I am a non-student working through the first edition of Yates and Goodman's text, Probability and Stochastic Processes. On page $115$, question $3.6.9$ goes like this:

Each millisecond at a telephone switch, a call independently arrives with probability $p$. Each call is either a data call $(d)$ with probability $q$ or a voice call $(v)$. Each data is a fax call with probability $r$. Let $N$ equal the number of milliseconds required to observe the first $100$ fax calls. Let $T$ equal the number of milliseconds you observe the switch waiting for the first fax call. Find the marginal PMF $P_T(t)$ and the conditional PMF $P_{N|T}(n,t)$. Lastly, find the conditional PMF $P_{T|N}(t,n).$

Now, I seem to have two events with a random chance of happening: getting a call, and then having that call specifically yield a fax call.

From the perspective of getting a fax call (given getting a call), I don't think the PMF is difficult to figure out: $$\text{PMF} =\begin{cases} r(1-r)^{t-1},\quad t=1,2,... \\ 0,\quad \text{otherwise} \end{cases}$$ Of course, now we have to consider that we might not get a phone call altogether, and here is where my (personal) first question enters:

(a) If we have two (or more!) random variables, nested within each other (in this context, a getting a fax call assuming that we've gotten a call), how do we factor in the second variable into the PMF? My initial guess was this:

$$\text{PMF} =\begin{cases} q \cdot r(1-r)^{t-1},\quad t=1,2,... \\ 0,\quad \text{otherwise} \end{cases},$$ essentially multiplying the first PMF I showed by the probability of getting a call at all. My reasoning is that the first PMF (with only factor $r$) that I showed is inherently the probability given that a call was received; multiply that by the probability that a call was received, and you should get $P_T(t)$. However, should I be including a $(1-q)$ term in there? If so, to what power do I raise it to? It can't be $t-1$.

(b) How would I go about computing the joint PMF $P_{N,T}(n,t)$? This is necessary, as I understand it, for calculating the conditional PMFs. My guess is that $T$ is geometrically distributed, and $N$ is a Negative Binomial/"Pascal" distributed random variable.


Let $F_n\stackrel{\sf{iid}}\sim\sf{Geo}(pqr)$. Then $S_n=\sum_{k=1}^n F_k$ is the time of the $n^{\sf{th}}$ fax call, and $N=S_{100}$. By induction we may show that $N\sim\sf{NegBin}(100,pqr)$ (cf. How to compute the sum of random variables of geometric distribution) and so, $$\mathbb P(N=n) = \binom{n-1}{99} (pqr)^{100}(1-pqr)^{n-100},\ n=100, 101,\ldots\tag{1} $$ Further $T=F_1$ so $$\mathbb P(T=t)=pqr(1-pqr)^{t-1},\ t=1,2,\ldots \tag{2} $$ By the memoryless property of the geometric distribution, we have $$\mathbb P(N=n\mid T=t) = \mathbb P(F_{99}=n-t)=\binom{n-t-1}{98}(pqr)^{99}(1-pqr)^{n-t-99}\tag{3}$$ when $n-t\geqslant 99$. Bayes' rule then yields \begin{align} \mathbb P(T=t\mid N=n) &= \frac{\mathbb P(N=n\mid T=t)\mathbb P(T=t)}{\mathbb P(N=n)} \tag{4}\\ &= \frac{\binom{n-t-1}{98}(pqr)^{99}(1-pqr)^{n-t-99}pqr(1-pqr)^{t-1}}{\binom{n-1}{99} (pqr)^{100}(1-pqr)^{n-100}}\tag{5}\\ &= \frac{\binom{n-t-1}{98}}{\binom{n-1}{99}}\tag{6}. \end{align}

  • $\begingroup$ Thank you. I did have one question, though (and took the liberty to label your question lines with numbers, hope you don't mind)- I don't understand how you got equation (3). I have an intuitive understanding of what the memoryless property is, but I'm not really sure why you decided to compute $F_99=n-t$...? $\endgroup$ – daOnlyBG Mar 30 '16 at 20:39
  • $\begingroup$ $N=T+\sum_{k=2}^{100} F_k$ so $N-T=\sum_{k=2}^{100}F_k$ has the same distribution as $F_{99}=\sum_{k=1}^{99} F_k$. $\endgroup$ – Math1000 Mar 30 '16 at 20:55
  • $\begingroup$ Oh no, I understood $how$ you got $P(F_{99})$, I'm just not sure $why$ you computed it. Does it relate to $\mathbb P(T=t\mid N=n)$ in that we compute the probability of $N$ happening given that $T$ has already happened, hence why we subtract one $F$? (Once again, thank you for helping) $\endgroup$ – daOnlyBG Mar 30 '16 at 20:59
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    $\begingroup$ Continuing from my last comment, $$\mathbb P(N=n\mid T=t) = \frac{\mathbb P(N=n,T=t)}{\mathbb P(T=t)} = \frac{\mathbb P(N-T=n-t)\mathbb P(T=t)}{\mathbb P(T=t)}=\mathbb P(N-T=n-t). $$ $\endgroup$ – Math1000 Mar 30 '16 at 21:06
  • $\begingroup$ You're welcome! $\endgroup$ – Math1000 Mar 30 '16 at 21:11

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