Almost sure convergence of a martingale sum
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Consider the senquence of iid r.v. $(Y_k)_{kgeq1}$ such that $mathbb{P}(Y_k=1)=mathbb{P}(Y_k=-1)=frac{1}{2}$ and then consider the process $X=(X_k)_{ngeq1}$ such that $X_n=sum_{k=1}^nfrac{Y_k}{k}$. It's very easy to see that $X$ is a martingale w.r.t. the filtration that it generate itself. However i tried proving that it is almost surely convergent using the convergence theorem for martingales but i failed. Is this really almost surely convergent? if yes is possible to prove this using martingales theory or is better to use other techniques like Borel-Cantelli lemmas?
probability martingales almost-everywhere
asked Jan 7 at 12:12
StabiloBossStabiloBoss
1016
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add a comment |
$begingroup$
Consider the senquence of iid r.v. $(Y_k)_{kgeq1}$ such that $mathbb{P}(Y_k=1)=mathbb{P}(Y_k=-1)=frac{1}{2}$ and then consider the process $X=(X_k)_{ngeq1}$ such that $X_n=sum_{k=1}^nfrac{Y_k}{k}$. It's very easy to see that $X$ is a martingale w.r.t. the filtration that it generate itself. However i tried proving that it is almost surely convergent using the convergence theorem for martingales but i failed. Is this really almost surely convergent? if yes is possible to prove this using martingales theory or is better to use other techniques like Borel-Cantelli lemmas?
probability martingales almost-everywhere
asked Jan 7 at 12:12
StabiloBossStabiloBoss
1016
$endgroup$
add a comment |
$begingroup$
Consider the senquence of iid r.v. $(Y_k)_{kgeq1}$ such that $mathbb{P}(Y_k=1)=mathbb{P}(Y_k=-1)=frac{1}{2}$ and then consider the process $X=(X_k)_{ngeq1}$ such that $X_n=sum_{k=1}^nfrac{Y_k}{k}$. It's very easy to see that $X$ is a martingale w.r.t. the filtration that it generate itself. However i tried proving that it is almost surely convergent using the convergence theorem for martingales but i failed. Is this really almost surely convergent? if yes is possible to prove this using martingales theory or is better to use other techniques like Borel-Cantelli lemmas?
probability martingales almost-everywhere
asked Jan 7 at 12:12
StabiloBossStabiloBoss
1016
$endgroup$
Consider the senquence of iid r.v. $(Y_k)_{kgeq1}$ such that $mathbb{P}(Y_k=1)=mathbb{P}(Y_k=-1)=frac{1}{2}$ and then consider the process $X=(X_k)_{ngeq1}$ such that $X_n=sum_{k=1}^nfrac{Y_k}{k}$. It's very easy to see that $X$ is a martingale w.r.t. the filtration that it generate itself. However i tried proving that it is almost surely convergent using the convergence theorem for martingales but i failed. Is this really almost surely convergent? if yes is possible to prove this using martingales theory or is better to use other techniques like Borel-Cantelli lemmas?
probability martingales almost-everywhere
probability martingales almost-everywhere
asked Jan 7 at 12:12
StabiloBossStabiloBoss
1016
asked Jan 7 at 12:12
StabiloBossStabiloBoss
1016
edited Jan 7 at 12:21
StabiloBoss
asked Jan 7 at 12:12
StabiloBossStabiloBoss
1016
asked Jan 7 at 12:12
StabiloBossStabiloBoss
1016
asked Jan 7 at 12:12
StabiloBossStabiloBoss
1016
1016
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
One can see that
$$
mathbb Eleft[X_n^2right]=sum_{k=1}^nfrac 1{k^2}mathbb Eleft[Y_k^2right]
$$
hence that the martingale $left(X_nright)_{ngeqslant 1}$ is bounded in $mathbb L^2$.
answered Jan 7 at 12:28
Davide GiraudoDavide Giraudo
128k17156268
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add a comment |
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$EX_n^{2}=sum_{k=1}^{n} var (frac {Y_k} k)=sum_{k=1}^{n} frac 1 {k^{2}}$ which is bounded. It is well known that this implies uniform integrability and any uniformly integrable martingale converges almost surely.
answered Jan 7 at 12:15
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
$begingroup$
Sorry I forgot to divide se term in the sum please check the edited question
$endgroup$
– StabiloBoss
Jan 7 at 12:22
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@StabiloBoss I have answered the new version of your question also.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 12:28
add a comment |
$begingroup$
For a « martingale » proof, the convergence theorem for martingales states that if $(X_n)$ is a martingale wrt some filtration, and $mathbb{E}[|X_n|]$ is bounded, then $X_n$ converges pointwise.
Now note that $mathbb{E}[|X_n|^2]$ is clearly bounded and is not lower than $mathbb{E}[|X_n|]^2$.
answered Jan 7 at 12:19
MindlackMindlack
4,910211
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One can see that
$$
mathbb Eleft[X_n^2right]=sum_{k=1}^nfrac 1{k^2}mathbb Eleft[Y_k^2right]
$$
hence that the martingale $left(X_nright)_{ngeqslant 1}$ is bounded in $mathbb L^2$.
answered Jan 7 at 12:28
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
add a comment |
$begingroup$
One can see that
$$
mathbb Eleft[X_n^2right]=sum_{k=1}^nfrac 1{k^2}mathbb Eleft[Y_k^2right]
$$
hence that the martingale $left(X_nright)_{ngeqslant 1}$ is bounded in $mathbb L^2$.
answered Jan 7 at 12:28
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
add a comment |
$begingroup$
One can see that
$$
mathbb Eleft[X_n^2right]=sum_{k=1}^nfrac 1{k^2}mathbb Eleft[Y_k^2right]
$$
hence that the martingale $left(X_nright)_{ngeqslant 1}$ is bounded in $mathbb L^2$.
answered Jan 7 at 12:28
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
One can see that
$$
mathbb Eleft[X_n^2right]=sum_{k=1}^nfrac 1{k^2}mathbb Eleft[Y_k^2right]
$$
hence that the martingale $left(X_nright)_{ngeqslant 1}$ is bounded in $mathbb L^2$.
answered Jan 7 at 12:28
Davide GiraudoDavide Giraudo
128k17156268
answered Jan 7 at 12:28
Davide GiraudoDavide Giraudo
128k17156268
answered Jan 7 at 12:28
Davide GiraudoDavide Giraudo
128k17156268
answered Jan 7 at 12:28
Davide GiraudoDavide Giraudo
128k17156268
128k17156268
add a comment |
add a comment |
$begingroup$
$EX_n^{2}=sum_{k=1}^{n} var (frac {Y_k} k)=sum_{k=1}^{n} frac 1 {k^{2}}$ which is bounded. It is well known that this implies uniform integrability and any uniformly integrable martingale converges almost surely.
answered Jan 7 at 12:15
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
$begingroup$
Sorry I forgot to divide se term in the sum please check the edited question
$endgroup$
– StabiloBoss
Jan 7 at 12:22
$begingroup$
@StabiloBoss I have answered the new version of your question also.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 12:28
add a comment |
$begingroup$
$EX_n^{2}=sum_{k=1}^{n} var (frac {Y_k} k)=sum_{k=1}^{n} frac 1 {k^{2}}$ which is bounded. It is well known that this implies uniform integrability and any uniformly integrable martingale converges almost surely.
answered Jan 7 at 12:15
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
$begingroup$
Sorry I forgot to divide se term in the sum please check the edited question
$endgroup$
– StabiloBoss
Jan 7 at 12:22
$begingroup$
@StabiloBoss I have answered the new version of your question also.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 12:28
add a comment |
$begingroup$
$EX_n^{2}=sum_{k=1}^{n} var (frac {Y_k} k)=sum_{k=1}^{n} frac 1 {k^{2}}$ which is bounded. It is well known that this implies uniform integrability and any uniformly integrable martingale converges almost surely.
answered Jan 7 at 12:15
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
$endgroup$
$EX_n^{2}=sum_{k=1}^{n} var (frac {Y_k} k)=sum_{k=1}^{n} frac 1 {k^{2}}$ which is bounded. It is well known that this implies uniform integrability and any uniformly integrable martingale converges almost surely.
answered Jan 7 at 12:15
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
edited Jan 7 at 12:33
answered Jan 7 at 12:15
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
answered Jan 7 at 12:15
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
answered Jan 7 at 12:15
Kavi Rama MurthyKavi Rama Murthy
74.6k53270
74.6k53270
$begingroup$
Sorry I forgot to divide se term in the sum please check the edited question
$endgroup$
– StabiloBoss
Jan 7 at 12:22
$begingroup$
@StabiloBoss I have answered the new version of your question also.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 12:28
add a comment |
$begingroup$
Sorry I forgot to divide se term in the sum please check the edited question
$endgroup$
– StabiloBoss
Jan 7 at 12:22
$begingroup$
@StabiloBoss I have answered the new version of your question also.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 12:28
$begingroup$
Sorry I forgot to divide se term in the sum please check the edited question
$endgroup$
– StabiloBoss
Jan 7 at 12:22
$begingroup$
Sorry I forgot to divide se term in the sum please check the edited question
$endgroup$
– StabiloBoss
Jan 7 at 12:22
$begingroup$
@StabiloBoss I have answered the new version of your question also.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 12:28
$begingroup$
@StabiloBoss I have answered the new version of your question also.
$endgroup$
– Kavi Rama Murthy
Jan 7 at 12:28
add a comment |
$begingroup$
For a « martingale » proof, the convergence theorem for martingales states that if $(X_n)$ is a martingale wrt some filtration, and $mathbb{E}[|X_n|]$ is bounded, then $X_n$ converges pointwise.
Now note that $mathbb{E}[|X_n|^2]$ is clearly bounded and is not lower than $mathbb{E}[|X_n|]^2$.
answered Jan 7 at 12:19
MindlackMindlack
4,910211
$endgroup$
add a comment |
$begingroup$
For a « martingale » proof, the convergence theorem for martingales states that if $(X_n)$ is a martingale wrt some filtration, and $mathbb{E}[|X_n|]$ is bounded, then $X_n$ converges pointwise.
Now note that $mathbb{E}[|X_n|^2]$ is clearly bounded and is not lower than $mathbb{E}[|X_n|]^2$.
answered Jan 7 at 12:19
MindlackMindlack
4,910211
$endgroup$
add a comment |
$begingroup$
For a « martingale » proof, the convergence theorem for martingales states that if $(X_n)$ is a martingale wrt some filtration, and $mathbb{E}[|X_n|]$ is bounded, then $X_n$ converges pointwise.
Now note that $mathbb{E}[|X_n|^2]$ is clearly bounded and is not lower than $mathbb{E}[|X_n|]^2$.
answered Jan 7 at 12:19
MindlackMindlack
4,910211
$endgroup$
For a « martingale » proof, the convergence theorem for martingales states that if $(X_n)$ is a martingale wrt some filtration, and $mathbb{E}[|X_n|]$ is bounded, then $X_n$ converges pointwise.
Now note that $mathbb{E}[|X_n|^2]$ is clearly bounded and is not lower than $mathbb{E}[|X_n|]^2$.
answered Jan 7 at 12:19
MindlackMindlack
4,910211
edited Jan 7 at 12:29
answered Jan 7 at 12:19
MindlackMindlack
4,910211
answered Jan 7 at 12:19
MindlackMindlack
4,910211
answered Jan 7 at 12:19
MindlackMindlack
4,910211
4,910211
add a comment |
add a comment |
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