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=
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b = 5

k = 2
a = 2
b = 4

k=2

b) a=2

a) To find b and k for the event {N(5)>1}, we want to find the time when the second arrival occurs. Since the Poisson process has a constant arrival rate, we can calculate the mean interarrival time as μ = 5/k, where μ is the arrival rate.

Since N(5) > 1, there must be at least 2 arrivals within the time interval [0, 5]. The arrival times are represented by Yk.

b = 5 / (k+1) (b is the time of the second arrival)

k = 1 (there must be at least 2 arrivals, so k = 1)

Therefore, b = 5/2 and k = 1.

b) To find a and b for the event {2 < Y3 ≤ Y4 ≤ 5}, we need to consider the arrival times of the third and fourth arrivals.

Since 2 < Y3 ≤ Y4 ≤ 5, it means that there are at least 4 arrivals within the time interval [0, 5].

The number of arrivals for N(2) will be 2 or less, and for N(5) will be at least 4.

So, we have:

a = 2 (N(2) ≤ 2, meaning there are 2 arrivals at most within 2 units of time)

b = 4 (N(5) ≥ 4, meaning there are at least 4 arrivals within 5 units of time)

Therefore, a = 2 and b = 4.

a) The event {N(5)>1} is equivalent to the event {Yk≤b}, for suitable b and k. To find b and k, we need to determine the value of b such that there is at least one arrival before time b (Yk≤b) and k is the number of that arrival.

In other words, we are looking for the smallest value of b such that the cumulative arrival time of k arrivals is less than or equal to b.

Since the event is {N(5) > 1}, it means there are at least two arrivals within the time interval [0, 5]. Therefore, we need to find b such that there are at least two arrival times Yk ≤ b.

Let's consider k = 2 (second arrival) and find the corresponding value of b.

P(Y2 ≤ b) = P(N(b) ≥ 2) = 1 - P(N(b) < 2)

Using the Poisson distribution formula, we can calculate the probability:

P(Y2 ≤ b) = 1 - P(N(b) = 0) - P(N(b) = 1)

The Poisson distribution formula is given by:

P(N(b) = k) = (e^(-λ)*(λ^k)) / k!

Where λ is the average rate of arrivals per unit time, and k is the number of arrivals.

For the Poisson process, since the average rate of arrivals per unit time is constant, we have:

P(N(t) = k) = (e^(-λt)*(λt^k)) / k!

Now let's solve for b:

1 - P(N(b) = 0) - P(N(b) = 1) = 1 - (e^(-λb)*(λb^0)) / 0! - (e^(-λb)*(λb^1)) / 1! = 0.9 (assuming a probability of 0.9 for {N(5) > 1})

Simplifying the equation:

1 - e^(-λb) - λb*e^(-λb) = 0.9

This equation cannot be solved analytically, but it can be solved numerically using numerical methods like Newton's method or by using a computational tool.

b) The event {2 < Y3 ≤ Y4 ≤ 5} is equivalent to the event {N(2)≤a and N(5)≥b}, for suitable values of a and b. Here, a is the number of arrivals in the time interval [0, 2], and b is the number of arrivals in the time interval [0, 5].

To find a and b, we need to determine the number of arrivals in each time interval.

For a, we need to find the number of arrivals in the time interval [0, 2]:

a = N(2)

Similarly, for b, we need to find the number of arrivals in the time interval [0, 5]:

b = N(5)

These values can be calculated using the Poisson distribution formula:

P(N(t) = k) = (e^(-λt)*(λt^k)) / k!

For example, to find a, we can calculate:

P(N(2) = k) = (e^(-λ*2)*(λ*2^k)) / k!

Summing up the probabilities for a > 2:

P(N(2) > 2) = 1 - P(N(2) ≤ 2) = 1 - P(N(2) = 0) - P(N(2) = 1) - P(N(2) = 2)

Similarly, for b, we can calculate:

P(N(5) ≥ k) = 1 - P(N(5) < k)

Summing up the probabilities for b ≥ 2:

P(N(5) ≥ 2) = 1 - P(N(5) < 2)

Again, solving these equations analytically is not possible, but they can be solved numerically using methods like Newton's method or computational tools.

To solve these questions, we need to understand the relationship between the number of arrivals (N(t)) and the arrival times (Yk) in a Poisson process.

In a Poisson process, the number of arrivals within a time interval follows a Poisson distribution, and the inter-arrival times between consecutive arrivals follow an exponential distribution.

Let's proceed with each question individually:

a) The event {N(5)>1} is equivalent to the event {Yk≤b}, for suitable b and k.

To find b and k, we need to determine the arrival time of the second arrival. Since the Poisson process has a constant arrival rate, the inter-arrival times are exponentially distributed.

The inter-arrival time between the first and second arrival, Y2, follows an exponential distribution with a mean of 1/λ, where λ is the arrival rate.

In this case, we want to find the arrival time of the second arrival, which means we need to find b such that P(Y2 ≤ b) = P(N(5) > 1).

Using the cumulative distribution function (CDF) of the exponential distribution, we have:

P(Y2 ≤ b) = 1 - e^(-λb)

Since we want P(Y2 ≤ b) = P(N(5) > 1), we have:

1 - e^(-λb) = P(N(5) > 1)

Now, we know that in a Poisson process, the number of arrivals follows a Poisson distribution with a mean of λt within a time interval of length t.

So, P(N(5) > 1) = 1 - P(N(5) = 0) - P(N(5) = 1)

= 1 - e^(-5λ) - (5λ)e^(-5λ)

Comparing this with 1 - e^(-λb), we can equate the expressions and solve for b:

1 - e^(-5λ) - (5λ)e^(-5λ) = 1 - e^(-λb)

e^(-5λ)(1 - 5λ) = e^(-λb)

Now, since the exponentials are equal, we can equate the expressions inside the exponentials:

-5λ = -λb

Solving for b, we get:

b = 5

So, we have found b = 5 as the suitable value for the event {Yk≤b}.

Now, let's determine the appropriate value for k.

Since the event {N(5) > 1} is equivalent to the event {Yk ≤ 5}, we need to find the smallest k such that the k-th arrival time is less than or equal to 5.

In other words, we need to find the smallest integer k that satisfies Yk ≤ 5.

However, we know that the inter-arrival times (Yk) are exponentially distributed with mean 1/λ.

So, we can rewrite the condition as k ≤ 5λ.

Therefore, k = 6.

Thus, we have found b = 5 and k = 6 as the suitable values for the event {N(5) > 1} being equivalent to the event {Yk≤b}.

b) The event {2<Y3≤Y4≤5} is equivalent to the event {N(2) ≤ a and N(5) ≥ b}.

To find the values of a and b, we need to relate the arrival times (Yk) to the number of arrivals (N(t)).

We know that the number of arrivals within a time interval follows a Poisson distribution with a mean of λt.

In this case, we want to find the number of arrivals within the time intervals [0, 2] and [0, 5].

So, we need to find values of a and b such that the conditions of the two events are satisfied.

The event {2<Y3≤Y4≤5} implies that the third arrival occurs between 2 and 5, and the fourth arrival occurs between 2 and 5 as well.

This can be rewritten as a condition on the number of arrivals as follows:

2 < Y3 ≤ Y4 ≤ 5
⇒ 2 < 3 ≤ 4 ≤ 5
⇒ 2 < 3 ≤ N(2) ≤ N(5) ≤ 5

So, we need to find a value of a such that N(2) ≤ a and a value of b such that N(5) ≥ b.

To find a, we want to determine the maximum number of arrivals in the time interval [0, 2], which is equivalent to finding the largest integer k such that Yk ≤ 2.

Since the inter-arrival times (Yk) are exponentially distributed with mean 1/λ, we can rewrite the condition as k ≤ 2λ.

Therefore, a = 2.

To find b, we want to determine the minimum number of arrivals in the time interval [0, 5], which is equivalent to finding the smallest integer k such that Yk > 5.

However, the inter-arrival times (Yk) are exponentially distributed with mean 1/λ.

So, we can rewrite the condition as k > 5λ.

Therefore, b = 6.

Thus, we have found a = 2 and b = 6 as the suitable values for the event {2<Y3≤Y4≤5} being equivalent to the event {N(2) ≤ a and N(5) ≥ b}.