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Waiting time in Poisson process
Average waiting time in a Poisson processprobability (waiting time = infinity) for a poisson processHow to calculate the average waiting time from an in-homogeneous Poisson process.Waiting time for two independent poisson processesAre these facts about the Poisson process correct?Number of arrivals of a compound Poisson process?Poisson process with dead timePoisson process: waiting time probabilitiesConditional waiting time in Poisson processNumber of Events Registered in a Poisson Process
$begingroup$
Let $X(t) : t geq 0$ be a Poisson process with rate $lambda$, and let $W_n$ denote the waiting time for the $n$-th event. For $s geq 0$, determine $P( W_X(t) leq t+s)$ and $P( W_X(t)+2 leq t+s)$.
Here's what I've tried.
For $W_X(t)$ to be less than $t+s$, the $X(t)$-th event must occur before time $t+s$. This implies that at least $X(t)$ many events must occur before time $t+s$, so we have
$$
P( W_X(t) leq t+s)
= P( X(t+s) geq X(t))
= P( X(t+s) - X(t) geq 0)
= 1.
$$
Applying similar logic to the second one,
beginalign*
P(W_X(t)+2 leq t + s)
&= P(X(t+s) geq X(t) + 2) \
&= 1 - P(X(t+s) - X(t) < 2) \
&= 1 - P(X(t+s) - X(t) = 0) - P(X(t+s) - X(t) = 1) \
&= 1 - e^-lambda t - lambda t e^-lambda t.
endalign*
Is what I've done ok? I'm getting the niggling feeling that I should be using the uniform distribution somehow. Please advise, thanks!
proof-verification stochastic-processes poisson-process
asked Mar 22 at 8:44
jessicajessica
7719
$endgroup$
add a comment |
$begingroup$
Let $X(t) : t geq 0$ be a Poisson process with rate $lambda$, and let $W_n$ denote the waiting time for the $n$-th event. For $s geq 0$, determine $P( W_X(t) leq t+s)$ and $P( W_X(t)+2 leq t+s)$.
Here's what I've tried.
For $W_X(t)$ to be less than $t+s$, the $X(t)$-th event must occur before time $t+s$. This implies that at least $X(t)$ many events must occur before time $t+s$, so we have
$$
P( W_X(t) leq t+s)
= P( X(t+s) geq X(t))
= P( X(t+s) - X(t) geq 0)
= 1.
$$
Applying similar logic to the second one,
beginalign*
P(W_X(t)+2 leq t + s)
&= P(X(t+s) geq X(t) + 2) \
&= 1 - P(X(t+s) - X(t) < 2) \
&= 1 - P(X(t+s) - X(t) = 0) - P(X(t+s) - X(t) = 1) \
&= 1 - e^-lambda t - lambda t e^-lambda t.
endalign*
Is what I've done ok? I'm getting the niggling feeling that I should be using the uniform distribution somehow. Please advise, thanks!
proof-verification stochastic-processes poisson-process
asked Mar 22 at 8:44
jessicajessica
7719
$endgroup$
$begingroup$
Frankly, my understanding of poisson processes is really weak. Have I understood $W_X(t)$ correctly? I've only ever encountered things like $W_2$ and $W_n$ so I'm really at a loss
$endgroup$
– jessica
Mar 22 at 9:05
add a comment |
$begingroup$
Let $X(t) : t geq 0$ be a Poisson process with rate $lambda$, and let $W_n$ denote the waiting time for the $n$-th event. For $s geq 0$, determine $P( W_X(t) leq t+s)$ and $P( W_X(t)+2 leq t+s)$.
Here's what I've tried.
For $W_X(t)$ to be less than $t+s$, the $X(t)$-th event must occur before time $t+s$. This implies that at least $X(t)$ many events must occur before time $t+s$, so we have
$$
P( W_X(t) leq t+s)
= P( X(t+s) geq X(t))
= P( X(t+s) - X(t) geq 0)
= 1.
$$
Applying similar logic to the second one,
beginalign*
P(W_X(t)+2 leq t + s)
&= P(X(t+s) geq X(t) + 2) \
&= 1 - P(X(t+s) - X(t) < 2) \
&= 1 - P(X(t+s) - X(t) = 0) - P(X(t+s) - X(t) = 1) \
&= 1 - e^-lambda t - lambda t e^-lambda t.
endalign*
Is what I've done ok? I'm getting the niggling feeling that I should be using the uniform distribution somehow. Please advise, thanks!
proof-verification stochastic-processes poisson-process
asked Mar 22 at 8:44
jessicajessica
7719
$endgroup$
Let $X(t) : t geq 0$ be a Poisson process with rate $lambda$, and let $W_n$ denote the waiting time for the $n$-th event. For $s geq 0$, determine $P( W_X(t) leq t+s)$ and $P( W_X(t)+2 leq t+s)$.
Here's what I've tried.
For $W_X(t)$ to be less than $t+s$, the $X(t)$-th event must occur before time $t+s$. This implies that at least $X(t)$ many events must occur before time $t+s$, so we have
$$
P( W_X(t) leq t+s)
= P( X(t+s) geq X(t))
= P( X(t+s) - X(t) geq 0)
= 1.
$$
Applying similar logic to the second one,
beginalign*
P(W_X(t)+2 leq t + s)
&= P(X(t+s) geq X(t) + 2) \
&= 1 - P(X(t+s) - X(t) < 2) \
&= 1 - P(X(t+s) - X(t) = 0) - P(X(t+s) - X(t) = 1) \
&= 1 - e^-lambda t - lambda t e^-lambda t.
endalign*
Is what I've done ok? I'm getting the niggling feeling that I should be using the uniform distribution somehow. Please advise, thanks!
proof-verification stochastic-processes poisson-process
proof-verification stochastic-processes poisson-process
asked Mar 22 at 8:44
jessicajessica
7719
asked Mar 22 at 8:44
jessicajessica
7719
asked Mar 22 at 8:44
jessicajessica
7719
asked Mar 22 at 8:44
jessicajessica
7719
asked Mar 22 at 8:44
jessicajessica
7719
7719
$begingroup$
Frankly, my understanding of poisson processes is really weak. Have I understood $W_X(t)$ correctly? I've only ever encountered things like $W_2$ and $W_n$ so I'm really at a loss
$endgroup$
– jessica
Mar 22 at 9:05
add a comment |
$begingroup$
Frankly, my understanding of poisson processes is really weak. Have I understood $W_X(t)$ correctly? I've only ever encountered things like $W_2$ and $W_n$ so I'm really at a loss
$endgroup$
– jessica
Mar 22 at 9:05
$begingroup$
Frankly, my understanding of poisson processes is really weak. Have I understood $W_X(t)$ correctly? I've only ever encountered things like $W_2$ and $W_n$ so I'm really at a loss
$endgroup$
– jessica
Mar 22 at 9:05
$begingroup$
Frankly, my understanding of poisson processes is really weak. Have I understood $W_X(t)$ correctly? I've only ever encountered things like $W_2$ and $W_n$ so I'm really at a loss
$endgroup$
– jessica
Mar 22 at 9:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
After much discussion with another student, we've determined that what I did is correct except for a typo in the last line. The answer is
$$
P(W_X(t)+2 leq t + s)
= 1 - e^-lambda s - lambda s e^-lambda s.
$$
Rather than focusing on the Poisson process on its own, it's better to think of things as a renewal process and consider the special case of Poisson processes.
answered Mar 27 at 10:01
jessicajessica
7719
$endgroup$
1
$begingroup$
As you pointed out earlier, $P(W_X(t) + 2 leq t+s) = P(X(t+s) geq X(t) + 2) = P(X(t+s) -X(t) geq 2)$. Since the process is Poisson, you can write $P(W_X(t) + 2 leq t+s) = P(X(s) - X(0) geq 2$. Note that the last statement is not true for a renewal process, except for the special case of the Poisson process. For a renewal process $X(t)$ counts the number of renewal events till time $t$ (including events if any at time t), so at time $t$, $W_X(t)$ tells you the last renewal epoch before time $t$ (again we account for renewals at $t$ if any).
$endgroup$
– Sudheer
Mar 27 at 12:55
add a comment |
Your Answer
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$begingroup$
After much discussion with another student, we've determined that what I did is correct except for a typo in the last line. The answer is
$$
P(W_X(t)+2 leq t + s)
= 1 - e^-lambda s - lambda s e^-lambda s.
$$
Rather than focusing on the Poisson process on its own, it's better to think of things as a renewal process and consider the special case of Poisson processes.
answered Mar 27 at 10:01
jessicajessica
7719
$endgroup$
1
$begingroup$
As you pointed out earlier, $P(W_X(t) + 2 leq t+s) = P(X(t+s) geq X(t) + 2) = P(X(t+s) -X(t) geq 2)$. Since the process is Poisson, you can write $P(W_X(t) + 2 leq t+s) = P(X(s) - X(0) geq 2$. Note that the last statement is not true for a renewal process, except for the special case of the Poisson process. For a renewal process $X(t)$ counts the number of renewal events till time $t$ (including events if any at time t), so at time $t$, $W_X(t)$ tells you the last renewal epoch before time $t$ (again we account for renewals at $t$ if any).
$endgroup$
– Sudheer
Mar 27 at 12:55
add a comment |
$begingroup$
After much discussion with another student, we've determined that what I did is correct except for a typo in the last line. The answer is
$$
P(W_X(t)+2 leq t + s)
= 1 - e^-lambda s - lambda s e^-lambda s.
$$
Rather than focusing on the Poisson process on its own, it's better to think of things as a renewal process and consider the special case of Poisson processes.
answered Mar 27 at 10:01
jessicajessica
7719
$endgroup$
1
$begingroup$
As you pointed out earlier, $P(W_X(t) + 2 leq t+s) = P(X(t+s) geq X(t) + 2) = P(X(t+s) -X(t) geq 2)$. Since the process is Poisson, you can write $P(W_X(t) + 2 leq t+s) = P(X(s) - X(0) geq 2$. Note that the last statement is not true for a renewal process, except for the special case of the Poisson process. For a renewal process $X(t)$ counts the number of renewal events till time $t$ (including events if any at time t), so at time $t$, $W_X(t)$ tells you the last renewal epoch before time $t$ (again we account for renewals at $t$ if any).
$endgroup$
– Sudheer
Mar 27 at 12:55
add a comment |
$begingroup$
After much discussion with another student, we've determined that what I did is correct except for a typo in the last line. The answer is
$$
P(W_X(t)+2 leq t + s)
= 1 - e^-lambda s - lambda s e^-lambda s.
$$
Rather than focusing on the Poisson process on its own, it's better to think of things as a renewal process and consider the special case of Poisson processes.
answered Mar 27 at 10:01
jessicajessica
7719
$endgroup$
After much discussion with another student, we've determined that what I did is correct except for a typo in the last line. The answer is
$$
P(W_X(t)+2 leq t + s)
= 1 - e^-lambda s - lambda s e^-lambda s.
$$
Rather than focusing on the Poisson process on its own, it's better to think of things as a renewal process and consider the special case of Poisson processes.
answered Mar 27 at 10:01
jessicajessica
7719
answered Mar 27 at 10:01
jessicajessica
7719
answered Mar 27 at 10:01
jessicajessica
7719
answered Mar 27 at 10:01
jessicajessica
7719
7719
1
$begingroup$
As you pointed out earlier, $P(W_X(t) + 2 leq t+s) = P(X(t+s) geq X(t) + 2) = P(X(t+s) -X(t) geq 2)$. Since the process is Poisson, you can write $P(W_X(t) + 2 leq t+s) = P(X(s) - X(0) geq 2$. Note that the last statement is not true for a renewal process, except for the special case of the Poisson process. For a renewal process $X(t)$ counts the number of renewal events till time $t$ (including events if any at time t), so at time $t$, $W_X(t)$ tells you the last renewal epoch before time $t$ (again we account for renewals at $t$ if any).
$endgroup$
– Sudheer
Mar 27 at 12:55
add a comment |
1
$begingroup$
As you pointed out earlier, $P(W_X(t) + 2 leq t+s) = P(X(t+s) geq X(t) + 2) = P(X(t+s) -X(t) geq 2)$. Since the process is Poisson, you can write $P(W_X(t) + 2 leq t+s) = P(X(s) - X(0) geq 2$. Note that the last statement is not true for a renewal process, except for the special case of the Poisson process. For a renewal process $X(t)$ counts the number of renewal events till time $t$ (including events if any at time t), so at time $t$, $W_X(t)$ tells you the last renewal epoch before time $t$ (again we account for renewals at $t$ if any).
$endgroup$
– Sudheer
Mar 27 at 12:55
1
1
$begingroup$
As you pointed out earlier, $P(W_X(t) + 2 leq t+s) = P(X(t+s) geq X(t) + 2) = P(X(t+s) -X(t) geq 2)$. Since the process is Poisson, you can write $P(W_X(t) + 2 leq t+s) = P(X(s) - X(0) geq 2$. Note that the last statement is not true for a renewal process, except for the special case of the Poisson process. For a renewal process $X(t)$ counts the number of renewal events till time $t$ (including events if any at time t), so at time $t$, $W_X(t)$ tells you the last renewal epoch before time $t$ (again we account for renewals at $t$ if any).
$endgroup$
– Sudheer
Mar 27 at 12:55
$begingroup$
As you pointed out earlier, $P(W_X(t) + 2 leq t+s) = P(X(t+s) geq X(t) + 2) = P(X(t+s) -X(t) geq 2)$. Since the process is Poisson, you can write $P(W_X(t) + 2 leq t+s) = P(X(s) - X(0) geq 2$. Note that the last statement is not true for a renewal process, except for the special case of the Poisson process. For a renewal process $X(t)$ counts the number of renewal events till time $t$ (including events if any at time t), so at time $t$, $W_X(t)$ tells you the last renewal epoch before time $t$ (again we account for renewals at $t$ if any).
$endgroup$
– Sudheer
Mar 27 at 12:55
add a comment |
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$begingroup$
Frankly, my understanding of poisson processes is really weak. Have I understood $W_X(t)$ correctly? I've only ever encountered things like $W_2$ and $W_n$ so I'm really at a loss
$endgroup$
– jessica
Mar 22 at 9:05