Question

Suppose, as in Exercise 52, that a stream of customers arrives at an $\mathrm{r}$-way junction, so that the time between successive arrivals has an exponential distribution with mean $1 / \lambda$. Then, by Theorem $3.2 .1(\mathrm{~g})$, the number of arrivals per unit of time has a Poisson distribution with mean $\lambda$. Suppose the branch selected by each arrival is chosen independently with the probability that an arrival takes path $i$ equal to $p_i$ for $i=1,2, \cdots, r$. We can imagine a random number generator that chooses 1 with probability $p_1, 2$ with probability $p_2, \ldots, r$ with probability $p_r$, where $\sum_{i=1}^r p_r=1$. Each customer then takes the path chosen by the random number generator and the generator makes a new choice for each arriving customer. Prove that the $i$ th output stream has a Poisson pattern with mean rate $p_i \lambda$. [Hint: Let $N(t)$ be the number of customer arrivals to the junction in $t$ time units. (We assume the observations begin when $t=0$.) Let $N_i(t)$ be the number of these arrivals that take the $i$ th path. Then the conditional joint distribution of $N_i(t)(i=1,2, \cdots, r)$ given that $N(t)=n$, $$ P\left[N_1(t)=k_1, N_2(t)=k_2, \ldots, N_r(t)=k_r \mid N(t)=n\right], $$ has a multinomial distribution (see Exercise 8 where event $E_i$ is the event that a customer takes path $i$ ). Multiplying this probability by the probability that $N(t)=n$, which has a Poisson distribution with mean $\lambda t$ by Theorem $3.2 .1(\mathrm{~g})$, we obtain the joint probability distribution $P\left(k_1, k_2, \ldots, k_r\right) . P\left(k_1, k_2, \ldots, k_r\right)$ expresses the probability that $k_1$ customers take the first path, $k_2$ take the second path, etc. Show that the joint probability factors into the product of $r$ Poisson probabilities.]

   Suppose, as in Exercise 52, that a stream of customers arrives at an $\mathrm{r}$-way junction, so that the time between successive arrivals has an exponential distribution with mean $1 / \lambda$. Then, by Theorem $3.2 .1(\mathrm{~g})$, the number of arrivals per unit of time has a Poisson distribution with mean $\lambda$. Suppose the branch selected by each arrival is chosen independently with the probability that an arrival takes path $i$ equal to $p_i$ for $i=1,2, \cdots, r$. We can imagine a random number generator that chooses 1 with probability $p_1, 2$ with probability $p_2, \ldots, r$ with probability $p_r$, where $\sum_{i=1}^r p_r=1$. Each customer then takes the path chosen by the random number generator and the generator makes a new choice for each arriving customer. Prove that the $i$ th output stream has a Poisson pattern with mean rate $p_i \lambda$. [Hint: Let $N(t)$ be the number of customer arrivals to the junction in $t$ time units. (We assume the observations begin when $t=0$.) Let $N_i(t)$ be the number of these arrivals that take the $i$ th path. Then the conditional joint distribution of $N_i(t)(i=1,2, \cdots, r)$ given that $N(t)=n$,
$$
P\left[N_1(t)=k_1, N_2(t)=k_2, \ldots, N_r(t)=k_r \mid N(t)=n\right],
$$
has a multinomial distribution (see Exercise 8 where event $E_i$ is the event that a customer takes path $i$ ). Multiplying this probability by the probability that $N(t)=n$, which has a Poisson distribution with mean $\lambda t$ by Theorem $3.2 .1(\mathrm{~g})$, we obtain the joint probability distribution $P\left(k_1, k_2, \ldots, k_r\right) . P\left(k_1, k_2, \ldots, k_r\right)$ expresses the probability that $k_1$ customers take the first path, $k_2$ take the second path, etc. Show that the joint probability factors into the product of $r$ Poisson probabilities.]
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Probability, Statistics, and Queuing Theory with Computer Science Applications, Second Edition (Computer Science and Scientific Computing)
Probability, Statistics, and Queuing Theory with Computer Science Applications, Second Edition (Computer Science and Scientific Computing)
Arnold O. Allen 2nd Edition
Chapter 3, Problem 53 ↓

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- Let $N(t)$ be the total number of customers arriving at the junction in $t$ time units. - Let $N_i(t)$ be the number of customers who choose path $i$ in $t$ time units, for $i = 1, 2, \ldots, r$.  Show more…

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Suppose, as in Exercise 52, that a stream of customers arrives at an $\mathrm{r}$-way junction, so that the time between successive arrivals has an exponential distribution with mean $1 / \lambda$. Then, by Theorem $3.2 .1(\mathrm{~g})$, the number of arrivals per unit of time has a Poisson distribution with mean $\lambda$. Suppose the branch selected by each arrival is chosen independently with the probability that an arrival takes path $i$ equal to $p_i$ for $i=1,2, \cdots, r$. We can imagine a random number generator that chooses 1 with probability $p_1, 2$ with probability $p_2, \ldots, r$ with probability $p_r$, where $\sum_{i=1}^r p_r=1$. Each customer then takes the path chosen by the random number generator and the generator makes a new choice for each arriving customer. Prove that the $i$ th output stream has a Poisson pattern with mean rate $p_i \lambda$. [Hint: Let $N(t)$ be the number of customer arrivals to the junction in $t$ time units. (We assume the observations begin when $t=0$.) Let $N_i(t)$ be the number of these arrivals that take the $i$ th path. Then the conditional joint distribution of $N_i(t)(i=1,2, \cdots, r)$ given that $N(t)=n$, $$ P\left[N_1(t)=k_1, N_2(t)=k_2, \ldots, N_r(t)=k_r \mid N(t)=n\right], $$ has a multinomial distribution (see Exercise 8 where event $E_i$ is the event that a customer takes path $i$ ). Multiplying this probability by the probability that $N(t)=n$, which has a Poisson distribution with mean $\lambda t$ by Theorem $3.2 .1(\mathrm{~g})$, we obtain the joint probability distribution $P\left(k_1, k_2, \ldots, k_r\right) . P\left(k_1, k_2, \ldots, k_r\right)$ expresses the probability that $k_1$ customers take the first path, $k_2$ take the second path, etc. Show that the joint probability factors into the product of $r$ Poisson probabilities.]
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Key Concepts

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Exponential Distribution
The exponential distribution describes the time between successive events in a Poisson process. It is memoryless, meaning that the probability of an event occurring in the future is independent of how much time has already elapsed, and its mean is the reciprocal of the rate parameter (1/?).
Thinning of a Poisson Process
Thinning is a technique where each event in a Poisson process is randomly classified into one of several types, independently and with fixed probabilities. This procedure results in independent Poisson processes for each type, with the mean rate of each process being the original rate multiplied by the probability of classification for that type.
Poisson Process
A Poisson process is a stochastic process that models the occurrence of events randomly over time, with the property that the number of events in disjoint time intervals are independent. It is characterized by a constant average rate (?) and has the property that, over a fixed interval, the number of events follows a Poisson distribution.
Multinomial Distribution
When conditioning on a fixed total number of events, if each event is independently assigned to one of several categories based on given probabilities, the counts in the categories follow a multinomial distribution. This result is instrumental in showing that the joint distribution of the counts factors into the product of independent Poisson distributions upon unconditioning.

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