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







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  • 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

















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1
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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







share|cite|improve this question


















  • 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















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







share|cite|improve this question













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






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asked 19 hours ago









boukkoun

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  • 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




    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












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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.






share|cite|improve this answer





















  • 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











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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.






share|cite|improve this answer





















  • 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















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.






share|cite|improve this answer





















  • 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













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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • 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


















 

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