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

I figured out the marginal PMFs, $P_N(n)$ and $P_T(t)$:

$\quad \quad \quad \quad \quad P_N(n) =\begin{cases} \binom{n-1}{99}pqr^{100}(1-pqr)^{n-100},\quad n=100, 101,... \\ 0,\quad \text{otherwise} \end{cases}$

$\quad \quad \quad \quad \quad \space P_T(t) =\begin{cases} pqr(1-pqr)^{t-1},\quad t=1,2... \\ 0,\quad \text{otherwise} \end{cases}$

How would I go about computing the joint PMF $P_{N,T}(n,t)$?

My understanding is that the joint PMF is the intersection of the probabilities of both random variables $N$ and $T$. Thus, my intuition tells me that the joint PMF will have a domain of $n,t \geq 100$.

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  • $\begingroup$ @Xi'an is this distribution you speak of geometric? $\endgroup$
    – daOnlyBG
    Mar 31, 2016 at 22:06
  • $\begingroup$ @Xi'an My thinking is that $T$ is still a geometrically distributed random variable, and we are searching for the PMF of the time required for the first call (in milliseconds), which here is analogous to the number of trials before a particular event happens. I'm really not sure where to go from here, though; perhaps we have to use a binomial constant, i.e., $\binom{n}{k}$? $\endgroup$
    – daOnlyBG
    Mar 31, 2016 at 22:25
  • $\begingroup$ @Xi'an If I understand you correctly, I should be looking for the PMF for $n-T_1 = \sum_{i=2}^{100} T_i$ (?). If so, we now have a fixed number of milliseconds (100) and have to calculate the PMF of an event happening $(T_i)$. Should we be using a binomial distribution? $\endgroup$
    – daOnlyBG
    Apr 1, 2016 at 15:32
  • $\begingroup$ @Xi'an I appreciate the time you've taken to comment on my question. Unfortunately, I don't seem to be any closer to finding the solution. I will add that I'm not even sure $T_i$ makes sense, since $T$ (and not $T_1$) is defined to be the time needed until the first fax call, not the $i^{th}$ fax call. $\endgroup$
    – daOnlyBG
    Apr 1, 2016 at 17:31

1 Answer 1

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Hint #1 You should start from first principles rather than trying at guessing among standard distributions: consider the joint distribution of the 100 $T_i$'s, which are (i) independent and (ii) geometric $\mathcal{G}(pqr)$. In other words, the $T_i$'s are iid: $$T_i\stackrel{\text{i.i.d.}}{\sim} \mathcal{G}(\rho)\qquad i=1,\ldots,100$$ with $\rho=pqr$. You can then deduce from this joint distribution on $(T_1,\ldots,T_{100})$, the distribution of $T_1$ conditional on $$N=∑^{100}_{i=1}T_i=n$$

Hint #2 The joint pmf of $(T_1,\ldots,T_{100})$ is provided by the product of their respective pmfs: $$\begin{align*} \mathbb{P}((T_1,\ldots,T_{100})&=(t_1,\ldots,t_{100}))=\prod_{i=1}^{100}\rho(1-\rho)^{t_i-1}\\&=\rho^{100}(1-\rho)^{-100}(1-\rho)^{ \sum_{i=1}^{100} t_i}\end{align*}$$ since they are independent

Hint #3 If $\sum_{i=1}^{100} t_i=n$, can you plug $n$ in the above joint pmf and deduce the distribution of $(T_1,\ldots,T_{100})$ given $N=n$? What are the consequences of the pmf being constant in $(t_1,\ldots,t_{100})$ given $n$?

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  • $\begingroup$ I went ahead and drew a graph (i.e., a tree) of the fax vs non-fax events, for small values (i.e., $N$ is the number of milliseconds needed to get $m=2$ faxes). I noticed that the odds of getting all $m$ faxes are $(pqr)^m (1-pqr)^{n-m}$; in the context of this problem, I suppose that would be $(pqr)^{100}(1-pqr)^{n-100}$. I suspect this is $P_{T|N}(t|n)$, if we were to add a binomial coefficient $\binom{100}{t}$? $\endgroup$
    – daOnlyBG
    Apr 2, 2016 at 22:07
  • $\begingroup$ I took your advice and (as you said) started from the earliest principles, hence why I drew the tree. I did this before considering your other hints because I genuinely wanted to figure out as much as possible independently. I proceeded to look at your other hints and I think you're trying to tell me that $P_{T|N}(t|n) = (pqr)^{100}(1-pqr)^{n-100}$. $\endgroup$
    – daOnlyBG
    Apr 4, 2016 at 0:30

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