# Optional sampling theorem St. Petersburg paradox

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.

• 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). – saz May 22 at 15:37
• Why not just take $\sigma = 1$? – 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 $$\mathbb{E}(X_{\tau} \mid \mathcal{F}_{\sigma}) \neq X_{\sigma}.$$