Optional sampling theorem with a.s. finite stopping time The 2019 Stack Overflow Developer Survey Results Are InDoob's stopping time theorem with unbounded stopping timeDose “optional stopping theorem” imply “optional sampling theorem”?Proof of the optional sampling theoremDiscrete stochastic integral and optional sampling theoremQuestion on the stopping time theorems for martingales and UI martingales, is finite stopping time required?“Converse” of optional stopping theoremIs the hitting time of a Geometric Brownian Motion a.s. finite?Doob's Optional stopping time TheoremMartingale and Stopping Time with Finite Expectationprove that $S_tau land n to S_tau$ in $L^1$ for a random walk with $Etau^1/2 < infty$
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Optional sampling theorem with a.s. finite stopping time
The 2019 Stack Overflow Developer Survey Results Are InDoob's stopping time theorem with unbounded stopping timeDose “optional stopping theorem” imply “optional sampling theorem”?Proof of the optional sampling theoremDiscrete stochastic integral and optional sampling theoremQuestion on the stopping time theorems for martingales and UI martingales, is finite stopping time required?“Converse” of optional stopping theoremIs the hitting time of a Geometric Brownian Motion a.s. finite?Doob's Optional stopping time TheoremMartingale and Stopping Time with Finite Expectationprove that $S_tau land n to S_tau$ in $L^1$ for a random walk with $Etau^1/2 < infty$
$begingroup$
I'm trying to prove the following generalized version of Doob's optional sampling theorem:
Let $X$ be a square integrable martingale with respect to a filtration $mathbb F = left mathcal F_n right_n in mathbb N$ with square variation process $langle X rangle$. Let $tau$ be a finite stopping time, and suppose $mathbb Eleft[langle X rangle_tau right] < infty$. Then $$mathbb Eleft[ left(X_tau - X_0 right)^2 right] = mathbb Eleft[langle X rangle_tau right] quad textrmand quad mathbb Eleft[ X_tau right] = mathbb Eleft[X_0 right]$$
I know that $langle X rangle$ is the unique predictable process for which $left(X_n^2 - langle X rangle_nright)_n in mathbb N$ is a martingale, and it can be expressed in equations as $$langle X rangle = sum_i=1^n mathbb Eleft[ left(X_i - X_i-1right)^2 bigg| mathcal F_i-1right] quad textrmand quad mathbb Eleft[langle Xrangle_nright] = mathbfVarleft[X_n - X_0 right].$$
And I know that if $tau$ is bounded, then $mathbb Eleft[X_tauright] = mathbb Eleft[X_0right]$. So I know in particular that $mathbb Eleft[X_tau wedge Tright] = mathbb Eleft[X_0right]$ for every $T in mathbb N$. Clearly $X_tau wedge T mathbb 1_tau = N to X_N mathbb 1_tau = N$ as $T to infty$, and $$left|X_tau wedge T mathbb 1_tau = Nright| leq max_1 leq t leq N |X_t|,$$
so by dominated convergence,
$$
mathbb Eleft[ X_tau wedge T mathbb 1_tau = Nright] xrightarrowT to infty mathbb Eleft[ X_N mathbb 1_tau = Nright]
$$
for each $N in mathbb N$. I think, therefore, it follows that
$$
mathbb Eleft[X_0right] = lim_T to infty mathbb Eleft[X_tau wedge Tright] = lim_T to infty sum_N = 0^inftymathbb Eleft[X_tau wedge T mathbb 1_tau = N right] = sum_N = 0^inftymathbb Eleft[X_tau mathbb 1_tau = N right] = mathbb Eleft[X_tauright].
$$
But I'm not sure if passing this limit through the sum is valid. It requires dominated convergence, i.e. that the terms $mathbb Eleft[X_tau wedge T mathbb 1_tau = N right]$ are uniformly bounded by some function, but I don't know what that function is. Is there a better approach?
probability probability-theory martingales stopping-times
asked Mar 23 at 2:16
D FordD Ford
675313
$endgroup$
|
show 1 more comment
$begingroup$
I'm trying to prove the following generalized version of Doob's optional sampling theorem:
Let $X$ be a square integrable martingale with respect to a filtration $mathbb F = left mathcal F_n right_n in mathbb N$ with square variation process $langle X rangle$. Let $tau$ be a finite stopping time, and suppose $mathbb Eleft[langle X rangle_tau right] < infty$. Then $$mathbb Eleft[ left(X_tau - X_0 right)^2 right] = mathbb Eleft[langle X rangle_tau right] quad textrmand quad mathbb Eleft[ X_tau right] = mathbb Eleft[X_0 right]$$
I know that $langle X rangle$ is the unique predictable process for which $left(X_n^2 - langle X rangle_nright)_n in mathbb N$ is a martingale, and it can be expressed in equations as $$langle X rangle = sum_i=1^n mathbb Eleft[ left(X_i - X_i-1right)^2 bigg| mathcal F_i-1right] quad textrmand quad mathbb Eleft[langle Xrangle_nright] = mathbfVarleft[X_n - X_0 right].$$
And I know that if $tau$ is bounded, then $mathbb Eleft[X_tauright] = mathbb Eleft[X_0right]$. So I know in particular that $mathbb Eleft[X_tau wedge Tright] = mathbb Eleft[X_0right]$ for every $T in mathbb N$. Clearly $X_tau wedge T mathbb 1_tau = N to X_N mathbb 1_tau = N$ as $T to infty$, and $$left|X_tau wedge T mathbb 1_tau = Nright| leq max_1 leq t leq N |X_t|,$$
so by dominated convergence,
$$
mathbb Eleft[ X_tau wedge T mathbb 1_tau = Nright] xrightarrowT to infty mathbb Eleft[ X_N mathbb 1_tau = Nright]
$$
for each $N in mathbb N$. I think, therefore, it follows that
$$
mathbb Eleft[X_0right] = lim_T to infty mathbb Eleft[X_tau wedge Tright] = lim_T to infty sum_N = 0^inftymathbb Eleft[X_tau wedge T mathbb 1_tau = N right] = sum_N = 0^inftymathbb Eleft[X_tau mathbb 1_tau = N right] = mathbb Eleft[X_tauright].
$$
But I'm not sure if passing this limit through the sum is valid. It requires dominated convergence, i.e. that the terms $mathbb Eleft[X_tau wedge T mathbb 1_tau = N right]$ are uniformly bounded by some function, but I don't know what that function is. Is there a better approach?
probability probability-theory martingales stopping-times
asked Mar 23 at 2:16
D FordD Ford
675313
$endgroup$
$begingroup$
By Doob's inequality (together with a uniform integrability argument), if a martingale is bounded in $L^p$ for some $p>1$, then it converges in $L^p$. Apply this to the martingale $M_n:=X_tau wedge n$ with $p=2$ and you can obtain the first result. Then, note that convergence in $L^2$ implies convergence in $L^1$ and you can obtain the second result as well.
$endgroup$
– Shalop
Mar 23 at 4:13
$begingroup$
Sorry, the link in my above comment went to the wrong set of notes. I actually meant to link to Theorem 4.8 in this noteset. Very useful fact to keep in mind.
$endgroup$
– Shalop
Mar 23 at 4:24
$begingroup$
Okay, I may give that a shot. Although I'd prefer not to use Doob's inequality or martingale convergence theorems since those haven't shown up at this point in my reference text.
$endgroup$
– D Ford
Mar 23 at 4:26
1
$begingroup$
Ah, actually it is not so difficult to prove the convergence theorem directly when $p=2$. Specifically, let $M_n = X_tau wedge n$ again. From the given hypothesis $E[langle X rangle_tau]<infty$ you can basically see (using orthogonality of the increments of $M$) that $sum_n E[(M_n+1-M_n)^2]<infty$. But, if $a_n$ is a sequence of orthogonal vectors in a Hilbert space such that $sum |a_n|^2<infty$, then $a_n$ is a Cauchy sequence in that Hilbert space. From this, one sees that $M_n$ converges in $L^2$ (and therefore also in $L^1$) which will then easily yield the results.
$endgroup$
– Shalop
Mar 23 at 5:01
$begingroup$
By orthogonality of increments of $M$, do you mean $mathbb Eleft[left(M_n+1 - M_nright)left(M_n - M_n-1right)right] = 0$? I'm having a hard time seeing how this with the hypothesis $mathbb Eleft[langle Xrangle_tauright]<infty$ implies $sum_n mathbb Eleft[left(M_n+1-M_nright)^2right] < infty$. Also, if $sum||a_n||^2 < infty$, would it not be the case that $a_n to 0$? In which case we have $(M_n)$ is $L^2$-Cauchy, which I guess is your main point.
$endgroup$
– D Ford
Mar 23 at 15:05
|
show 1 more comment
$begingroup$
I'm trying to prove the following generalized version of Doob's optional sampling theorem:
Let $X$ be a square integrable martingale with respect to a filtration $mathbb F = left mathcal F_n right_n in mathbb N$ with square variation process $langle X rangle$. Let $tau$ be a finite stopping time, and suppose $mathbb Eleft[langle X rangle_tau right] < infty$. Then $$mathbb Eleft[ left(X_tau - X_0 right)^2 right] = mathbb Eleft[langle X rangle_tau right] quad textrmand quad mathbb Eleft[ X_tau right] = mathbb Eleft[X_0 right]$$
I know that $langle X rangle$ is the unique predictable process for which $left(X_n^2 - langle X rangle_nright)_n in mathbb N$ is a martingale, and it can be expressed in equations as $$langle X rangle = sum_i=1^n mathbb Eleft[ left(X_i - X_i-1right)^2 bigg| mathcal F_i-1right] quad textrmand quad mathbb Eleft[langle Xrangle_nright] = mathbfVarleft[X_n - X_0 right].$$
And I know that if $tau$ is bounded, then $mathbb Eleft[X_tauright] = mathbb Eleft[X_0right]$. So I know in particular that $mathbb Eleft[X_tau wedge Tright] = mathbb Eleft[X_0right]$ for every $T in mathbb N$. Clearly $X_tau wedge T mathbb 1_tau = N to X_N mathbb 1_tau = N$ as $T to infty$, and $$left|X_tau wedge T mathbb 1_tau = Nright| leq max_1 leq t leq N |X_t|,$$
so by dominated convergence,
$$
mathbb Eleft[ X_tau wedge T mathbb 1_tau = Nright] xrightarrowT to infty mathbb Eleft[ X_N mathbb 1_tau = Nright]
$$
for each $N in mathbb N$. I think, therefore, it follows that
$$
mathbb Eleft[X_0right] = lim_T to infty mathbb Eleft[X_tau wedge Tright] = lim_T to infty sum_N = 0^inftymathbb Eleft[X_tau wedge T mathbb 1_tau = N right] = sum_N = 0^inftymathbb Eleft[X_tau mathbb 1_tau = N right] = mathbb Eleft[X_tauright].
$$
But I'm not sure if passing this limit through the sum is valid. It requires dominated convergence, i.e. that the terms $mathbb Eleft[X_tau wedge T mathbb 1_tau = N right]$ are uniformly bounded by some function, but I don't know what that function is. Is there a better approach?
probability probability-theory martingales stopping-times
asked Mar 23 at 2:16
D FordD Ford
675313
$endgroup$
I'm trying to prove the following generalized version of Doob's optional sampling theorem:
Let $X$ be a square integrable martingale with respect to a filtration $mathbb F = left mathcal F_n right_n in mathbb N$ with square variation process $langle X rangle$. Let $tau$ be a finite stopping time, and suppose $mathbb Eleft[langle X rangle_tau right] < infty$. Then $$mathbb Eleft[ left(X_tau - X_0 right)^2 right] = mathbb Eleft[langle X rangle_tau right] quad textrmand quad mathbb Eleft[ X_tau right] = mathbb Eleft[X_0 right]$$
I know that $langle X rangle$ is the unique predictable process for which $left(X_n^2 - langle X rangle_nright)_n in mathbb N$ is a martingale, and it can be expressed in equations as $$langle X rangle = sum_i=1^n mathbb Eleft[ left(X_i - X_i-1right)^2 bigg| mathcal F_i-1right] quad textrmand quad mathbb Eleft[langle Xrangle_nright] = mathbfVarleft[X_n - X_0 right].$$
And I know that if $tau$ is bounded, then $mathbb Eleft[X_tauright] = mathbb Eleft[X_0right]$. So I know in particular that $mathbb Eleft[X_tau wedge Tright] = mathbb Eleft[X_0right]$ for every $T in mathbb N$. Clearly $X_tau wedge T mathbb 1_tau = N to X_N mathbb 1_tau = N$ as $T to infty$, and $$left|X_tau wedge T mathbb 1_tau = Nright| leq max_1 leq t leq N |X_t|,$$
so by dominated convergence,
$$
mathbb Eleft[ X_tau wedge T mathbb 1_tau = Nright] xrightarrowT to infty mathbb Eleft[ X_N mathbb 1_tau = Nright]
$$
for each $N in mathbb N$. I think, therefore, it follows that
$$
mathbb Eleft[X_0right] = lim_T to infty mathbb Eleft[X_tau wedge Tright] = lim_T to infty sum_N = 0^inftymathbb Eleft[X_tau wedge T mathbb 1_tau = N right] = sum_N = 0^inftymathbb Eleft[X_tau mathbb 1_tau = N right] = mathbb Eleft[X_tauright].
$$
But I'm not sure if passing this limit through the sum is valid. It requires dominated convergence, i.e. that the terms $mathbb Eleft[X_tau wedge T mathbb 1_tau = N right]$ are uniformly bounded by some function, but I don't know what that function is. Is there a better approach?
probability probability-theory martingales stopping-times
probability probability-theory martingales stopping-times
asked Mar 23 at 2:16
D FordD Ford
675313
asked Mar 23 at 2:16
D FordD Ford
675313
asked Mar 23 at 2:16
D FordD Ford
675313
asked Mar 23 at 2:16
D FordD Ford
675313
asked Mar 23 at 2:16
D FordD Ford
675313
675313
$begingroup$
By Doob's inequality (together with a uniform integrability argument), if a martingale is bounded in $L^p$ for some $p>1$, then it converges in $L^p$. Apply this to the martingale $M_n:=X_tau wedge n$ with $p=2$ and you can obtain the first result. Then, note that convergence in $L^2$ implies convergence in $L^1$ and you can obtain the second result as well.
$endgroup$
– Shalop
Mar 23 at 4:13
$begingroup$
Sorry, the link in my above comment went to the wrong set of notes. I actually meant to link to Theorem 4.8 in this noteset. Very useful fact to keep in mind.
$endgroup$
– Shalop
Mar 23 at 4:24
$begingroup$
Okay, I may give that a shot. Although I'd prefer not to use Doob's inequality or martingale convergence theorems since those haven't shown up at this point in my reference text.
$endgroup$
– D Ford
Mar 23 at 4:26
1
$begingroup$
Ah, actually it is not so difficult to prove the convergence theorem directly when $p=2$. Specifically, let $M_n = X_tau wedge n$ again. From the given hypothesis $E[langle X rangle_tau]<infty$ you can basically see (using orthogonality of the increments of $M$) that $sum_n E[(M_n+1-M_n)^2]<infty$. But, if $a_n$ is a sequence of orthogonal vectors in a Hilbert space such that $sum |a_n|^2<infty$, then $a_n$ is a Cauchy sequence in that Hilbert space. From this, one sees that $M_n$ converges in $L^2$ (and therefore also in $L^1$) which will then easily yield the results.
$endgroup$
– Shalop
Mar 23 at 5:01
$begingroup$
By orthogonality of increments of $M$, do you mean $mathbb Eleft[left(M_n+1 - M_nright)left(M_n - M_n-1right)right] = 0$? I'm having a hard time seeing how this with the hypothesis $mathbb Eleft[langle Xrangle_tauright]<infty$ implies $sum_n mathbb Eleft[left(M_n+1-M_nright)^2right] < infty$. Also, if $sum||a_n||^2 < infty$, would it not be the case that $a_n to 0$? In which case we have $(M_n)$ is $L^2$-Cauchy, which I guess is your main point.
$endgroup$
– D Ford
Mar 23 at 15:05
|
show 1 more comment
$begingroup$
By Doob's inequality (together with a uniform integrability argument), if a martingale is bounded in $L^p$ for some $p>1$, then it converges in $L^p$. Apply this to the martingale $M_n:=X_tau wedge n$ with $p=2$ and you can obtain the first result. Then, note that convergence in $L^2$ implies convergence in $L^1$ and you can obtain the second result as well.
$endgroup$
– Shalop
Mar 23 at 4:13
$begingroup$
Sorry, the link in my above comment went to the wrong set of notes. I actually meant to link to Theorem 4.8 in this noteset. Very useful fact to keep in mind.
$endgroup$
– Shalop
Mar 23 at 4:24
$begingroup$
Okay, I may give that a shot. Although I'd prefer not to use Doob's inequality or martingale convergence theorems since those haven't shown up at this point in my reference text.
$endgroup$
– D Ford
Mar 23 at 4:26
1
$begingroup$
Ah, actually it is not so difficult to prove the convergence theorem directly when $p=2$. Specifically, let $M_n = X_tau wedge n$ again. From the given hypothesis $E[langle X rangle_tau]<infty$ you can basically see (using orthogonality of the increments of $M$) that $sum_n E[(M_n+1-M_n)^2]<infty$. But, if $a_n$ is a sequence of orthogonal vectors in a Hilbert space such that $sum |a_n|^2<infty$, then $a_n$ is a Cauchy sequence in that Hilbert space. From this, one sees that $M_n$ converges in $L^2$ (and therefore also in $L^1$) which will then easily yield the results.
$endgroup$
– Shalop
Mar 23 at 5:01
$begingroup$
By orthogonality of increments of $M$, do you mean $mathbb Eleft[left(M_n+1 - M_nright)left(M_n - M_n-1right)right] = 0$? I'm having a hard time seeing how this with the hypothesis $mathbb Eleft[langle Xrangle_tauright]<infty$ implies $sum_n mathbb Eleft[left(M_n+1-M_nright)^2right] < infty$. Also, if $sum||a_n||^2 < infty$, would it not be the case that $a_n to 0$? In which case we have $(M_n)$ is $L^2$-Cauchy, which I guess is your main point.
$endgroup$
– D Ford
Mar 23 at 15:05
$begingroup$
By Doob's inequality (together with a uniform integrability argument), if a martingale is bounded in $L^p$ for some $p>1$, then it converges in $L^p$. Apply this to the martingale $M_n:=X_tau wedge n$ with $p=2$ and you can obtain the first result. Then, note that convergence in $L^2$ implies convergence in $L^1$ and you can obtain the second result as well.
$endgroup$
– Shalop
Mar 23 at 4:13
$begingroup$
By Doob's inequality (together with a uniform integrability argument), if a martingale is bounded in $L^p$ for some $p>1$, then it converges in $L^p$. Apply this to the martingale $M_n:=X_tau wedge n$ with $p=2$ and you can obtain the first result. Then, note that convergence in $L^2$ implies convergence in $L^1$ and you can obtain the second result as well.
$endgroup$
– Shalop
Mar 23 at 4:13
$begingroup$
Sorry, the link in my above comment went to the wrong set of notes. I actually meant to link to Theorem 4.8 in this noteset. Very useful fact to keep in mind.
$endgroup$
– Shalop
Mar 23 at 4:24
$begingroup$
Sorry, the link in my above comment went to the wrong set of notes. I actually meant to link to Theorem 4.8 in this noteset. Very useful fact to keep in mind.
$endgroup$
– Shalop
Mar 23 at 4:24
$begingroup$
Okay, I may give that a shot. Although I'd prefer not to use Doob's inequality or martingale convergence theorems since those haven't shown up at this point in my reference text.
$endgroup$
– D Ford
Mar 23 at 4:26
$begingroup$
Okay, I may give that a shot. Although I'd prefer not to use Doob's inequality or martingale convergence theorems since those haven't shown up at this point in my reference text.
$endgroup$
– D Ford
Mar 23 at 4:26
1
1
$begingroup$
Ah, actually it is not so difficult to prove the convergence theorem directly when $p=2$. Specifically, let $M_n = X_tau wedge n$ again. From the given hypothesis $E[langle X rangle_tau]<infty$ you can basically see (using orthogonality of the increments of $M$) that $sum_n E[(M_n+1-M_n)^2]<infty$. But, if $a_n$ is a sequence of orthogonal vectors in a Hilbert space such that $sum |a_n|^2<infty$, then $a_n$ is a Cauchy sequence in that Hilbert space. From this, one sees that $M_n$ converges in $L^2$ (and therefore also in $L^1$) which will then easily yield the results.
$endgroup$
– Shalop
Mar 23 at 5:01
$begingroup$
Ah, actually it is not so difficult to prove the convergence theorem directly when $p=2$. Specifically, let $M_n = X_tau wedge n$ again. From the given hypothesis $E[langle X rangle_tau]<infty$ you can basically see (using orthogonality of the increments of $M$) that $sum_n E[(M_n+1-M_n)^2]<infty$. But, if $a_n$ is a sequence of orthogonal vectors in a Hilbert space such that $sum |a_n|^2<infty$, then $a_n$ is a Cauchy sequence in that Hilbert space. From this, one sees that $M_n$ converges in $L^2$ (and therefore also in $L^1$) which will then easily yield the results.
$endgroup$
– Shalop
Mar 23 at 5:01
$begingroup$
By orthogonality of increments of $M$, do you mean $mathbb Eleft[left(M_n+1 - M_nright)left(M_n - M_n-1right)right] = 0$? I'm having a hard time seeing how this with the hypothesis $mathbb Eleft[langle Xrangle_tauright]<infty$ implies $sum_n mathbb Eleft[left(M_n+1-M_nright)^2right] < infty$. Also, if $sum||a_n||^2 < infty$, would it not be the case that $a_n to 0$? In which case we have $(M_n)$ is $L^2$-Cauchy, which I guess is your main point.
$endgroup$
– D Ford
Mar 23 at 15:05
$begingroup$
By orthogonality of increments of $M$, do you mean $mathbb Eleft[left(M_n+1 - M_nright)left(M_n - M_n-1right)right] = 0$? I'm having a hard time seeing how this with the hypothesis $mathbb Eleft[langle Xrangle_tauright]<infty$ implies $sum_n mathbb Eleft[left(M_n+1-M_nright)^2right] < infty$. Also, if $sum||a_n||^2 < infty$, would it not be the case that $a_n to 0$? In which case we have $(M_n)$ is $L^2$-Cauchy, which I guess is your main point.
$endgroup$
– D Ford
Mar 23 at 15:05
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Hints:
- Show that for any square-integrable martingale $(M_n,mathcalF_n)_n geq 1$ it holds that $$mathbbE(M_n M_m mid mathcalF_m) = M_m^2, qquad m leq n.$$ Conclude that $$mathbbE((M_n-M_m)^2) = mathbbE(M_n^2-M_m^2) = mathbbE(langle M rangle_n)- mathbbE(langle M rangle_m), qquad m leq n.$$
- Using Step 1 for the martingale $M_n := X_n wedge tau$ gives $$mathbbE((X_n wedge tau-X_m wedge tau)^2) = mathbbE(langle X rangle_tau wedge n - langle X rangle_tau wedge m), qquad m leq n.$$ Deduce from the monotonicity of $langle X rangle$ and the fact that $mathbbE(langle X rangle_tau)< infty$ that $$mathbbE((X_n wedge tau-X_m wedge tau)^2) xrightarrow[]m,n to infty 0.$$
- By the completeness of $L^2(mathbbP)$ it follows that $Z := lim_n to infty X_n wedge tau$ exists in $L^2(mathbbP)$. Show that $Z=X_tau$.
- Combine the identities $$mathbbE((X_n wedge tau-X_0)^2) = mathbbE(langle X rangle_tau) quad textand quad mathbbE(X_tau wedge n) = mathbbE(X_0)$$ with Step 3 to prove the assertion.
answered Mar 24 at 12:31
sazsaz
82.3k862131
$endgroup$
add a comment |
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$begingroup$
Hints:
- Show that for any square-integrable martingale $(M_n,mathcalF_n)_n geq 1$ it holds that $$mathbbE(M_n M_m mid mathcalF_m) = M_m^2, qquad m leq n.$$ Conclude that $$mathbbE((M_n-M_m)^2) = mathbbE(M_n^2-M_m^2) = mathbbE(langle M rangle_n)- mathbbE(langle M rangle_m), qquad m leq n.$$
- Using Step 1 for the martingale $M_n := X_n wedge tau$ gives $$mathbbE((X_n wedge tau-X_m wedge tau)^2) = mathbbE(langle X rangle_tau wedge n - langle X rangle_tau wedge m), qquad m leq n.$$ Deduce from the monotonicity of $langle X rangle$ and the fact that $mathbbE(langle X rangle_tau)< infty$ that $$mathbbE((X_n wedge tau-X_m wedge tau)^2) xrightarrow[]m,n to infty 0.$$
- By the completeness of $L^2(mathbbP)$ it follows that $Z := lim_n to infty X_n wedge tau$ exists in $L^2(mathbbP)$. Show that $Z=X_tau$.
- Combine the identities $$mathbbE((X_n wedge tau-X_0)^2) = mathbbE(langle X rangle_tau) quad textand quad mathbbE(X_tau wedge n) = mathbbE(X_0)$$ with Step 3 to prove the assertion.
answered Mar 24 at 12:31
sazsaz
82.3k862131
$endgroup$
add a comment |
$begingroup$
Hints:
- Show that for any square-integrable martingale $(M_n,mathcalF_n)_n geq 1$ it holds that $$mathbbE(M_n M_m mid mathcalF_m) = M_m^2, qquad m leq n.$$ Conclude that $$mathbbE((M_n-M_m)^2) = mathbbE(M_n^2-M_m^2) = mathbbE(langle M rangle_n)- mathbbE(langle M rangle_m), qquad m leq n.$$
- Using Step 1 for the martingale $M_n := X_n wedge tau$ gives $$mathbbE((X_n wedge tau-X_m wedge tau)^2) = mathbbE(langle X rangle_tau wedge n - langle X rangle_tau wedge m), qquad m leq n.$$ Deduce from the monotonicity of $langle X rangle$ and the fact that $mathbbE(langle X rangle_tau)< infty$ that $$mathbbE((X_n wedge tau-X_m wedge tau)^2) xrightarrow[]m,n to infty 0.$$
- By the completeness of $L^2(mathbbP)$ it follows that $Z := lim_n to infty X_n wedge tau$ exists in $L^2(mathbbP)$. Show that $Z=X_tau$.
- Combine the identities $$mathbbE((X_n wedge tau-X_0)^2) = mathbbE(langle X rangle_tau) quad textand quad mathbbE(X_tau wedge n) = mathbbE(X_0)$$ with Step 3 to prove the assertion.
answered Mar 24 at 12:31
sazsaz
82.3k862131
$endgroup$
add a comment |
$begingroup$
Hints:
- Show that for any square-integrable martingale $(M_n,mathcalF_n)_n geq 1$ it holds that $$mathbbE(M_n M_m mid mathcalF_m) = M_m^2, qquad m leq n.$$ Conclude that $$mathbbE((M_n-M_m)^2) = mathbbE(M_n^2-M_m^2) = mathbbE(langle M rangle_n)- mathbbE(langle M rangle_m), qquad m leq n.$$
- Using Step 1 for the martingale $M_n := X_n wedge tau$ gives $$mathbbE((X_n wedge tau-X_m wedge tau)^2) = mathbbE(langle X rangle_tau wedge n - langle X rangle_tau wedge m), qquad m leq n.$$ Deduce from the monotonicity of $langle X rangle$ and the fact that $mathbbE(langle X rangle_tau)< infty$ that $$mathbbE((X_n wedge tau-X_m wedge tau)^2) xrightarrow[]m,n to infty 0.$$
- By the completeness of $L^2(mathbbP)$ it follows that $Z := lim_n to infty X_n wedge tau$ exists in $L^2(mathbbP)$. Show that $Z=X_tau$.
- Combine the identities $$mathbbE((X_n wedge tau-X_0)^2) = mathbbE(langle X rangle_tau) quad textand quad mathbbE(X_tau wedge n) = mathbbE(X_0)$$ with Step 3 to prove the assertion.
answered Mar 24 at 12:31
sazsaz
82.3k862131
$endgroup$
Hints:
- Show that for any square-integrable martingale $(M_n,mathcalF_n)_n geq 1$ it holds that $$mathbbE(M_n M_m mid mathcalF_m) = M_m^2, qquad m leq n.$$ Conclude that $$mathbbE((M_n-M_m)^2) = mathbbE(M_n^2-M_m^2) = mathbbE(langle M rangle_n)- mathbbE(langle M rangle_m), qquad m leq n.$$
- Using Step 1 for the martingale $M_n := X_n wedge tau$ gives $$mathbbE((X_n wedge tau-X_m wedge tau)^2) = mathbbE(langle X rangle_tau wedge n - langle X rangle_tau wedge m), qquad m leq n.$$ Deduce from the monotonicity of $langle X rangle$ and the fact that $mathbbE(langle X rangle_tau)< infty$ that $$mathbbE((X_n wedge tau-X_m wedge tau)^2) xrightarrow[]m,n to infty 0.$$
- By the completeness of $L^2(mathbbP)$ it follows that $Z := lim_n to infty X_n wedge tau$ exists in $L^2(mathbbP)$. Show that $Z=X_tau$.
- Combine the identities $$mathbbE((X_n wedge tau-X_0)^2) = mathbbE(langle X rangle_tau) quad textand quad mathbbE(X_tau wedge n) = mathbbE(X_0)$$ with Step 3 to prove the assertion.
answered Mar 24 at 12:31
sazsaz
82.3k862131
edited Mar 27 at 10:38
answered Mar 24 at 12:31
sazsaz
82.3k862131
answered Mar 24 at 12:31
sazsaz
82.3k862131
answered Mar 24 at 12:31
sazsaz
82.3k862131
82.3k862131
add a comment |
add a comment |
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$begingroup$
By Doob's inequality (together with a uniform integrability argument), if a martingale is bounded in $L^p$ for some $p>1$, then it converges in $L^p$. Apply this to the martingale $M_n:=X_tau wedge n$ with $p=2$ and you can obtain the first result. Then, note that convergence in $L^2$ implies convergence in $L^1$ and you can obtain the second result as well.
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– Shalop
Mar 23 at 4:13
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Sorry, the link in my above comment went to the wrong set of notes. I actually meant to link to Theorem 4.8 in this noteset. Very useful fact to keep in mind.
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– Shalop
Mar 23 at 4:24
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Okay, I may give that a shot. Although I'd prefer not to use Doob's inequality or martingale convergence theorems since those haven't shown up at this point in my reference text.
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– D Ford
Mar 23 at 4:26
1
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Ah, actually it is not so difficult to prove the convergence theorem directly when $p=2$. Specifically, let $M_n = X_tau wedge n$ again. From the given hypothesis $E[langle X rangle_tau]<infty$ you can basically see (using orthogonality of the increments of $M$) that $sum_n E[(M_n+1-M_n)^2]<infty$. But, if $a_n$ is a sequence of orthogonal vectors in a Hilbert space such that $sum |a_n|^2<infty$, then $a_n$ is a Cauchy sequence in that Hilbert space. From this, one sees that $M_n$ converges in $L^2$ (and therefore also in $L^1$) which will then easily yield the results.
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– Shalop
Mar 23 at 5:01
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By orthogonality of increments of $M$, do you mean $mathbb Eleft[left(M_n+1 - M_nright)left(M_n - M_n-1right)right] = 0$? I'm having a hard time seeing how this with the hypothesis $mathbb Eleft[langle Xrangle_tauright]<infty$ implies $sum_n mathbb Eleft[left(M_n+1-M_nright)^2right] < infty$. Also, if $sum||a_n||^2 < infty$, would it not be the case that $a_n to 0$? In which case we have $(M_n)$ is $L^2$-Cauchy, which I guess is your main point.
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– D Ford
Mar 23 at 15:05