# Using the correct discrete random distribution for a Conditional PMF problem: “Suppose you arrive at a bus stop at time 0…”

I am a non-student working through the first edition of Yates and Goodman's text, Probability and Stochastic Processes. On page 115, question 3.6.8 goes like this:

Suppose you arrive at a bus stop at time 0 and that at the end of each minute, with probability $p$ a bus arrives or with probability $1-p,$ no bus arrives. Whenever a bus arrives, you board that bus with probability $q$ and depart. Let $T$ equal the number of minutes you stand at a bus stop. Let $N$ be the number of buses that arrive while you wait at the bus stop.

Now, there are four parts to this problem; the second part is where I am struggling:

Find $P_{N,T}(n,t).$

I understand that $P_{N,T}(n,t) = P_{N|T}(n|t) \cdot P_T(t).$ However, I'm unsure how to proceed because I'm unsure of which discrete random distribution applies here. At first I thought this was a case where using a binomial random variable would suffice; however, my textbook also brings up the "Pascal random variable," which as I understand it, features a random variable that is the number of trials up to and including the $k^{th}$ success.

Is it fair to say that time $T$ (in this problem) is a Pascal random variable? Could $N$ be one as well?

It is correct that $T$ follows a geometric (or Pascal) distribution. For the joint distribution of $N$ and $T$ we could instead calculate this directly without worrying about conditioning.

For $1 \leq n \leq t$ to find $P(N = n \cap T = t)$ we know that one bus must have arrived at time $t$ which we boarded, and the remaining $n - 1$ buses arrived during the first $t - 1$ minutes and we didn't board any of them. On every other minute no bus arrived.

We can calculate the probabilities of each of these events in a straightforward way and so get the probability of any one sequence with $N = n$ and $T = t$. Then since we know the last event has to involve the arrival of a bus we just need to count the total the number of ways we can choose the positions of the first $n - 1$ buses among the first $t - 1$ minutes which is $\binom{t - 1}{n - 1}$. Putting everything together we get

$$P(N = n \cap T = t) = \binom{t - 1}{n - 1} p^n (1 - p)^{t-n} q (1 - q)^{n-1} .$$

• From this derivation you can deduce that $N$ has a geometric distribution with probability $q$. – Xi'an Mar 2 '16 at 4:49
• And that $T$ has a geometric distribution with probability $pq$, – Xi'an Mar 2 '16 at 5:00
• One thing I don't understand, @dsaxton- why do we have $\binom{t - 1}{n - 1}$ and not $\binom{t}{n}$? – daOnlyBG Mar 23 '16 at 21:45
• @daOnlyBG It's because the $n^\text{th}$ bus always arrives at time $t$, so the way we count valid sequences involving $n$ arrivals is by fixing the last arrival time and permuting the remaining $n - 1$ among the previous $t - 1$ times. – dsaxton Mar 23 '16 at 22:46
• @dsaxton I am sorry, but I still do not understand why it's done that way. I believe you're correct (I saw the solution for this problem somewhere) but I'm not sure why it's so. If we say $q(1-q)^{n-1}$ is a geometric distribution, then $\binom{t-1}{n-1} p^n (1-p)^{t-n}$ is a Negative Binomial computation, where the random variable is $t$, the number of minutes at the bus stop- but why? – daOnlyBG Mar 25 '16 at 19:39