I want to show that the optional sampling theorem does not hold for unbounded stopping times using the example of the St. Petersburg paradox/St. Petersburg game. We have a consecutive (fair) coin toss and play until we win for the first time. In round one you bet $1$ unit of money. You lose it if you lose the game and you keep it if you win the game. In the next rounds the stakes are always doubled.

I want to consider the total win/loss process which will be a martingale, namely let $(\xi_k)_{k\in\mathbb{N}}$ with $\xi_k\in\{-1,1\}$ an i.i.d. sequence of random variables corresponding to losing/winning a round. The bet amount will be $(b_k)_{k\in\mathbb{N}}$ with $b_1=1$ and $b_k=2^{k-1}$. Then the total process is $X_k=\sum_{j=1}^k b_k \xi_k$. One can easily check that this is a martingale (with respect to $\mathcal{F}_k^X=\sigma(\{X_1,...,X_k\})$).

The optional sampling theorem says that for two bounded $\mathcal{F}^X$-stopping times $\sigma\le\tau$ one has $\mathbb{E}(X_{\tau}|\mathcal{F}_{\sigma}^X)=X_{\sigma}$.

Now I want to use the time of the first win $\tau=\inf\{k\in\mathbb{N}|\xi_k=1\}$ (which is an unbounded $\mathcal{F}^X$-stopping time) to show that the optional sampling theorem does not hold.

The problem is that I don't know how to choose my second stopping time. While $\sigma=\tau-1$ would work nicely, it is not allowed because it is not a $\mathcal{F}^X$-stopping time. I have tried numerous possibilities for $\sigma$ but it does not work.

The general idea however is to calculate $$\mathbb{E}(X_{\tau}|\mathcal{F}_{\sigma}^X)=\sum_{k\in\mathbb{N}} \mathbb{P}(\tau=k)\mathbb{E}(X_{k}|\mathcal{F}_{\sigma}^X)= \sum_{k\in\mathbb{N}} 2^{-k}\mathbb{E}(X_{k}|\mathcal{F}_{\sigma}^X)$$ and then using the fact that $(X_k)_{k\in\mathbb{N}}$ is a martingale so $\mathbb{E}(X_{k}|\mathcal{F}_{s}^X)=X_s$ if $s\le t$. After that I would plug in the definition of $X_k$ and use the fact that $\xi_j=-1$ for all $0\le j\le k-1$ since we assumed $\tau=k$ in that case.

  • $\begingroup$ Moreover, your general idea for the computation of the conditional expectation does not work (the first "=" is wrong; you cannot pull out the probability like this). $\endgroup$ – saz May 22 at 15:37
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    $\begingroup$ Why not just take $\sigma = 1$? $\endgroup$ – John Dawkins May 22 at 17:15

If $\tau(\omega)=k$, then you have lost the first $(k-1)$ rounds and you have won the $k$-th round, i.e.

$$X_{\tau}(\omega)= -\sum_{j=1}^{k-1} 2^{j-1} + 2^{k-1} = (1-2^{k-1}) + 2^{k-1}=1 $$

for any such $\omega$. Since this holds for arbitrary $k \geq 1$, this shows $X_{\tau}=1$ almost surely. In particular, $\mathbb{E}(X_{\tau})=1$.

Now set $\sigma:=1$, then $\sigma$ is a stopping time satisfying $\sigma \leq \tau$ and $\mathbb{E}(X_{\sigma})=0$. In particular, $$\mathbb{E}(X_{\tau}) \neq \mathbb{E}(X_{\sigma})$$ implying $$\mathbb{E}(X_{\tau} \mid \mathcal{F}_{\sigma}) \neq X_{\sigma}.$$

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    $\begingroup$ I don't understand, why it would not make sense. We play the game until we win for the first time. If we lose in the first round, we play on until we win. In this setup the time of the first loss must be always 1, if not you win the first round and the game will be over. Maybe I was a bit unclear. We bet 1 unit of money in the first round. If we win, we get the money back and the game is over and if we lose, we will lose the money and play the next round betting twice as much money as in the round before. $\endgroup$ – mathemagician99 May 22 at 16:53
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    $\begingroup$ @mathemagician99 Thanks for the clarification. I edited my answer accordingly. $\endgroup$ – saz May 22 at 17:53

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