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: $$\mathbb{E}[(H\cdot X)_{n+1} | \mathcal{F}_n ] = (H\cdot X)_{n} + \mathbb{E}[H_{n+1}(X_{n+1}-X_{n}) | \mathcal{F}_n ] = (H\cdot X)_{n} + H_{n+1}\mathbb{E}[(X_{n+1}-X_{n}) | \mathcal{F}_n ] \leq (H\cdot X)_{n} $$ 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 $(H\cdot X)_{n+1}$ 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 start with one dollar, we double the stakes whenever we loose and stop playing once we have won our dollar back, has each $H_n$ and $(H \cdot X)_{n}$ integrable, even though the $H_n$ are not bounded.

The example I mean comes from Durrett (slightly modified, see comment at the end):

Let $X_n = \sum_{i=1}^{n} \chi_i$ where $\chi_i = 1$ with probability $p$ and $-1$ with probability $1-p$. Let $H_n = 2H_{n-1}$ if $X_{n-1} = - 1$, $H_n = 0$ if $X_{n-1} = 1$. Then $\mathbb{P}(H_n = 2^k) = p^k$ so $H_n$ is unbounded, but for each $n$, $H_n$ and $(H \cdot X)_{n}$ are integrable. Now since $\mathbb{P}(H_n < \infty)=1$, $\mathbb{E}[ \lim_{n\to \infty} (H \cdot X)_{n}]=1$, i.e. we are sure to win our dollar back. So eventhough we started with a strict supermartingale (if $p<\frac{1}{2}$) we ended up not loosing money, i.e. we kind of cheated the system. (I modified the example a little by setting $H_n=0$ after we win so that we do end up really winning the dollar back. In Durrett's example we start again betting 1 dollar once we have a net gain of 1, i.e. $H_n = 2H_{n-1}$ if $X_{n-1} = - 1$, $H_n = 1$ if $X_{n-1} = 1$.)

I now realize that this does not contradict the theorem. I was led to think that because after stating this example Durrett says "This system seems to provide us with a "sure thing" as long as $\mathbb{P}(\chi_m=1)>0$. However the next results says there is no system for beating an unfavorable game." So my new question is: can a gambling system prevent us from loosing while playing a strict supermartingale?

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    $\begingroup$ It is just a strong condition to avoid integrability problems. $\endgroup$ – RRL Nov 19 '18 at 6:00
  • $\begingroup$ @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. $\endgroup$ – Michael Nov 19 '18 at 6:06
  • $\begingroup$ @Michael: I added the example, and needed to modify my question subsequently. Sorry about that. I still think that I got confused somewhere and would be really happy if you could clear that up! $\endgroup$ – boukkoun Nov 21 '18 at 6:57
  • $\begingroup$ @boukkoun : I couldn't follow your given example. What if $X_{n-1}=2$? Or $-2$? Perhaps you are mixing $X_n$ and $\chi_n$ in one or more places. I suspect your example can be simplified by removing $X_n$ altogether. Intuitively I think you are just observing that double-or-nothing will almost surely (eventually) win. However, it requires an unbounded amount of money already in your account, an unbounded amount of time, it only wins one dollar, and risks a lot. If you already have an infinite number of dollars, there is no reason to try to win one more dollar. $\endgroup$ – Michael Nov 21 '18 at 17:29
  • $\begingroup$ If you repeat the process by starting over when you win (perhaps Durrett's original example) and sample at times you win, the winnings goes to $\infty$ (so the $\limsup$ of winnings is $\infty$). However if the win probability is $\leq 1/2$ it can be shown the $\liminf$ of winnings is $-\infty$ and you are often in (huge) debt. $\endgroup$ – Michael Nov 21 '18 at 17:47

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.

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    $\begingroup$ 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. $\endgroup$ – Aaron Montgomery Nov 19 '18 at 16:08

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