Question

Let \(\\{N(t), t \geqslant 0\\}\) be a Poisson process with rate \(\lambda\). For \(sN(s))\) (b) \(P(N(s)=0, N(t)=3)\); (c) \(E[N(t) \mid N(s)=4] ;\) (d) \(E[N(s) \mid N(t)=4]\).

Short answer

(a) \(P(N(t)>N(s)) = 1 - e^{-\lambda(t-s)}\) (b) \(P(N(s)=0, N(t)=3) = e^{-\lambda s} \times \frac{(\lambda(t-s))^3 e^{-\lambda(t-s)}}{3!}\) (c) \(E[N(t) | N(s)=4] = 4 + \lambda(t-s)\) (d) \(E[N(s) | N(t)=4] = 4 - \lambda(t-s)\)

Step-by-step solution

Recall the Poisson Process properties

A Poisson process with rate λ has the following properties: 1. N(0) = 0 2. It has independent increments. That is, for t >= s, the random variable N(t) - N(s) is independent of N(u) for u ≤ s. 3. For any interval [s, t), the number of events in the interval has a Poisson distribution with parameter λ(t - s). Now we'll go through each part of the question.

Find (a) P(N(t) > N(s))

Let's find the probability \(P(N(t) - N(s) > 0)\). Since the increments are independent, we know that \(N(t) - N(s)\) has a Poisson distribution with parameter \(\lambda(t-s)\). So, \[P(N(t) > N(s)) = P(N(t) - N(s) > 0) = 1 - P(N(t) - N(s) = 0)\] Now, we know that a Poisson distribution with parameter \(\lambda(t-s)\) will have the probability mass function given by: \[P(N(t) - N(s) = k) = \frac{(\lambda(t - s))^k e^{-\lambda(t-s)}}{k!}\] So, for k = 0: \[P(N(t) - N(s) = 0) = \frac{(\lambda(t - s))^0 e^{-\lambda(t-s)}}{0!} = e^{-\lambda(t-s)}\] Thus, \[P(N(t) > N(s)) = 1 - e^{-\lambda(t-s)}\]

Find (b) P(N(s)=0, N(t)=3)

We know that \(N(t) - N(s)\) is independent of \(N(s)\) and has a Poisson distribution with parameter \(\lambda(t-s)\). Therefore, \[P(N(t) - N(s) = 3) = \frac{(\lambda(t-s))^3 e^{-\lambda(t-s)}}{3!}\] Now, we'll find the probability of \(N(s) = 0\): \[P(N(s) = 0) = \frac{(\lambda s)^0 e^{-\lambda s }}{0!} = e^{-\lambda s}\] And so, the joint probability is given by multiplying the two probabilities, since they are independent: \[P(N(s)=0, N(t)=3) = P(N(s)=0)P(N(t) - N(s)=3) = e^{-\lambda s} \times \frac{(\lambda(t-s))^3 e^{-\lambda(t-s)}}{3!}\]

Find (c) E[N(t) | N(s)=4]

Given that the Poisson process has independent increments, we know that \(N(t) - N(s)\) is independent of \(N(s)\). Since \(N(t) - N(s)\) has a Poisson distribution with parameter \(\lambda(t-s)\), the expected value is given by \(E[N(t) - N(s)] = \lambda(t-s)\). Now, conditioning on \(N(s) = 4\), we have: \[E[N(t) | N(s) = 4] = E[N(s) + (N(t) - N(s)) | N(s) = 4] = E[4 + N(t) - N(s)] = 4 + E[N(t) - N(s)] = 4 + \lambda(t-s)\]

Find (d) E[N(s) | N(t)=4]

Similarly, since \(N(t) - N(s)\) is independent of \(N(s)\), it's also independent of \(N(t)\). We can write \(N(s) = N(t) - (N(t) - N(s))\). Now, conditioning on \(N(t) = 4\), we have: \[E[N(s) | N(t) = 4] = E[N(t) - (N(t) - N(s)) | N(t) = 4] = E[4 - (N(t) - N(s))] = 4 - E[N(t) - N(s)] = 4 - \lambda(t-s)\] So, we have the final answers for each part of the question: (a) \(P(N(t)>N(s)) = 1 - e^{-\lambda(t-s)}\) (b) \(P(N(s)=0, N(t)=3) = e^{-\lambda s} \times \frac{(\lambda(t-s))^3 e^{-\lambda(t-s)}}{3!}\) (c) \(E[N(t) | N(s)=4] = 4 + \lambda(t-s)\) (d) \(E[N(s) | N(t)=4] = 4 - \lambda(t-s)\)

Key concepts

Independent Increments

Understanding the concept of independent increments is critical when studying random processes like the Poisson process. In this context, independent increments signify that the number of events occurring in disjoint time intervals is independent of one another. This implies that what happens in one period does not influence the probability of events in another.

For example, if you're observing a process where you count the number of emails you receive per hour, the number of emails received in one hour is unaffected by the number received in the previous hour, assuming this is a Poisson process. This specific property is what allows us to solve various probability questions related to the Poisson process as we can treat segments of time as distinct and separate in our calculations.

Probability Mass Function

The probability mass function (PMF) is an essential concept for discrete random variables, like the number of occurrences in a Poisson process. A PMF assigns a probability to each possible outcome. For a Poisson process with a given rate \( \lambda \), the PMF tells us the likelihood of observing exactly \( k \) events in a set time frame. Formally, it's expressed as:
\[ P(N(t) - N(s) = k) = \frac{(\lambda(t - s))^k e^{-\lambda(t-s)}}{k!} \]
Understanding and using the PMF is pivotal in working out specific probabilities, as we see in the textbook exercise where it's used to calculate the probability of a certain number of events in different intervals. Having this function at hand enables students to compute probabilities swiftly for any given number of occurrences within the process.

Conditional Expectation

Conditional expectation is a profound concept in probability theory, which deals with the expected value of a random variable given that another random variable or an event has occurred. In the realm of the Poisson process, this allows us to determine the expected number of events in one time frame knowing the number of events in another.

For instance, if we want to find out the expected number of emails we will receive by the end of the day, given that we've already received a certain amount by lunchtime, we would use the concept of conditional expectation. The formula would look something like this: \[ E[N(t) | N(s) = x] = x + \lambda(t-s) \]
By incorporating the rate of the process and the known count of occurrences up to time \( s \) (in this case, lunchtime), we're able to predict the average total count by time \( t \) (end of the day).