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?

Thank you in advance.


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

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

  • $\begingroup$ From this derivation you can deduce that $N$ has a geometric distribution with probability $q$. $\endgroup$ – Xi'an Mar 2 '16 at 4:49
  • $\begingroup$ And that $T$ has a geometric distribution with probability $pq$, $\endgroup$ – Xi'an Mar 2 '16 at 5:00
  • $\begingroup$ One thing I don't understand, @dsaxton- why do we have $\binom{t - 1}{n - 1}$ and not $\binom{t}{n}$? $\endgroup$ – daOnlyBG Mar 23 '16 at 21:45
  • $\begingroup$ @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. $\endgroup$ – dsaxton Mar 23 '16 at 22:46
  • $\begingroup$ @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? $\endgroup$ – daOnlyBG Mar 25 '16 at 19:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.