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!

• Please add self-study as a tag and explain which part of the explanation remains confusing to you. Apr 19, 2021 at 6:10
• Cross-post: math.stackexchange.com/q/4107638/321264. Apr 19, 2021 at 10:51

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