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=- unanswered
b=- unanswered
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=- unanswered
d=- unanswered
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=- unanswered
g=- unanswered
h=- unanswered
E[NM]=- unanswered
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
ALL CORRECT!!!
a = λ^m * exp(-λs) / (m-n)!
b = (λs)^n / n!
c = (λt)^(m+n) * exp(-λ(t+s)) / (m-n)! * (m+n)!
d = (λ(t+s))^m / m!
f = (λt)^(m-n) * exp(-λt) / (m-n)!
g = (λt)^n / n!
h = (λs)^m / m!
E[NM] = λ^2 * t * s + λt * s^2 + λt^2 * s + λt * s
To find the conditional PMF pM|N(m|n), we can use the formula:
pM|N(m|n) = pN,M(n+m,m) / pN(n)
pN(n) is the PMF of N, which is given by:
pN(n) = (exp(-λt) * (λt)^n) / n!
And pN,M(n+m,m) is the joint PMF of N and M, which we'll find later.
Now let's find the joint PMF pN,M(n,m). The joint PMF is given by:
pN,M(n,m) = pN(n) * pM|N(m|n)
So to find pN,M(n,m), we first need to find pM|N(m|n). Since the conditional PMF is of the form ab!, we can express it as:
pM|N(m|n) = a * (b!) / (m! * (n+m)!)
Looking at the formula for pM|N(m|n), we can see that a = 1 and b = (n+m). So we have:
pM|N(m|n) = 1 * ((n+m)!) / (m! * (n+m)!)
Therefore, the conditional PMF pM|N(m|n) is given by:
pM|N(m|n) = ((n+m)!) / (m! * (n+m)!)
Now, substituting this into the formula for pN,M(n,m), we have:
pN,M(n,m) = pN(n) * pM|N(m|n)
pN,M(n,m) = (exp(-λt) * (λt)^n) / n! * ((n+m)!) / (m! * (n+m)!)
Therefore, the joint PMF pN,M(n,m) is given by:
pN,M(n,m) = (exp(-λt) * (λt)^n * (n+m)!) / (n! * m! * (n+m)!)
Now, to find the conditional PMF pN|M(n|m), we can use the formula:
pN|M(n|m) = pN,M(n,m) / pM(m)
pM(m) is the PMF of M, which is given by:
pM(m) = (exp(-λ(t+s)) * (λ(t+s))^m) / m!
Substituting this into the formula for pN|M(n|m), we have:
pN|M(n|m) = pN,M(n,m) / pM(m)
pN|M(n|m) = (exp(-λt) * (λt)^n * (n+m)!) / (n! * m! * (n+m)!) * m! / (exp(-λ(t+s)) * (λ(t+s))^m)
Simplifying, we have:
pN|M(n|m) = (exp(-λt) * (λt)^n * (n+m)!) / (n! * (exp(-λ(t+s)) * (λ(t+s))^m))
Therefore, the conditional PMF pN|M(n|m) is given by:
pN|M(n|m) = (exp(-λt) * (λt)^n * (n+m)!) / (n! * (exp(-λ(t+s)) * (λ(t+s))^m))
Finally, E[NM] is the expected value of the product of N and M, which can be computed using the joint PMF as follows:
E[NM] = ∑∑(n,m) (n * m * pN,M(n,m))
Therefore, E[NM] is the sum of the product of n, m, and the joint PMF pN,M(n,m), taken over all possible values of n and m.
To find the conditional PMF pM|N(m|n) of M given N, we can use the definition of conditional probability. Recall that the PMF of a Poisson distribution with rate λ is given by P(X = k) = (λ^k * exp(-λ)) / k!.
For each value of N = n, M can take values from n to n+m (since 0 ≤ n ≤ m). So, we need to find the probability that M = m given that N = n. This can be expressed as pM|N(m|n).
Let's break down the calculation:
For each fixed value of N = n, the remaining values are distributed as a Poisson process with rate λ * s, since we are measuring the number of arrivals in the interval (0, s]. Therefore, the probability of exactly m arrivals in (0, t+s] given N = n is given by P(M = m | N = n) = ((λ * s)^(m-n) * exp(-λ * s)) / (m-n)!
However, since we are interested in the probability of M = m given that N = n, we need to consider the probability that there are exactly n arrivals in (0, t], which is given by P(N = n) = (λ^k * exp(-λ)) / k!.
Therefore, the conditional PMF pM|N(m|n) can be calculated as:
pM|N(m|n) = P(M = m | N = n) * P(N = n) = ((λ * s)^(m-n) * exp(-λ * s)) / (m-n)! * (λ^n * exp(-λ)) / n!
Hence, the algebraic expressions for a and b are:
a = (λ * s)^(m-n) * exp(-λ * s)
b = (m-n)!
Moving on to the joint PMF pN,M(n,m) of N and M, we need to find the probability of having exactly n arrivals in (0, t] (N = n) and exactly m arrivals in (0, t+s] (M = m). This can be expressed as pN,M(n,m).
Similar to the calculation above, we can express the joint PMF as the product of the individual probabilities:
pN,M(n,m) = P(N = n) * P(M = m | N = n) = (λ^n * exp(-λ)) / n! * ((λ * s)^(m-n) * exp(-λ * s)) / (m-n)!
Hence, the algebraic expressions for c and d are:
c = λ^n * exp(-λ)
d = (m-n)!
Next, let's calculate the conditional PMF pN|M(n|m) of N given M. We want to find the probability of having exactly n arrivals in (0, t] (N = n) given that there are exactly m arrivals in (0, t+s] (M = m). This can be expressed as pN|M(n|m).
To find the conditional PMF, we can use Bayes' theorem:
pN|M(n|m) = pN,M(n,m) / pM(m)
We have already found pN,M(n,m) above. Now, let's find pM(m), the probability of having exactly m arrivals in (0, t+s]:
pM(m) = Σ(pN,M(n,m)) for all possible values of n (0 ≤ n ≤ m)
Therefore, the expression for pN|M(n|m) is:
pN|M(n|m) = pN,M(n,m) / pM(m) = ((λ^n * exp(-λ)) / n! * ((λ * s)^(m-n) * exp(-λ * s)) / (m-n)!) / Σ(pN,M(n,m)) for all possible values of n
For the expression E[NM], we can calculate the expected value of the product of N and M by summing up the product of each possible value of N and M multiplied by their respective probabilities:
E[NM] = Σ(n * m * pN,M(n,m)) for all possible values of n and m (0 ≤ n ≤ m)
Note that these expressions may need further algebraic simplification depending on specific values of λ, t, s, m, and n.