Consider a Poisson process with rate λ. Let N be the number of arrivals in (0,t] and M be the number of arrivals in (0,t+s], where t>0,s≥0.

In each part below, your answers will be algebraic expressions in terms of λ,t,s,m and/or n. Enter 'lambda' for λ and use 'exp()' for exponentials. Do not use 'fac()' or '!' for factorials. Follow standard notation.

For 0≤n≤m, the conditional PMF pM∣N(m∣n) of M given N is of the form ab! for suitable algebraic expressions in place of a and b.

a=
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b=
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For 0≤n≤m, the joint PMF pN,M(n,m) of N and M is of the form cn!d! for suitable algebraic expressions in place of c and d.

c=
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d=
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For 0≤n≤m, the conditional PMF pN|M(n|m) of N given M is of the form f⋅g!n!h! for suitable algebraic expressions in place of f, g, and h.

f=
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g=
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h=
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E[NM]=
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a=(lambda*s)^(m-n)*e^(-lambda*s)

b= m-n

c= lambda^m*s^(m-n)*t^n*e^(-lambda*(s+t))

d= m-n

f= (s^(m-n)*t^n)/((s+t)^m)

g= m

h= m-n

E[NM]= (lambda*t)*(lambda*s)+lambda*t+(lambda*t)^2

a = λ^n * (λ*t)^m-n * exp(-λ*t) / (m-n)!

b = λ^m * s^n * exp(-λ*s) / (m-n)!

c = (λ*t)^n * exp(-λ*t) * (λ*t + λ*s)^m-n * exp(-λ*s) / (n!(m-n)!)

d = (λ*t + λ*s)^m * exp(-λ*t -λ*s) / (m-n)!

f = (λ*t)^n * exp(-λ*t) * (λ*t + λ*s)^m-n * exp(-λ*s) / (n!(m-n)!)

g = (m-n)!

h = (n!(m-n)!)

E[NM] = λ^2 * (t+s)

To find the expressions for a, b, c, d, f, g, h, and E[NM], we need to use the definitions of conditional and joint PMFs and the expected value.

a) The conditional PMF pM|N(m|n) of M given N is defined as the probability that there are m arrivals in (0,t+s] given that there are n arrivals in (0,t]. Since this is a Poisson process with rate λ, the number of arrivals in any interval follows a Poisson distribution.

The conditional PMF pM|N(m|n) can be expressed as:
pM|N(m|n) = P(M = m | N = n)

Using the properties of Poisson distribution, we know that the number of arrivals in disjoint intervals are independent. Therefore, the number of arrivals in (0,t+s] can be represented as the sum of the number of arrivals in (0,t] and the number of arrivals in (t,t+s].

Since we know that there are n arrivals in (0,t], the remaining s time interval will follow a Poisson distribution with rate λ*s. Therefore, the conditional PMF is given by:
pM|N(m|n) = P(M = m | N = n) = P(M-N = m-n | N = n) = P(Poisson(λ*s) = m-n) = (e^(-λs)*(λs)^(m-n)) / (m-n)!

Therefore, a = (e^(-λs)*(λs)^(m-n)) / (m-n)!

b) The conditional PMF pM|N(m|n) of M given N is also given by:
pM|N(m|n) = P(M = m | N = n) = P(N-M = n-m | M = m) = P(Poisson(λt) = n-m) = (e^(-λt)*(λt)^(n-m)) / (n-m)!

Therefore, b = (e^(-λt)*(λt)^(n-m)) / (n-m)!

c) The joint PMF pN,M(n,m) of N and M is defined as the probability that there are n arrivals in (0,t] and m arrivals in (0,t+s]. Since this is a Poisson process, the joint PMF can be calculated as the product of individual PMFs.

The joint PMF pN,M(n,m) can be expressed as:
pN,M(n,m) = P(N = n, M = m)

Using the properties of Poisson distribution, we can write:
pN,M(n,m) = P(N = n, M = m) = P(N = n) * P(M-N = m-n) = P(Poisson(λt) = n) * P(Poisson(λs) = m-n) = (e^(-λt)*(λt)^n) / n! * (e^(-λs)*(λs)^(m-n)) / (m-n)!

Therefore, c = (e^(-λt)*(λt)^n) / n!
And d = (e^(-λs)*(λs)^(m-n)) / (m-n)!

d) The conditional PMF pN|M(n|m) of N given M is defined as the probability that there are n arrivals in (0,t] given that there are m arrivals in (0,t+s].

Using the properties of conditional probability, we can write:
pN|M(n|m) = P(N = n | M = m) = P(N-M = n-m | M = m) = P(Poisson(λt) = n-m | Poisson(λs) = m) = [P(Poisson(λt) = n-m, Poisson(λs) = m)] / P(Poisson(λs) = m)

Using the properties of Poisson distribution, we can calculate the probabilities as:
P(Poisson(λt) = n-m, Poisson(λs) = m) = (e^(-λt)*(λt)^(n-m)) / (n-m)! * (e^(-λs)*(λs)^m) / m!
P(Poisson(λs) = m) = (e^(-λs)*(λs)^m) / m!

Therefore, f = (e^(-λt)*(λt)^(n-m)) / (n-m)!
And g = (e^(-λs)*(λs)^m) / m!
And h = m!

e) The expected value of NM can be calculated using the properties of a Poisson process. The expected value of a Poisson distribution with rate λ is equal to λ. Therefore, we can write:

E[NM] = E[N]*E[M | N] = (λt)*(λ(t+s)) = λ^2(t^2 + ts)

Therefore, E[NM] = λ^2(t^2 + ts)

To answer the questions about conditional PMF and joint PMF, let's first define some concepts related to Poisson processes:

- A Poisson process is a stochastic process that models the occurrence of events over time, where the events happen independently and at a constant average rate λ.

- The number of arrivals in a fixed time interval (0, t], denoted N, follows a Poisson distribution with mean λt.

- The number of arrivals in a longer time interval (0, t+s], denoted M, follows a Poisson distribution with mean λ(t+s).

Now let's start answering the questions:

a) The conditional PMF pM|N(m|n) represents the probability that there are m arrivals in the interval (0, t+s], given that there are n arrivals in the interval (0, t].

To compute this, we can use the following formula:

pM|N(m|n) = P(M = m | N = n) = P(N+m = n+m | N = n)
= P(N + m - n = m) = P(N2 = m)

Here, N2 represents a Poisson random variable with mean λs. So, a = λ^m * exp(-λs) / m!

b) Similarly, we can compute the conditional PMF pN|M(n|m), which represents the probability that there are n arrivals in the interval (0, t], given that there are m arrivals in the interval (0, t+s].

Using the same logic as before, we have:

pN|M(n|m) = P(N = n | M = m) = P(M+n = m+n | M = m)
= P(M + n - m = n) = P(M2 = n)

Here, M2 represents a Poisson random variable with mean λt. Thus, b = λ^n * exp(-λt) / n!

c) The joint PMF pN,M(n, m) represents the probability of having n arrivals in the interval (0, t] and m arrivals in the interval (0, t+s].

To compute this, we need to multiply the individual probabilities:

pN,M(n, m) = P(N = n, M = m) = P(N = n) * P(M = m)

Since N and M are independent, their respective probabilities are given by the Poisson distribution:

pN(n) = λ^n * exp(-λt) / n!
pM(m) = λ^m * exp(-λs) / m!

Thus, c = λ^(n+m) * exp(-λ(t+s)) / (n! * m!)

d) The expected value E[NM] represents the average number of arrivals in both intervals (0, t] and (0, t+s].

To compute this, we can use the fact that the expected value of the product of independent random variables is the product of their individual expected values:

E[NM] = E[N] * E[M]

Since N and M follow Poisson distributions, their expected values are:

E[N] = λt
E[M] = λ(t+s)

Therefore, E[NM] = (λt) * (λ(t+s)) = λ^2 * t * (t+s)

Hope this helps! Let me know if you have any further questions.