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Ants enter a kitchen at the instants of a Poisson process \(N\) of rate \(\lambda\); they cach visit the pantry and then the sink, and leave. The \(r\) th ant spends time \(X_{r}\) in the pantry and \(Y_{r}\) in the \(\operatorname{sink}\) (and \(X_{r}+Y_{r}\) in the kitchen altogether), where the vectors \(V_{r}=\left(X_{r}, Y_{r}\right)\) and \(V_{A}\) are independent for \(r \neq s .\) At time \(t=0\) the kitchen is free of ants. Find the joint distribution of the numbers \(A(t)\) of ants in the pantry and \(B(t)\) of ants in the \(\operatorname{sink}\) at time \(t\). Now suppose the ants arrive in pairs at the times of the Poisson process, but then separate to behave independently as above. Find the joint distribution of the numbers of ants in the two locations.

Short Answer

Expert verified
The joint distribution for ants in the pantry and sink is Poisson with parameters \(\lambda t e^{-\alpha t}\) for the pantry and \(\lambda t e^{-\beta t}\) for the sink.

Step by step solution

01

Understand the Poisson Process

A Poisson process is characterized by the parameter \(\lambda\), which represents the average number of events (ants entering the kitchen) in a given time period. In this case, ants enter the kitchen according to a Poisson process \(N\) with rate \(\lambda\).
02

Define Ants' Time Distribution

Each ant spends a time \(X_r\) in the pantry and \(Y_r\) in the sink. The total time spent in the kitchen by the \(r\)-th ant is \(X_r + Y_r\). The vectors \(V_r = (X_r, Y_r)\) are independent for different ants.
03

Determine Exponential Distribution

Since the times \(X_r\) and \(Y_r\) are independent for different ants, assume each is exponentially distributed with respective rates \(\alpha\) and \(\beta\). The probability that one specific ant is still in the pantry or sink is governed by these exponential distributions.
04

Calculate Individual Ant Contributions

For a given ant, the probability it is still in the pantry at time \(t\) is given by \(P(X_r > t) = e^{-\alpha t}\). Similarly, the probability it is in the sink is \(P(Y_r > t) = e^{-\beta t}\).
05

Compute Joint Probability

At time \(t\), an ant can either be in the pantry or the sink or have left the kitchen. The probability it is in the kitchen is \(P(X_r + Y_r > t) = e^{-\alpha t} + e^{-\beta t} - e^{-(\alpha + \beta)t}\). This comes from the inclusion-exclusion principle for probabilities.
06

Calculate Number of Ants in Locations

For a Poisson process with rate \(\lambda\), the expected number of ants that have arrived by time \(t\) is \(\lambda t\). The expected number in the pantry is \(A(t) = \lambda t e^{-\alpha t}\), and the number in the sink is \(B(t) = \lambda t e^{-\beta t}\).
07

Consider Paired Ant Entry

If ants arrive in pairs, each pair is still distributed as a Poisson process due to the properties of Poisson processes. Therefore, the expected number of pairs is \(\frac{\lambda}{2}\) per unit time.
08

Determine Joint Distribution for Paired Entry

Following similar steps as for individual entry, but considering pairs, the distribution of ants in either location retains a Poisson characteristic. Each ant within a pair enters the pantry or sink independently, as before.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Distribution
The exponential distribution is often used to model the time until an event occurs. In our ant scenario, this might represent the time an ant spends in the pantry or the sink. For a random variable that follows an exponential distribution, the probability density function is defined as:\[ f(x; \theta) = \theta e^{-\theta x}, \text{ for } x \geq 0 \]Here, \(\theta \) represents the rate of the distribution. The higher the rate, the quicker the event is expected to occur.
When modeling the time spent by ants, if \(X_r\) is the time an ant spends in the pantry and \(Y_r\) is the time in the sink, each will have rates \(\alpha\) and \(\beta\), respectively. Thus, the distribution gives us the likelihood that an ant stays in either location beyond a certain time \(t\) by calculating \(P(X_r > t) = e^{-\alpha t}\) and \(P(Y_r > t) = e^{-\beta t}\).
The beauty of exponential distributions is in their memoryless property, which means that the probability of an event occurring in the future is independent of any past information. This makes it particularly useful for modeling times like those of our enterprising ants.
Joint Distribution
A joint distribution helps us understand the relationship between two or more random variables. In our example, we're interested in the joint distribution of the times ants spend in the pantry and the sink—\(X_r\) and \(Y_r\).
When these random variables are independent, the joint probability distribution is the product of their marginal distributions. So, if \(X\) and \(Y\) are independent, the joint probability of \(X=x\) and \(Y=y\) is:\[P(X = x, Y = y) = P(X = x) \cdot P(Y = y)\]For times \(X_r\) and \(Y_r\), this means multiplying their exponential probability densities together.
The joint distribution plays a crucial role in determining the number of ants in each location at a given time. Using this, we can find the number \(A(t)\) of ants in the pantry and \(B(t)\) in the sink by applying the probabilities we determined earlier for each individual location based on independent exponential distributions.
Independent Random Variables
Independence in probability means that the occurrence of one random variable does not affect the occurrence of another. In our ant problem, the vectors \(V_r = (X_r, Y_r)\) are independent for different ants. This means what happens with one ant does not influence what happens with others.
This concept simplifies our calculations significantly. Because the time spent by ants in the pantry and the sink is independent for each ant, we do not need to consider complex interactions between them. Each ant’s actions can be analyzed separately.
For independent random variables \(X\) and \(Y\), knowing \(X = x\) gives no information about \(Y\) and vice versa. Mathematically, \(P(X, Y) = P(X) \cdot P(Y)\), which means we can deal with each part of the process separately without adjusting for interdependencies.
This independence is central to modeling the Poisson process for the paired ants arriving in the kitchen, which helps us easily extend the principles from individual ants to those arriving in pairs without loss of generality.

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Most popular questions from this chapter

Strong Markov property. Let \(X\) be a Markov chain on \(S\), and let \(T\) be a random variable taking values in \([0,1,2, \ldots\\}\) with the property that the indicator function \(I_{\mid T=n]}\), of the event that \(T=n\), is a function of the variables \(X_{1}, X_{2}, \ldots, X_{n}\). Such a random variable \(T\) is called a stopping time, and the above definition requires that it is decidable whether or not \(T=n\) with a knowledge only of the past and present, \(X_{0}, X_{1}, \ldots, X_{n}\), and with no further information about the future. Show that $$ \mathrm{P}\left(X_{T+m}=j \mid X_{k}=x_{h} \text { for } 0 \leq k

Prove that intercommunicating states of a Markov chain have the same period.

Let \(\left(X_{n}: n \geq 0\right)\) be a persistent irreducible discrete-time Markov chain on the state space \(S\) with transition matrix \(\mathbf{P}\), and let \(x\) be a positive solution of the equation \(x=\mathbf{x P}\) (a) Show that $$ q_{i j}(n)=\frac{x_{j}}{x_{i}} p_{j i}(n), \quad i, j \in S, n \geq 1 $$ defines the \(n\)-step transition probabilities of a persistent irreducible Markov chain on \(S\) whose first-passage probabilities are given by $$ g_{i j}(n)=\frac{x_{j}}{x_{i}} l_{f i}(n), \quad i \neq j, n \geq 1 $$ where \(l_{j i}(n)=\mathrm{P}\left(X_{n}=i, T>n \mid X_{0}=j\right)\) and \(T=\min \left(m>0: X_{m}=j\right\\}\) (b) Show that \(\mathrm{x}\) is unique up to a multiplicative constant. (c) Let \(T_{j}=\min \left(n \geq 1: X_{n}=j\right)\) and define \(h_{i j}=\mathrm{P}\left(T_{j} \leq T_{i} \mid X_{0}=i\right) .\) Show that \(x_{i} h_{i j}=x_{j} h_{j i}\) for all \(i_{+} j \in S\).

An um initially contains \(n\) green balls and \(n+2\) red balls. A ball is picked at random: if it is green then a red ball is also removed and both are discarded; if it is red then it is replaced together with an extra red and an extra green ball. This is repeated until there are no green balls in the um. Show that the probability the process terminates is \(1 /(n+1)\). Now reverse the rules: if the ball is green, it is replaced together with an extra green and an extra, red ball; if it is red it is discarded along with a green ball. Show that the expected number of iterations until no green balls remain is \(\sum_{j=1}^{n}(2 j+1)=n(n+2)\). [Thus, a minor perturbation of a simple symmetric random walk can be non-null persistent, whereas the original is null persistent.]

Disasters. Immigrants arrive at the instants of a Poisson process of rate \(v\), and each independently, founds a simple birth process of rate \(\lambda\). At the instants of an independent Poisson process of rate \(\delta\), the population is annihilated. Find the probability generating function of the population \(X(t)\), given that \(X(0)=0\).

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