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bounded gambling systems (Theorem 4.2.8 in Durrett: Probability Theory and Examples)
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Theorem 4.2.8 in Durrett: Probability Theory and Examples states
Let $X_n, n geq 0$ be a supermartingale. If $H_n geq 0$, is predictable and each $H_n$ is bounded then $(H cdot X)_n := sum_{m=1}^{n}H_m(X_m-X_{m-1})$ is a supermartingale.
While I know an example showing that this statement does not hold when $H_n$ is not bounded, I cannot see where we use boundedness in the proof. The proof is based on this equation:
$$mathbb{E}[(Hcdot X)_{n+1} | mathcal{F}_n ] = (Hcdot X)_{n} + mathbb{E}[H_{n+1}(X_{n+1}-X_{n}) | mathcal{F}_n ] = (Hcdot X)_{n} + H_{n+1}mathbb{E}[(X_{n+1}-X_{n}) | mathcal{F}_n ] leq (Hcdot X)_{n}
$$
since $H_{n+1} geq 0$ and $mathbb{E}[(X_{n+1}-X_{n}) | mathcal{F}_n ] leq 0$.
Now the properties of conditional expectation we use in this equation (linearity and that we can factor $H_{n+1}$ out because it is $mathcal{F}_n$ measurable) do require $(Hcdot X)_{n+1}$ and $H_{n+1}$ to be integrable. Is boundedness just a way to guarantee that these two conditions hold, or is boundedness actually necessary?
I believe that boundedness must appear somewhere else since the gambling system where we double the stakes whenever we loose and stop playing when we have a net win of 1, has each $H_n$ and $Hcdot(X)_{n}$ integrable, even though the $H_n$ are not bounded.
probability martingales gambling
asked 19 hours ago
boukkoun
625
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Theorem 4.2.8 in Durrett: Probability Theory and Examples states
Let $X_n, n geq 0$ be a supermartingale. If $H_n geq 0$, is predictable and each $H_n$ is bounded then $(H cdot X)_n := sum_{m=1}^{n}H_m(X_m-X_{m-1})$ is a supermartingale.
While I know an example showing that this statement does not hold when $H_n$ is not bounded, I cannot see where we use boundedness in the proof. The proof is based on this equation:
$$mathbb{E}[(Hcdot X)_{n+1} | mathcal{F}_n ] = (Hcdot X)_{n} + mathbb{E}[H_{n+1}(X_{n+1}-X_{n}) | mathcal{F}_n ] = (Hcdot X)_{n} + H_{n+1}mathbb{E}[(X_{n+1}-X_{n}) | mathcal{F}_n ] leq (Hcdot X)_{n}
$$
since $H_{n+1} geq 0$ and $mathbb{E}[(X_{n+1}-X_{n}) | mathcal{F}_n ] leq 0$.
Now the properties of conditional expectation we use in this equation (linearity and that we can factor $H_{n+1}$ out because it is $mathcal{F}_n$ measurable) do require $(Hcdot X)_{n+1}$ and $H_{n+1}$ to be integrable. Is boundedness just a way to guarantee that these two conditions hold, or is boundedness actually necessary?
I believe that boundedness must appear somewhere else since the gambling system where we double the stakes whenever we loose and stop playing when we have a net win of 1, has each $H_n$ and $Hcdot(X)_{n}$ integrable, even though the $H_n$ are not bounded.
probability martingales gambling
asked 19 hours ago
boukkoun
625
1
It is just a strong condition to avoid integrability problems.
– RRL
16 hours ago
@boukkoun : I did not understand the nature of your counter-example. You are suggesting a case when everything is integrable and still there are problems pulling out what is known, can you give details on that case? Your gamble example does not say what $H_n$ and $X_n$ are.
– Michael
16 hours ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Theorem 4.2.8 in Durrett: Probability Theory and Examples states
Let $X_n, n geq 0$ be a supermartingale. If $H_n geq 0$, is predictable and each $H_n$ is bounded then $(H cdot X)_n := sum_{m=1}^{n}H_m(X_m-X_{m-1})$ is a supermartingale.
While I know an example showing that this statement does not hold when $H_n$ is not bounded, I cannot see where we use boundedness in the proof. The proof is based on this equation:
$$mathbb{E}[(Hcdot X)_{n+1} | mathcal{F}_n ] = (Hcdot X)_{n} + mathbb{E}[H_{n+1}(X_{n+1}-X_{n}) | mathcal{F}_n ] = (Hcdot X)_{n} + H_{n+1}mathbb{E}[(X_{n+1}-X_{n}) | mathcal{F}_n ] leq (Hcdot X)_{n}
$$
since $H_{n+1} geq 0$ and $mathbb{E}[(X_{n+1}-X_{n}) | mathcal{F}_n ] leq 0$.
Now the properties of conditional expectation we use in this equation (linearity and that we can factor $H_{n+1}$ out because it is $mathcal{F}_n$ measurable) do require $(Hcdot X)_{n+1}$ and $H_{n+1}$ to be integrable. Is boundedness just a way to guarantee that these two conditions hold, or is boundedness actually necessary?
I believe that boundedness must appear somewhere else since the gambling system where we double the stakes whenever we loose and stop playing when we have a net win of 1, has each $H_n$ and $Hcdot(X)_{n}$ integrable, even though the $H_n$ are not bounded.
probability martingales gambling
asked 19 hours ago
boukkoun
625
Theorem 4.2.8 in Durrett: Probability Theory and Examples states
Let $X_n, n geq 0$ be a supermartingale. If $H_n geq 0$, is predictable and each $H_n$ is bounded then $(H cdot X)_n := sum_{m=1}^{n}H_m(X_m-X_{m-1})$ is a supermartingale.
While I know an example showing that this statement does not hold when $H_n$ is not bounded, I cannot see where we use boundedness in the proof. The proof is based on this equation:
$$mathbb{E}[(Hcdot X)_{n+1} | mathcal{F}_n ] = (Hcdot X)_{n} + mathbb{E}[H_{n+1}(X_{n+1}-X_{n}) | mathcal{F}_n ] = (Hcdot X)_{n} + H_{n+1}mathbb{E}[(X_{n+1}-X_{n}) | mathcal{F}_n ] leq (Hcdot X)_{n}
$$
since $H_{n+1} geq 0$ and $mathbb{E}[(X_{n+1}-X_{n}) | mathcal{F}_n ] leq 0$.
Now the properties of conditional expectation we use in this equation (linearity and that we can factor $H_{n+1}$ out because it is $mathcal{F}_n$ measurable) do require $(Hcdot X)_{n+1}$ and $H_{n+1}$ to be integrable. Is boundedness just a way to guarantee that these two conditions hold, or is boundedness actually necessary?
I believe that boundedness must appear somewhere else since the gambling system where we double the stakes whenever we loose and stop playing when we have a net win of 1, has each $H_n$ and $Hcdot(X)_{n}$ integrable, even though the $H_n$ are not bounded.
probability martingales gambling
probability martingales gambling
asked 19 hours ago
boukkoun
625
asked 19 hours ago
boukkoun
625
asked 19 hours ago
boukkoun
625
asked 19 hours ago
boukkoun
625
asked 19 hours ago
boukkoun
625
625
1
It is just a strong condition to avoid integrability problems.
– RRL
16 hours ago
@boukkoun : I did not understand the nature of your counter-example. You are suggesting a case when everything is integrable and still there are problems pulling out what is known, can you give details on that case? Your gamble example does not say what $H_n$ and $X_n$ are.
– Michael
16 hours ago
add a comment |
1
It is just a strong condition to avoid integrability problems.
– RRL
16 hours ago
@boukkoun : I did not understand the nature of your counter-example. You are suggesting a case when everything is integrable and still there are problems pulling out what is known, can you give details on that case? Your gamble example does not say what $H_n$ and $X_n$ are.
– Michael
16 hours ago
1
1
It is just a strong condition to avoid integrability problems.
– RRL
16 hours ago
It is just a strong condition to avoid integrability problems.
– RRL
16 hours ago
@boukkoun : I did not understand the nature of your counter-example. You are suggesting a case when everything is integrable and still there are problems pulling out what is known, can you give details on that case? Your gamble example does not say what $H_n$ and $X_n$ are.
– Michael
16 hours ago
@boukkoun : I did not understand the nature of your counter-example. You are suggesting a case when everything is integrable and still there are problems pulling out what is known, can you give details on that case? Your gamble example does not say what $H_n$ and $X_n$ are.
– Michael
16 hours ago
add a comment |
1 Answer
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Since $X_n$ is an (integrable) supermartingale, boundedness of $H_n$ guarantees that $H_nX_n$ is integrable.
Note that $x^{-1/2}$ is integrable on $[0,1]$ with respect to Lebesgue measure , but $x^{-1/2} cdot x^{-1/2}$ is not.
answered 16 hours ago
RRL
46.6k42366
I think the crux of boukkon's question is: could we have imposed a weaker condition instead? I think the answer to that is yes, and that such a condition might have looked like, "Suppose $H_n$ is such that $mathbb E|H_n| < infty$ and $mathbb E|H_n X_n| < infty$." My best guess for why the theorem doesn't say that is that it's somewhat convoluted, and that the collection of cases for which that condition will be true and $H_n$ will also be unbounded is quite small in practice.
– Aaron Montgomery
6 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Since $X_n$ is an (integrable) supermartingale, boundedness of $H_n$ guarantees that $H_nX_n$ is integrable.
Note that $x^{-1/2}$ is integrable on $[0,1]$ with respect to Lebesgue measure , but $x^{-1/2} cdot x^{-1/2}$ is not.
answered 16 hours ago
RRL
46.6k42366
I think the crux of boukkon's question is: could we have imposed a weaker condition instead? I think the answer to that is yes, and that such a condition might have looked like, "Suppose $H_n$ is such that $mathbb E|H_n| < infty$ and $mathbb E|H_n X_n| < infty$." My best guess for why the theorem doesn't say that is that it's somewhat convoluted, and that the collection of cases for which that condition will be true and $H_n$ will also be unbounded is quite small in practice.
– Aaron Montgomery
6 hours ago
add a comment |
up vote
1
down vote
Since $X_n$ is an (integrable) supermartingale, boundedness of $H_n$ guarantees that $H_nX_n$ is integrable.
Note that $x^{-1/2}$ is integrable on $[0,1]$ with respect to Lebesgue measure , but $x^{-1/2} cdot x^{-1/2}$ is not.
answered 16 hours ago
RRL
46.6k42366
I think the crux of boukkon's question is: could we have imposed a weaker condition instead? I think the answer to that is yes, and that such a condition might have looked like, "Suppose $H_n$ is such that $mathbb E|H_n| < infty$ and $mathbb E|H_n X_n| < infty$." My best guess for why the theorem doesn't say that is that it's somewhat convoluted, and that the collection of cases for which that condition will be true and $H_n$ will also be unbounded is quite small in practice.
– Aaron Montgomery
6 hours ago
add a comment |
up vote
1
down vote
up vote
1
down vote
Since $X_n$ is an (integrable) supermartingale, boundedness of $H_n$ guarantees that $H_nX_n$ is integrable.
Note that $x^{-1/2}$ is integrable on $[0,1]$ with respect to Lebesgue measure , but $x^{-1/2} cdot x^{-1/2}$ is not.
answered 16 hours ago
RRL
46.6k42366
Since $X_n$ is an (integrable) supermartingale, boundedness of $H_n$ guarantees that $H_nX_n$ is integrable.
Note that $x^{-1/2}$ is integrable on $[0,1]$ with respect to Lebesgue measure , but $x^{-1/2} cdot x^{-1/2}$ is not.
answered 16 hours ago
RRL
46.6k42366
answered 16 hours ago
RRL
46.6k42366
answered 16 hours ago
RRL
46.6k42366
answered 16 hours ago
RRL
46.6k42366
46.6k42366
I think the crux of boukkon's question is: could we have imposed a weaker condition instead? I think the answer to that is yes, and that such a condition might have looked like, "Suppose $H_n$ is such that $mathbb E|H_n| < infty$ and $mathbb E|H_n X_n| < infty$." My best guess for why the theorem doesn't say that is that it's somewhat convoluted, and that the collection of cases for which that condition will be true and $H_n$ will also be unbounded is quite small in practice.
– Aaron Montgomery
6 hours ago
add a comment |
I think the crux of boukkon's question is: could we have imposed a weaker condition instead? I think the answer to that is yes, and that such a condition might have looked like, "Suppose $H_n$ is such that $mathbb E|H_n| < infty$ and $mathbb E|H_n X_n| < infty$." My best guess for why the theorem doesn't say that is that it's somewhat convoluted, and that the collection of cases for which that condition will be true and $H_n$ will also be unbounded is quite small in practice.
– Aaron Montgomery
6 hours ago
I think the crux of boukkon's question is: could we have imposed a weaker condition instead? I think the answer to that is yes, and that such a condition might have looked like, "Suppose $H_n$ is such that $mathbb E|H_n| < infty$ and $mathbb E|H_n X_n| < infty$." My best guess for why the theorem doesn't say that is that it's somewhat convoluted, and that the collection of cases for which that condition will be true and $H_n$ will also be unbounded is quite small in practice.
– Aaron Montgomery
6 hours ago
I think the crux of boukkon's question is: could we have imposed a weaker condition instead? I think the answer to that is yes, and that such a condition might have looked like, "Suppose $H_n$ is such that $mathbb E|H_n| < infty$ and $mathbb E|H_n X_n| < infty$." My best guess for why the theorem doesn't say that is that it's somewhat convoluted, and that the collection of cases for which that condition will be true and $H_n$ will also be unbounded is quite small in practice.
– Aaron Montgomery
6 hours ago
add a comment |
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It is just a strong condition to avoid integrability problems.
– RRL
16 hours ago
@boukkoun : I did not understand the nature of your counter-example. You are suggesting a case when everything is integrable and still there are problems pulling out what is known, can you give details on that case? Your gamble example does not say what $H_n$ and $X_n$ are.
– Michael
16 hours ago