How can I show that the stochastic integral of a jump process w.r.t. Brownian motion is a local martingale by...
Suppose that $Y$ is a pure jump process with $N_t$ jumps in $(0,t]$ and $E[N_t]<infty$. Denote the jump times by $T_i$. Let $W$ be a Brownian motion. If $T_0=0$, then
begin{equation}
M_t=int_0^t Y_s,dW_s=sum_{i=0}^{N_t-1} Y_{T_i}(W_{T_{i+1}}-W_{T_i})+Y_t(W_t-W_{T_{N_t}}).
end{equation}
The process $M$ is continuous. I want to show that $M$ is a local martingale by using the sequence $tau_n=inf{t: |M_t|>n}$. That is, I have to show that ${M_{tau_nwedge t}}$ is a martingale. I have read that one can use the above sequence of stopping times for continuous local martingales. But why or how?
brownian-motion martingales stochastic-integrals stochastic-analysis local-martingales
asked Nov 13 '18 at 16:48
Sofia
83
add a comment |
Suppose that $Y$ is a pure jump process with $N_t$ jumps in $(0,t]$ and $E[N_t]<infty$. Denote the jump times by $T_i$. Let $W$ be a Brownian motion. If $T_0=0$, then
begin{equation}
M_t=int_0^t Y_s,dW_s=sum_{i=0}^{N_t-1} Y_{T_i}(W_{T_{i+1}}-W_{T_i})+Y_t(W_t-W_{T_{N_t}}).
end{equation}
The process $M$ is continuous. I want to show that $M$ is a local martingale by using the sequence $tau_n=inf{t: |M_t|>n}$. That is, I have to show that ${M_{tau_nwedge t}}$ is a martingale. I have read that one can use the above sequence of stopping times for continuous local martingales. But why or how?
brownian-motion martingales stochastic-integrals stochastic-analysis local-martingales
asked Nov 13 '18 at 16:48
Sofia
83
add a comment |
Suppose that $Y$ is a pure jump process with $N_t$ jumps in $(0,t]$ and $E[N_t]<infty$. Denote the jump times by $T_i$. Let $W$ be a Brownian motion. If $T_0=0$, then
begin{equation}
M_t=int_0^t Y_s,dW_s=sum_{i=0}^{N_t-1} Y_{T_i}(W_{T_{i+1}}-W_{T_i})+Y_t(W_t-W_{T_{N_t}}).
end{equation}
The process $M$ is continuous. I want to show that $M$ is a local martingale by using the sequence $tau_n=inf{t: |M_t|>n}$. That is, I have to show that ${M_{tau_nwedge t}}$ is a martingale. I have read that one can use the above sequence of stopping times for continuous local martingales. But why or how?
brownian-motion martingales stochastic-integrals stochastic-analysis local-martingales
asked Nov 13 '18 at 16:48
Sofia
83
Suppose that $Y$ is a pure jump process with $N_t$ jumps in $(0,t]$ and $E[N_t]<infty$. Denote the jump times by $T_i$. Let $W$ be a Brownian motion. If $T_0=0$, then
begin{equation}
M_t=int_0^t Y_s,dW_s=sum_{i=0}^{N_t-1} Y_{T_i}(W_{T_{i+1}}-W_{T_i})+Y_t(W_t-W_{T_{N_t}}).
end{equation}
The process $M$ is continuous. I want to show that $M$ is a local martingale by using the sequence $tau_n=inf{t: |M_t|>n}$. That is, I have to show that ${M_{tau_nwedge t}}$ is a martingale. I have read that one can use the above sequence of stopping times for continuous local martingales. But why or how?
brownian-motion martingales stochastic-integrals stochastic-analysis local-martingales
brownian-motion martingales stochastic-integrals stochastic-analysis local-martingales
asked Nov 13 '18 at 16:48
Sofia
83
asked Nov 13 '18 at 16:48
Sofia
83
edited Nov 14 '18 at 10:10
asked Nov 13 '18 at 16:48
Sofia
83
asked Nov 13 '18 at 16:48
Sofia
83
asked Nov 13 '18 at 16:48
Sofia
83
83
add a comment |
add a comment |
1 Answer
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I guess you could use the sequence $tau_n=inf{t,:, N_t=n}$ and then Corollary 6.14 (or 7.14 depending on the edition, it is called "martingale transforms") from Kallenberg's Foundations of Modern Probability yields the result.
The corollary states that $Y_t(W_t-W_{T_{N_i}})$ and
$$sum_{i=0}^{n}Y_{T_i}(W_{T_{i+1}}-W_{T_i})$$
are martingales for all (fixed / deterministic) $nin mathbb N$.
But in your case, you need that
$$sum_{i=0}^{N_{t-1}}Y_{T_i}(W_{T_{i+1}}-W_{T_i})$$
is a local martingale for random $N_{t-1}$. This is why you need to stop the process.
With the stopping time suggested above, you obtain
$$M_t^{T^n}=sum_{i=0}^{n-1}Y_{T_i}(W^{T_{i+1}}_t-W^{T_i}_t)$$
which is a martingale.
Of course, you could also use the well known fact, that a stochastic integral with respect to continuous local martingale is always a local martingale. Then you still need to show, that stopping at $tau_n=inf{t,;,|M_t|>n}$ yields a uniformly integrable martingale. Because of the continuity of the process $M_t$, you get that $M_t^{tau_n}$ is bounded by $n$ and any bounded local martingale is a uniformaly integrable martingale. Hence the result follows.
answered Nov 14 '18 at 11:10
Agnetha Timara
986
This would imply that $M$ is even a martingale, not just a local martingale, which makes me sceptical about this answer.
– Sofia
Nov 26 '18 at 18:05
You need to stop the process $M$ in order to make the sum consist of a fix number of summands. Without stopping the number of summands is random and thus you can't apply the corollary from Kallenberg. Thus you have only shown that $M$ is a local martingale. I don't know whether $M$ might even be a martingale. I can imagine that you need more information about the distribution of $N$ to determine whether it is only a local martingale or even a true martingale.
– Agnetha Timara
Nov 26 '18 at 18:45
Then I didn't understand how you would use the corollary. Could you please explain that a bit more precisely?
– Sofia
Nov 27 '18 at 9:37
I edited my answer above.
– Agnetha Timara
Nov 27 '18 at 10:08
add a comment |
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I guess you could use the sequence $tau_n=inf{t,:, N_t=n}$ and then Corollary 6.14 (or 7.14 depending on the edition, it is called "martingale transforms") from Kallenberg's Foundations of Modern Probability yields the result.
The corollary states that $Y_t(W_t-W_{T_{N_i}})$ and
$$sum_{i=0}^{n}Y_{T_i}(W_{T_{i+1}}-W_{T_i})$$
are martingales for all (fixed / deterministic) $nin mathbb N$.
But in your case, you need that
$$sum_{i=0}^{N_{t-1}}Y_{T_i}(W_{T_{i+1}}-W_{T_i})$$
is a local martingale for random $N_{t-1}$. This is why you need to stop the process.
With the stopping time suggested above, you obtain
$$M_t^{T^n}=sum_{i=0}^{n-1}Y_{T_i}(W^{T_{i+1}}_t-W^{T_i}_t)$$
which is a martingale.
Of course, you could also use the well known fact, that a stochastic integral with respect to continuous local martingale is always a local martingale. Then you still need to show, that stopping at $tau_n=inf{t,;,|M_t|>n}$ yields a uniformly integrable martingale. Because of the continuity of the process $M_t$, you get that $M_t^{tau_n}$ is bounded by $n$ and any bounded local martingale is a uniformaly integrable martingale. Hence the result follows.
answered Nov 14 '18 at 11:10
Agnetha Timara
986
This would imply that $M$ is even a martingale, not just a local martingale, which makes me sceptical about this answer.
– Sofia
Nov 26 '18 at 18:05
You need to stop the process $M$ in order to make the sum consist of a fix number of summands. Without stopping the number of summands is random and thus you can't apply the corollary from Kallenberg. Thus you have only shown that $M$ is a local martingale. I don't know whether $M$ might even be a martingale. I can imagine that you need more information about the distribution of $N$ to determine whether it is only a local martingale or even a true martingale.
– Agnetha Timara
Nov 26 '18 at 18:45
Then I didn't understand how you would use the corollary. Could you please explain that a bit more precisely?
– Sofia
Nov 27 '18 at 9:37
I edited my answer above.
– Agnetha Timara
Nov 27 '18 at 10:08
add a comment |
I guess you could use the sequence $tau_n=inf{t,:, N_t=n}$ and then Corollary 6.14 (or 7.14 depending on the edition, it is called "martingale transforms") from Kallenberg's Foundations of Modern Probability yields the result.
The corollary states that $Y_t(W_t-W_{T_{N_i}})$ and
$$sum_{i=0}^{n}Y_{T_i}(W_{T_{i+1}}-W_{T_i})$$
are martingales for all (fixed / deterministic) $nin mathbb N$.
But in your case, you need that
$$sum_{i=0}^{N_{t-1}}Y_{T_i}(W_{T_{i+1}}-W_{T_i})$$
is a local martingale for random $N_{t-1}$. This is why you need to stop the process.
With the stopping time suggested above, you obtain
$$M_t^{T^n}=sum_{i=0}^{n-1}Y_{T_i}(W^{T_{i+1}}_t-W^{T_i}_t)$$
which is a martingale.
Of course, you could also use the well known fact, that a stochastic integral with respect to continuous local martingale is always a local martingale. Then you still need to show, that stopping at $tau_n=inf{t,;,|M_t|>n}$ yields a uniformly integrable martingale. Because of the continuity of the process $M_t$, you get that $M_t^{tau_n}$ is bounded by $n$ and any bounded local martingale is a uniformaly integrable martingale. Hence the result follows.
answered Nov 14 '18 at 11:10
Agnetha Timara
986
This would imply that $M$ is even a martingale, not just a local martingale, which makes me sceptical about this answer.
– Sofia
Nov 26 '18 at 18:05
You need to stop the process $M$ in order to make the sum consist of a fix number of summands. Without stopping the number of summands is random and thus you can't apply the corollary from Kallenberg. Thus you have only shown that $M$ is a local martingale. I don't know whether $M$ might even be a martingale. I can imagine that you need more information about the distribution of $N$ to determine whether it is only a local martingale or even a true martingale.
– Agnetha Timara
Nov 26 '18 at 18:45
Then I didn't understand how you would use the corollary. Could you please explain that a bit more precisely?
– Sofia
Nov 27 '18 at 9:37
I edited my answer above.
– Agnetha Timara
Nov 27 '18 at 10:08
add a comment |
I guess you could use the sequence $tau_n=inf{t,:, N_t=n}$ and then Corollary 6.14 (or 7.14 depending on the edition, it is called "martingale transforms") from Kallenberg's Foundations of Modern Probability yields the result.
The corollary states that $Y_t(W_t-W_{T_{N_i}})$ and
$$sum_{i=0}^{n}Y_{T_i}(W_{T_{i+1}}-W_{T_i})$$
are martingales for all (fixed / deterministic) $nin mathbb N$.
But in your case, you need that
$$sum_{i=0}^{N_{t-1}}Y_{T_i}(W_{T_{i+1}}-W_{T_i})$$
is a local martingale for random $N_{t-1}$. This is why you need to stop the process.
With the stopping time suggested above, you obtain
$$M_t^{T^n}=sum_{i=0}^{n-1}Y_{T_i}(W^{T_{i+1}}_t-W^{T_i}_t)$$
which is a martingale.
Of course, you could also use the well known fact, that a stochastic integral with respect to continuous local martingale is always a local martingale. Then you still need to show, that stopping at $tau_n=inf{t,;,|M_t|>n}$ yields a uniformly integrable martingale. Because of the continuity of the process $M_t$, you get that $M_t^{tau_n}$ is bounded by $n$ and any bounded local martingale is a uniformaly integrable martingale. Hence the result follows.
answered Nov 14 '18 at 11:10
Agnetha Timara
986
I guess you could use the sequence $tau_n=inf{t,:, N_t=n}$ and then Corollary 6.14 (or 7.14 depending on the edition, it is called "martingale transforms") from Kallenberg's Foundations of Modern Probability yields the result.
The corollary states that $Y_t(W_t-W_{T_{N_i}})$ and
$$sum_{i=0}^{n}Y_{T_i}(W_{T_{i+1}}-W_{T_i})$$
are martingales for all (fixed / deterministic) $nin mathbb N$.
But in your case, you need that
$$sum_{i=0}^{N_{t-1}}Y_{T_i}(W_{T_{i+1}}-W_{T_i})$$
is a local martingale for random $N_{t-1}$. This is why you need to stop the process.
With the stopping time suggested above, you obtain
$$M_t^{T^n}=sum_{i=0}^{n-1}Y_{T_i}(W^{T_{i+1}}_t-W^{T_i}_t)$$
which is a martingale.
Of course, you could also use the well known fact, that a stochastic integral with respect to continuous local martingale is always a local martingale. Then you still need to show, that stopping at $tau_n=inf{t,;,|M_t|>n}$ yields a uniformly integrable martingale. Because of the continuity of the process $M_t$, you get that $M_t^{tau_n}$ is bounded by $n$ and any bounded local martingale is a uniformaly integrable martingale. Hence the result follows.
answered Nov 14 '18 at 11:10
Agnetha Timara
986
edited Nov 27 '18 at 10:08
answered Nov 14 '18 at 11:10
Agnetha Timara
986
answered Nov 14 '18 at 11:10
Agnetha Timara
986
answered Nov 14 '18 at 11:10
Agnetha Timara
986
986
This would imply that $M$ is even a martingale, not just a local martingale, which makes me sceptical about this answer.
– Sofia
Nov 26 '18 at 18:05
You need to stop the process $M$ in order to make the sum consist of a fix number of summands. Without stopping the number of summands is random and thus you can't apply the corollary from Kallenberg. Thus you have only shown that $M$ is a local martingale. I don't know whether $M$ might even be a martingale. I can imagine that you need more information about the distribution of $N$ to determine whether it is only a local martingale or even a true martingale.
– Agnetha Timara
Nov 26 '18 at 18:45
Then I didn't understand how you would use the corollary. Could you please explain that a bit more precisely?
– Sofia
Nov 27 '18 at 9:37
I edited my answer above.
– Agnetha Timara
Nov 27 '18 at 10:08
add a comment |
This would imply that $M$ is even a martingale, not just a local martingale, which makes me sceptical about this answer.
– Sofia
Nov 26 '18 at 18:05
You need to stop the process $M$ in order to make the sum consist of a fix number of summands. Without stopping the number of summands is random and thus you can't apply the corollary from Kallenberg. Thus you have only shown that $M$ is a local martingale. I don't know whether $M$ might even be a martingale. I can imagine that you need more information about the distribution of $N$ to determine whether it is only a local martingale or even a true martingale.
– Agnetha Timara
Nov 26 '18 at 18:45
Then I didn't understand how you would use the corollary. Could you please explain that a bit more precisely?
– Sofia
Nov 27 '18 at 9:37
I edited my answer above.
– Agnetha Timara
Nov 27 '18 at 10:08
This would imply that $M$ is even a martingale, not just a local martingale, which makes me sceptical about this answer.
– Sofia
Nov 26 '18 at 18:05
This would imply that $M$ is even a martingale, not just a local martingale, which makes me sceptical about this answer.
– Sofia
Nov 26 '18 at 18:05
You need to stop the process $M$ in order to make the sum consist of a fix number of summands. Without stopping the number of summands is random and thus you can't apply the corollary from Kallenberg. Thus you have only shown that $M$ is a local martingale. I don't know whether $M$ might even be a martingale. I can imagine that you need more information about the distribution of $N$ to determine whether it is only a local martingale or even a true martingale.
– Agnetha Timara
Nov 26 '18 at 18:45
You need to stop the process $M$ in order to make the sum consist of a fix number of summands. Without stopping the number of summands is random and thus you can't apply the corollary from Kallenberg. Thus you have only shown that $M$ is a local martingale. I don't know whether $M$ might even be a martingale. I can imagine that you need more information about the distribution of $N$ to determine whether it is only a local martingale or even a true martingale.
– Agnetha Timara
Nov 26 '18 at 18:45
Then I didn't understand how you would use the corollary. Could you please explain that a bit more precisely?
– Sofia
Nov 27 '18 at 9:37
Then I didn't understand how you would use the corollary. Could you please explain that a bit more precisely?
– Sofia
Nov 27 '18 at 9:37
I edited my answer above.
– Agnetha Timara
Nov 27 '18 at 10:08
I edited my answer above.
– Agnetha Timara
Nov 27 '18 at 10:08
add a comment |
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