(N_t = 0) = e^{- \lambda t}$$.

Therefore, the cdf of $$S_{N_{T}}$$ is: $$P(S_{N_{T}} \leq x) = \begin{array}{cc} \{ & \begin{array}{cc} 0 & x < 0 \\ e^{- \lambda (x-t)} & 0\leq x\leq t \\ 1 & x \geq t \end{array} \end{array}$$

I don't really understand the part of creating the variable M of the maximum of k i.i.d. random variables in order to solve the problem. Any help would be greatly appreciated, thank you!

In the future please be more careful when asking your question. The screenshot of the problem is missing so much context; how do you expect others to know how $$N_t$$ and $$S_{N_t}$$ are defined? Furthermore there are careless typos throughout your professor's solution (writing $$T$$ sometimes instead of $$t$$, missing equals signs, misplaced parentheses, mathematical typos like writing $$(\lambda t)^k$$ instead of $$(\lambda x)^k$$, and $$e^{-\lambda t} e^{-\lambda x} = e^{\lambda (x-t)}$$, etc.). I guess if these were transcriptions of your professor's handwritten notes these things might happen, but you should try your best to catch these things when studying.

I will assume that you have a Poisson process with rate $$\lambda$$, and $$N_t$$ is defined as the number of arrivals in time interval $$[0, t]$$, and $$S_i$$ is the time of the $$i$$th arrival.

$$P(S_{N_t} \le x) = \sum_{k=0}^\infty P(S_{N_t} \le x \mid N_t = k) P(N_t = k).$$ We know $$P(N_t = k) = e^{-\lambda t} (\lambda t)^k/k!$$.

The other term is the probability that the last arrival in $$[0, t]$$ happens before time $$x$$, given that there are $$k$$ arrivals in $$[0, t]$$. Your professor uses a nontrivial result about Poisson processes to compute this term.

Conditioned on the event $$N_t = k$$ (i.e. there are $$k$$ arrivals in $$[0, t]$$), the distribution of arrival times $$(S_1, S_2, \ldots, S_k)$$ are the order statistics of $$k$$ i.i.d. $$\text{Uniform}[0,t]$$ random variables.

See here for a reference, although if your professor is using this result, you may have already encountered it in class somewhere.

In particular, conditioned on $$N_t = k$$, the last arrival $$S_{N_t}$$ has the same distribution as the maximum of $$k$$ i.i.d. $$\text{Uniform}[0,t]$$ random variables $$U_1, \ldots, U_k$$. (I use $$U_i \sim \text{Uniform}[0,t]$$ to not conflate with the $$S_i$$ which are already defined as the $$i$$th arrival in the Poisson process.)

So, the term $$P(S_{N_t} \le x \mid N_t = k) = P(M \le x)$$ where $$M = \max\{U_1, \ldots, U_k\}$$.

Thus, $$P(S_{N_t} \le x) = \sum_{k=0}^\infty (x/t)^k e^{-\lambda t} (\lambda t)^k/k! = e^{-\lambda t} \sum_{k=0}^\infty (x \lambda)^k/k! = e^{-\lambda t} e^{x \lambda}.$$