Finding expected value of a stopping time dependent on a Poisson process and a variable $n$
$begingroup$
Situation:
We have that ${W_t,t geq 0}$ is a Brownian motion and ${N_t,tgeq 0}$ is a Poisson process such that $N_t$ follows a Poisson
distribution with parameter $t$. This process is independent from our
Brownian motion. We define the following stopping time: $tau_n =inf{tgeq 0:N_t geq n}$.
Question: We are given the hint that we must use the fact that $N_t - t$ and $(N_t - t)^{2}-t$ are martingales, and that we must use Doob's optional sampling theorem. The exercise is to find:
$mathbb{E}[(tau_{n} - n)^2]$ and $mathbb{E}[tau_{n}]$
My attempt: I'm getting confused with the way the stopping time is defined, more specifically the subscript $n$ in there. I figured that from the optional sampling theorem follows that:
$mathbb{E}[N_{t wedge tau_{n}}]=mathbb{E}[twedgetau_{n}]$
And then using the dominated convergence theorem and the monotone convergence theorem to find $mathbb{E}[tau_n]$, but this doesn't give correct results. This was pretty much my only idea on solving this exercise so I was going to use the same approach for finding the other expectation except then using the other martingale.
Tips on solving this exercise are much appreciated (rather than a full answer). Thanks in advance!
probability-theory stochastic-processes stopping-times poisson-process expected-value
asked Jan 2 at 18:55
S. CrimS. Crim
35612
$endgroup$
|
show 7 more comments
$begingroup$
Situation:
We have that ${W_t,t geq 0}$ is a Brownian motion and ${N_t,tgeq 0}$ is a Poisson process such that $N_t$ follows a Poisson
distribution with parameter $t$. This process is independent from our
Brownian motion. We define the following stopping time: $tau_n =inf{tgeq 0:N_t geq n}$.
Question: We are given the hint that we must use the fact that $N_t - t$ and $(N_t - t)^{2}-t$ are martingales, and that we must use Doob's optional sampling theorem. The exercise is to find:
$mathbb{E}[(tau_{n} - n)^2]$ and $mathbb{E}[tau_{n}]$
My attempt: I'm getting confused with the way the stopping time is defined, more specifically the subscript $n$ in there. I figured that from the optional sampling theorem follows that:
$mathbb{E}[N_{t wedge tau_{n}}]=mathbb{E}[twedgetau_{n}]$
And then using the dominated convergence theorem and the monotone convergence theorem to find $mathbb{E}[tau_n]$, but this doesn't give correct results. This was pretty much my only idea on solving this exercise so I was going to use the same approach for finding the other expectation except then using the other martingale.
Tips on solving this exercise are much appreciated (rather than a full answer). Thanks in advance!
probability-theory stochastic-processes stopping-times poisson-process expected-value
asked Jan 2 at 18:55
S. CrimS. Crim
35612
$endgroup$
$begingroup$
"[...] but this doesn't give correct results" It would be helpful if you could state a) the correct results and b) the results of your calculations...
$endgroup$
– saz
Jan 2 at 19:42
$begingroup$
@saz You're right. My result for a) was that $mathbb{E}[tau_{n}]=mathbb{E}[N_{tau_{n}}]$, where I don't know what $mathbb{E}[N_{tau_{n}}]$ is. I would suspect that we have that $mathbb{E}[N_{tau_{n}}]=tau_{n}$, because $N_{t}$ is a Poisson process. The correct answer for both question a) and b) is that the expectations equal $n$.
$endgroup$
– S. Crim
Jan 3 at 11:52
$begingroup$
I see. Note that $tau_n$ is a random mapping, and therefore $mathbb{E}(N_{tau_n}) = tau_n$ cannot hold true (the left-hand side is a "deterministic" constant whereas the right-hand side is a random variable). To prove $mathbb{E}(N_{tau_n})=n$ you have to show that $$N_{tau_n} = n quad text{a.s.}$$ to this end, you can use that $(N_t)_{t geq 0}$ has non-decreasing sample paths with jumps of height 1.
$endgroup$
– saz
Jan 3 at 12:39
1
$begingroup$
Yeah, sounds okay. Regarding the exact proof: Denote by $sigma_n$ the time of the $n$-th arrival. By the very definition of the $n$-th arrival we have $N_{sigma_n}=n$, and thus $tau_n leq sigma_n$ by the very definition of $tau_n$. On the other hand, we have $N_t < n$ for $t<sigma_n$ (again by the definition of $sigma_n$) and thus $tau_n geq sigma_n$. Hence, $sigma_n = tau_n$ and combining this with $N_{sigma_n}=n$ this gives $N_{tau_n}=n$.
$endgroup$
– saz
Jan 10 at 9:50
1
$begingroup$
If $sigma_n$ is some stopping time such that $N_{sigma_n}=n$ a.s. then $sigma_n = tau_n$ almost surely... however, that's not exactly obvious from the definitiion of $tau_n$ (at least not as far as I can see). Re the "a.s.": Many authors allow for exceptional null sets in the very definition of the Poisson process, e.g. they assume only that $(N_t)_{t geq 0}$ has càdlàg non-decreasing sample paths with probability $1$... that's where the exceptoinal null sets come into play.
$endgroup$
– saz
Jan 10 at 11:32
|
show 7 more comments
$begingroup$
Situation:
We have that ${W_t,t geq 0}$ is a Brownian motion and ${N_t,tgeq 0}$ is a Poisson process such that $N_t$ follows a Poisson
distribution with parameter $t$. This process is independent from our
Brownian motion. We define the following stopping time: $tau_n =inf{tgeq 0:N_t geq n}$.
Question: We are given the hint that we must use the fact that $N_t - t$ and $(N_t - t)^{2}-t$ are martingales, and that we must use Doob's optional sampling theorem. The exercise is to find:
$mathbb{E}[(tau_{n} - n)^2]$ and $mathbb{E}[tau_{n}]$
My attempt: I'm getting confused with the way the stopping time is defined, more specifically the subscript $n$ in there. I figured that from the optional sampling theorem follows that:
$mathbb{E}[N_{t wedge tau_{n}}]=mathbb{E}[twedgetau_{n}]$
And then using the dominated convergence theorem and the monotone convergence theorem to find $mathbb{E}[tau_n]$, but this doesn't give correct results. This was pretty much my only idea on solving this exercise so I was going to use the same approach for finding the other expectation except then using the other martingale.
Tips on solving this exercise are much appreciated (rather than a full answer). Thanks in advance!
probability-theory stochastic-processes stopping-times poisson-process expected-value
asked Jan 2 at 18:55
S. CrimS. Crim
35612
$endgroup$
Situation:
We have that ${W_t,t geq 0}$ is a Brownian motion and ${N_t,tgeq 0}$ is a Poisson process such that $N_t$ follows a Poisson
distribution with parameter $t$. This process is independent from our
Brownian motion. We define the following stopping time: $tau_n =inf{tgeq 0:N_t geq n}$.
Question: We are given the hint that we must use the fact that $N_t - t$ and $(N_t - t)^{2}-t$ are martingales, and that we must use Doob's optional sampling theorem. The exercise is to find:
$mathbb{E}[(tau_{n} - n)^2]$ and $mathbb{E}[tau_{n}]$
My attempt: I'm getting confused with the way the stopping time is defined, more specifically the subscript $n$ in there. I figured that from the optional sampling theorem follows that:
$mathbb{E}[N_{t wedge tau_{n}}]=mathbb{E}[twedgetau_{n}]$
And then using the dominated convergence theorem and the monotone convergence theorem to find $mathbb{E}[tau_n]$, but this doesn't give correct results. This was pretty much my only idea on solving this exercise so I was going to use the same approach for finding the other expectation except then using the other martingale.
Tips on solving this exercise are much appreciated (rather than a full answer). Thanks in advance!
probability-theory stochastic-processes stopping-times poisson-process expected-value
probability-theory stochastic-processes stopping-times poisson-process expected-value
asked Jan 2 at 18:55
S. CrimS. Crim
35612
asked Jan 2 at 18:55
S. CrimS. Crim
35612
edited Jan 9 at 20:27
S. Crim
asked Jan 2 at 18:55
S. CrimS. Crim
35612
asked Jan 2 at 18:55
S. CrimS. Crim
35612
asked Jan 2 at 18:55
S. CrimS. Crim
35612
35612
$begingroup$
"[...] but this doesn't give correct results" It would be helpful if you could state a) the correct results and b) the results of your calculations...
$endgroup$
– saz
Jan 2 at 19:42
$begingroup$
@saz You're right. My result for a) was that $mathbb{E}[tau_{n}]=mathbb{E}[N_{tau_{n}}]$, where I don't know what $mathbb{E}[N_{tau_{n}}]$ is. I would suspect that we have that $mathbb{E}[N_{tau_{n}}]=tau_{n}$, because $N_{t}$ is a Poisson process. The correct answer for both question a) and b) is that the expectations equal $n$.
$endgroup$
– S. Crim
Jan 3 at 11:52
$begingroup$
I see. Note that $tau_n$ is a random mapping, and therefore $mathbb{E}(N_{tau_n}) = tau_n$ cannot hold true (the left-hand side is a "deterministic" constant whereas the right-hand side is a random variable). To prove $mathbb{E}(N_{tau_n})=n$ you have to show that $$N_{tau_n} = n quad text{a.s.}$$ to this end, you can use that $(N_t)_{t geq 0}$ has non-decreasing sample paths with jumps of height 1.
$endgroup$
– saz
Jan 3 at 12:39
1
$begingroup$
Yeah, sounds okay. Regarding the exact proof: Denote by $sigma_n$ the time of the $n$-th arrival. By the very definition of the $n$-th arrival we have $N_{sigma_n}=n$, and thus $tau_n leq sigma_n$ by the very definition of $tau_n$. On the other hand, we have $N_t < n$ for $t<sigma_n$ (again by the definition of $sigma_n$) and thus $tau_n geq sigma_n$. Hence, $sigma_n = tau_n$ and combining this with $N_{sigma_n}=n$ this gives $N_{tau_n}=n$.
$endgroup$
– saz
Jan 10 at 9:50
1
$begingroup$
If $sigma_n$ is some stopping time such that $N_{sigma_n}=n$ a.s. then $sigma_n = tau_n$ almost surely... however, that's not exactly obvious from the definitiion of $tau_n$ (at least not as far as I can see). Re the "a.s.": Many authors allow for exceptional null sets in the very definition of the Poisson process, e.g. they assume only that $(N_t)_{t geq 0}$ has càdlàg non-decreasing sample paths with probability $1$... that's where the exceptoinal null sets come into play.
$endgroup$
– saz
Jan 10 at 11:32
|
show 7 more comments
$begingroup$
"[...] but this doesn't give correct results" It would be helpful if you could state a) the correct results and b) the results of your calculations...
$endgroup$
– saz
Jan 2 at 19:42
$begingroup$
@saz You're right. My result for a) was that $mathbb{E}[tau_{n}]=mathbb{E}[N_{tau_{n}}]$, where I don't know what $mathbb{E}[N_{tau_{n}}]$ is. I would suspect that we have that $mathbb{E}[N_{tau_{n}}]=tau_{n}$, because $N_{t}$ is a Poisson process. The correct answer for both question a) and b) is that the expectations equal $n$.
$endgroup$
– S. Crim
Jan 3 at 11:52
$begingroup$
I see. Note that $tau_n$ is a random mapping, and therefore $mathbb{E}(N_{tau_n}) = tau_n$ cannot hold true (the left-hand side is a "deterministic" constant whereas the right-hand side is a random variable). To prove $mathbb{E}(N_{tau_n})=n$ you have to show that $$N_{tau_n} = n quad text{a.s.}$$ to this end, you can use that $(N_t)_{t geq 0}$ has non-decreasing sample paths with jumps of height 1.
$endgroup$
– saz
Jan 3 at 12:39
1
$begingroup$
Yeah, sounds okay. Regarding the exact proof: Denote by $sigma_n$ the time of the $n$-th arrival. By the very definition of the $n$-th arrival we have $N_{sigma_n}=n$, and thus $tau_n leq sigma_n$ by the very definition of $tau_n$. On the other hand, we have $N_t < n$ for $t<sigma_n$ (again by the definition of $sigma_n$) and thus $tau_n geq sigma_n$. Hence, $sigma_n = tau_n$ and combining this with $N_{sigma_n}=n$ this gives $N_{tau_n}=n$.
$endgroup$
– saz
Jan 10 at 9:50
1
$begingroup$
If $sigma_n$ is some stopping time such that $N_{sigma_n}=n$ a.s. then $sigma_n = tau_n$ almost surely... however, that's not exactly obvious from the definitiion of $tau_n$ (at least not as far as I can see). Re the "a.s.": Many authors allow for exceptional null sets in the very definition of the Poisson process, e.g. they assume only that $(N_t)_{t geq 0}$ has càdlàg non-decreasing sample paths with probability $1$... that's where the exceptoinal null sets come into play.
$endgroup$
– saz
Jan 10 at 11:32
$begingroup$
"[...] but this doesn't give correct results" It would be helpful if you could state a) the correct results and b) the results of your calculations...
$endgroup$
– saz
Jan 2 at 19:42
$begingroup$
"[...] but this doesn't give correct results" It would be helpful if you could state a) the correct results and b) the results of your calculations...
$endgroup$
– saz
Jan 2 at 19:42
$begingroup$
@saz You're right. My result for a) was that $mathbb{E}[tau_{n}]=mathbb{E}[N_{tau_{n}}]$, where I don't know what $mathbb{E}[N_{tau_{n}}]$ is. I would suspect that we have that $mathbb{E}[N_{tau_{n}}]=tau_{n}$, because $N_{t}$ is a Poisson process. The correct answer for both question a) and b) is that the expectations equal $n$.
$endgroup$
– S. Crim
Jan 3 at 11:52
$begingroup$
@saz You're right. My result for a) was that $mathbb{E}[tau_{n}]=mathbb{E}[N_{tau_{n}}]$, where I don't know what $mathbb{E}[N_{tau_{n}}]$ is. I would suspect that we have that $mathbb{E}[N_{tau_{n}}]=tau_{n}$, because $N_{t}$ is a Poisson process. The correct answer for both question a) and b) is that the expectations equal $n$.
$endgroup$
– S. Crim
Jan 3 at 11:52
$begingroup$
I see. Note that $tau_n$ is a random mapping, and therefore $mathbb{E}(N_{tau_n}) = tau_n$ cannot hold true (the left-hand side is a "deterministic" constant whereas the right-hand side is a random variable). To prove $mathbb{E}(N_{tau_n})=n$ you have to show that $$N_{tau_n} = n quad text{a.s.}$$ to this end, you can use that $(N_t)_{t geq 0}$ has non-decreasing sample paths with jumps of height 1.
$endgroup$
– saz
Jan 3 at 12:39
$begingroup$
I see. Note that $tau_n$ is a random mapping, and therefore $mathbb{E}(N_{tau_n}) = tau_n$ cannot hold true (the left-hand side is a "deterministic" constant whereas the right-hand side is a random variable). To prove $mathbb{E}(N_{tau_n})=n$ you have to show that $$N_{tau_n} = n quad text{a.s.}$$ to this end, you can use that $(N_t)_{t geq 0}$ has non-decreasing sample paths with jumps of height 1.
$endgroup$
– saz
Jan 3 at 12:39
1
1
$begingroup$
Yeah, sounds okay. Regarding the exact proof: Denote by $sigma_n$ the time of the $n$-th arrival. By the very definition of the $n$-th arrival we have $N_{sigma_n}=n$, and thus $tau_n leq sigma_n$ by the very definition of $tau_n$. On the other hand, we have $N_t < n$ for $t<sigma_n$ (again by the definition of $sigma_n$) and thus $tau_n geq sigma_n$. Hence, $sigma_n = tau_n$ and combining this with $N_{sigma_n}=n$ this gives $N_{tau_n}=n$.
$endgroup$
– saz
Jan 10 at 9:50
$begingroup$
Yeah, sounds okay. Regarding the exact proof: Denote by $sigma_n$ the time of the $n$-th arrival. By the very definition of the $n$-th arrival we have $N_{sigma_n}=n$, and thus $tau_n leq sigma_n$ by the very definition of $tau_n$. On the other hand, we have $N_t < n$ for $t<sigma_n$ (again by the definition of $sigma_n$) and thus $tau_n geq sigma_n$. Hence, $sigma_n = tau_n$ and combining this with $N_{sigma_n}=n$ this gives $N_{tau_n}=n$.
$endgroup$
– saz
Jan 10 at 9:50
1
1
$begingroup$
If $sigma_n$ is some stopping time such that $N_{sigma_n}=n$ a.s. then $sigma_n = tau_n$ almost surely... however, that's not exactly obvious from the definitiion of $tau_n$ (at least not as far as I can see). Re the "a.s.": Many authors allow for exceptional null sets in the very definition of the Poisson process, e.g. they assume only that $(N_t)_{t geq 0}$ has càdlàg non-decreasing sample paths with probability $1$... that's where the exceptoinal null sets come into play.
$endgroup$
– saz
Jan 10 at 11:32
$begingroup$
If $sigma_n$ is some stopping time such that $N_{sigma_n}=n$ a.s. then $sigma_n = tau_n$ almost surely... however, that's not exactly obvious from the definitiion of $tau_n$ (at least not as far as I can see). Re the "a.s.": Many authors allow for exceptional null sets in the very definition of the Poisson process, e.g. they assume only that $(N_t)_{t geq 0}$ has càdlàg non-decreasing sample paths with probability $1$... that's where the exceptoinal null sets come into play.
$endgroup$
– saz
Jan 10 at 11:32
|
show 7 more comments
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$begingroup$
"[...] but this doesn't give correct results" It would be helpful if you could state a) the correct results and b) the results of your calculations...
$endgroup$
– saz
Jan 2 at 19:42
$begingroup$
@saz You're right. My result for a) was that $mathbb{E}[tau_{n}]=mathbb{E}[N_{tau_{n}}]$, where I don't know what $mathbb{E}[N_{tau_{n}}]$ is. I would suspect that we have that $mathbb{E}[N_{tau_{n}}]=tau_{n}$, because $N_{t}$ is a Poisson process. The correct answer for both question a) and b) is that the expectations equal $n$.
$endgroup$
– S. Crim
Jan 3 at 11:52
$begingroup$
I see. Note that $tau_n$ is a random mapping, and therefore $mathbb{E}(N_{tau_n}) = tau_n$ cannot hold true (the left-hand side is a "deterministic" constant whereas the right-hand side is a random variable). To prove $mathbb{E}(N_{tau_n})=n$ you have to show that $$N_{tau_n} = n quad text{a.s.}$$ to this end, you can use that $(N_t)_{t geq 0}$ has non-decreasing sample paths with jumps of height 1.
$endgroup$
– saz
Jan 3 at 12:39
1
$begingroup$
Yeah, sounds okay. Regarding the exact proof: Denote by $sigma_n$ the time of the $n$-th arrival. By the very definition of the $n$-th arrival we have $N_{sigma_n}=n$, and thus $tau_n leq sigma_n$ by the very definition of $tau_n$. On the other hand, we have $N_t < n$ for $t<sigma_n$ (again by the definition of $sigma_n$) and thus $tau_n geq sigma_n$. Hence, $sigma_n = tau_n$ and combining this with $N_{sigma_n}=n$ this gives $N_{tau_n}=n$.
$endgroup$
– saz
Jan 10 at 9:50
1
$begingroup$
If $sigma_n$ is some stopping time such that $N_{sigma_n}=n$ a.s. then $sigma_n = tau_n$ almost surely... however, that's not exactly obvious from the definitiion of $tau_n$ (at least not as far as I can see). Re the "a.s.": Many authors allow for exceptional null sets in the very definition of the Poisson process, e.g. they assume only that $(N_t)_{t geq 0}$ has càdlàg non-decreasing sample paths with probability $1$... that's where the exceptoinal null sets come into play.
$endgroup$
– saz
Jan 10 at 11:32