Consider a Poisson process with rate lambda = 2 and let T be the time of the first arrival.
1. Find the conditional PDF of T given that the second arrival came before time t = 1. Enter an expression in terms of lambda and t.
2. Find the conditional PDF of T given that the third arrival comes exactly at time t = 1.
1) (lambda*exp(-lambda*t) - lambda*exp(-lambda)) / (1-(lambda+1)*exp(-lambda))
2) 2*(1-t)
Both can be obtained using Bayes' rule P(A|B) = P(A)P(B|A) / P(B). I cannot say for sure they are right, but as a first test their integral from 0 to 1 (domain for t) is 1 so.. let's cross the fingers!
1) i am not sure
lambda*e^(-lambda)*(e^(-lambda*(t-1))-1)/(1-(1+lambda)*e^(-lambda))
I think the second part is 3t^2 - 6t + 3.
Is the first part supposed to be a constant? Is that what I should be looking for?
I think my calculation for the second part was wrong. If the third arrival comes at t=1, then the first two arrivals are uniformly distributed on [0, 1], so we should have E[T] = 1/3. Recalculating based on the order statistics gives 2 - 2t, which matches RocknRoll's answer above and gives the right expected value.
To find the conditional PDF of T given certain conditions, we need to use the properties of conditional probability and the memorylessness property of the Poisson process.
1. Find the conditional PDF of T given that the second arrival came before time t = 1:
To start, let's call the event "A" as the second arrival comes before time t = 1, and event "B" as T = t.
To find the conditional PDF of T given A, we need to calculate P(B|A) = P(T = t|A).
P(A) can be calculated using the complementary probability, as P(A) = 1 - P(second arrival after time t = 1).
The number of arrivals in a Poisson process with rate lambda = 2 in a time interval of length t follows a Poisson distribution with parameter lambda*t. Therefore, P(second arrival after time t = 1) = e^(-2(1)) * (2(1))^2 / 2! = e^(-2).
Therefore, P(A) = 1 - e^(-2).
Now, to find P(B|A), we need to find the joint probability P(A and B). Since the second arrival occurred before time t = 1, we can say that T must be less than 1. Therefore, P(B|A) = P(T = t|T < 1) = P(T = t)/P(T < 1).
The probability density function (PDF) of a continuous random variable T in a Poisson process with rate lambda is given by f_T(t) = lambda * e^(-lambda*t).
Thus, P(T = t)/P(T < 1) = lambda * e^(-lambda*t) / ∫[0,1] lambda * e^(-lambda*u) du
= lambda * e^(-lambda*t) / (-e^(-lambda*u) | [0,1])
= lambda * e^(-lambda*t) / (e^(-lambda) - 1).
Therefore, the conditional PDF of T given that the second arrival came before time t = 1 is:
f_T|A(t) = P(T = t|A) = lambda * e^(-lambda*t) / (e^(-lambda) - 1).
2. Find the conditional PDF of T given that the third arrival comes exactly at time t = 1:
Let's call the event "C" as the third arrival occurs exactly at time t = 1, and event "D" as T = t.
To find the conditional PDF of T given C, we need to calculate P(D|C) = P(T = t|C).
Similar to the previous case, we need to find P(T = t)/P(C), where P(C) is the probability that the third arrival occurs exactly at time t = 1.
The probability of the third arrival occurring at time t = 1 is the probability of exactly two arrivals occurring before time t = 1 in a Poisson process with rate lambda = 2. This can be calculated using the Poisson distribution:
P(C) = e^(-2) * (2(1))^2 / 2! = 2e^(-2).
Now, P(D|C) = P(T = t|T = 1) = P(T = t)/P(C), which can be calculated similar to the previous case:
P(D|C) = lambda * e^(-lambda*t) / (2e^(-2)) = (lambda/2) * e^(-lambda*t + 2).
Therefore, the conditional PDF of T given that the third arrival comes exactly at time t = 1 is:
f_T|C(t) = P(T = t|C) = (lambda/2) * e^(-lambda*t + 2).
1. 2*t*(1-t)
Given P(A⋂B) = λtexp(-λt)*λ(1-t)exp(-λ(1-t))
P(B)=λ²exp(-λ)/2!
2. 3*t*(1-t)²
Given P(A⋂B) = λtexp(-λt)*λ²(1-t)^2exp(-λ(1-t))
P(B)=λ^3exp(-λ)/3!