Events related to the Poisson process can be often described in two equivalent ways: in terms of numbers of arrivals during certain intervals or in terms of arrival times. The first description involves discrete random variables, the second continuous random variables.
Let N(t) be the number of arrivals during the time interval [0,t] in a Poisson process. Let Yk be the time of the k th arrival.
a) The event {N(5)>1} is equivalent to the event {Yk≤b} , for suitable b and k . Find b and k .
b=
unanswered
k=
unanswered
b) The event {2<Y3≤Y4≤5} is equivalent to the event {N(2)≤a and N(5)≥b} . Find a and b .
a=
unanswered
b=
unanswered
b = 5
k = 2
a = 2
b = 4
k=2
b) a=2
a) In order for the event {N(5)>1} to be equivalent to the event {Yk≤b}, we need to find the suitable values for b and k.
Since N(5) represents the number of arrivals during the time interval [0, 5], N(5)>1 implies that there were more than one arrival in that time interval.
Similarly, Yk represents the time of the kth arrival. So, Yk≤b implies that the time of the kth arrival is less than or equal to b.
Since we want the event {N(5)>1} to be equivalent to {Yk≤b}, we can choose b=5 and k=2. This means that if there were at least two arrivals in the time interval [0, 5], then the time of the second arrival (Y2) would be less than or equal to 5.
Therefore, b=5 and k=2.
b) In order for the event {2<Y3≤Y4≤5} to be equivalent to the event {N(2)≤a and N(5)≥b}, we need to identify suitable values for a and b.
The event {2<Y3≤Y4≤5} means that the time of the third arrival is greater than 2, and the time of the fourth arrival is between the times of the third and fifth arrivals, which are all within the time interval [0, 5].
On the other hand, the event {N(2)≤a} means that there are at most a arrivals in the time interval [0, 2], and the event {N(5)≥b} means that there are at least b arrivals in the time interval [0, 5].
So, if we choose a=2 and b=4, then the conditions of the two events will be equivalent. It means that if there are at most 2 arrivals in the time interval [0, 2] and at least 4 arrivals in the time interval [0, 5], then the time of the third arrival (Y3) will be greater than 2 and the time of the fourth arrival (Y4) will be between the times of the third and fifth arrivals, which are all within the time interval [0, 5].
Therefore, a=2 and b=4.
a) To find the values of b and k, we need to determine the time interval in which there are more than one arrival (N(5)>1). In a Poisson process, the number of arrivals in a fixed interval follows a Poisson distribution with parameter λt, where λ is the average rate of arrival and t is the length of the interval.
For N(5)>1, we need to find the smallest value of b such that Yk≤b, where k is the number of arrivals. Since N(5)>1, it means there is at least one arrival within the interval [0,5]. Therefore, we need to find the smallest arrival time (Yk) such that it is less than or equal to b.
Let's assume k=1. This means we are looking for the first arrival time (Y1). In this case, b=5, because if the first arrival occurs at or before time 5, then the event N(5)>1 is satisfied.
So, b=5 and k=1.
b) To find the values of a and b, we need to determine the time intervals in which the third and fourth arrivals occur and the total number of arrivals in the time interval [0,5].
The event 2<Y3≤Y4≤5 means that the third arrival occurs after time 2 and before or at the same time as the fourth arrival, both of which occur before or at time 5.
To express this event in terms of N(2) and N(5), we need to consider the number of arrivals in the intervals [0,2], (2,5], and [0,5].
- If the third arrival occurs after time 2, then N(2) will be less than 2 (2<Y3).
- If the fourth arrival occurs before or at the same time as the fifth arrival, then N(5) will be greater than or equal to 2 (N(5)≥b).
Therefore, a=2 and b=2.
So, a=2 and b=2.
a) To find the values of b and k, we need to understand the relationship between N(t) (the number of arrivals during [0,t]) and Yk (the time of the kth arrival). In a Poisson process, the intervals between arrivals are exponential random variables with rate λ (the arrival rate). Therefore, the waiting time until the kth arrival, Yk, follows a gamma distribution with parameters k and λ.
Now, we want to find b and k such that the event {N(5) > 1} is equivalent to {Yk ≤ b}.
In other words, we're looking for the smallest arrival time Yk such that there are more than 1 arrival in the time interval [0,5].
In a Poisson process, the number of arrivals N(t) follows a Poisson distribution with mean λt. Therefore, to have N(5) > 1, we need the mean number of arrivals in the interval [0,5] to be greater than 1.
Given that the arrival rate λ is constant, λt is the expected number of arrivals in the time interval [0,t]. So, setting λt > 1, we have:
λt > 1
λ(5) > 1
5λ > 1
λ > 1/5
Hence, the event {N(5) > 1} is equivalent to the event {Yk ≤ b}, where λ > 1/5. Since the arrival rate is λ, we can choose b = 1/5 and k = 1.
Therefore, b = 1/5 and k = 1.
b) To find the values of a and b, we need to understand the relationship between the event {2 < Y3 ≤ Y4 ≤ 5} and the events {N(2) ≤ a} and {N(5) ≥ b}.
In a Poisson process, the arrival times Yk are independent, and the interarrival times are exponentially distributed. Therefore, we can consider the intervals between arrivals.
The event {2 < Y3 ≤ Y4 ≤ 5} means that the third arrival occurs after 2 units of time, and the fourth arrival occurs after the third and before or at 5 units of time. This implies that there are at least 2 arrivals in the interval [0,2] and at least 3 arrivals in the interval [0,5].
The number of arrivals N(t) follows a Poisson distribution with mean λt. So, we have:
N(2) > 1 (to have at least 2 arrivals in [0,2])
N(5) > 2 (to have at least 3 arrivals in [0,5])
The mean number of arrivals in the interval [0,t] is λt, so:
λ(2) > 1
2λ > 1
λ > 1/2
λ(5) > 2
5λ > 2
λ > 2/5
Therefore, the event {2 < Y3 ≤ Y4 ≤ 5} is equivalent to the event {N(2) ≤ a and N(5) ≥ b}, where λ > 1/2 and λ > 2/5.
Hence, a = 1/2 and b = 2/5.