Difference between $\mathcal F_\tau$ and $\sigma(\tau)$

suppose we have a probability space $(\Omega,\mathcal{F},\mathbb{P})$ with a filtration $(\mathcal{F}_n)_{n \in \mathbb N}$. A stopping time is random variable $\tau$ such that $$\{\tau \leq n\} \in \mathcal F_n, \quad n \in \mathbb N.$$ For a given stopping time $\tau$ we can define a $\sigma$-algebra as $$\mathcal{F}_\tau= \big\{A \in \mathcal F\mid A \cap \{\tau \leq n\} \in \mathcal F_n, \ n \in \mathbb{N} \big\}.$$

The usual interpretation is that for $n \in \mathbb{N}, \$ $\mathcal F_n$ contains all the information up to time $n$ and $\mathcal F_\tau$ contains all the information up to time $\tau$.

My question is: what is the intuitive difference between $\mathcal F_\tau$ and $\sigma(\tau)$, i.e. the $\sigma$-algebra generated by $\tau$. For example, if $X=(X_n)_{n =\{0,\dots, N\}}$ denotes a gamblers stack and $\tau$ is a rule which the gambler uses to end the game, what is the difference between $$\mathbb{E}[X_N\mid \mathcal F_\tau ] \quad \text{and} \quad \mathbb{E}[X_N\mid \tau ]?$$ Thank you

As you say, $\mathcal{F}_\tau$ contains all the information up to the stopping time, while $\sigma(\tau)$ only contains information that is relevant to the stopping.

For example, take $X_1,X_2,\ldots,X_N$ iid with continuous distribution, $\tau=\inf\{i:X_{i-1}>0,X_i>0\}\wedge N$, and $S_k=\sum_{i=1}^{k\wedge \tau}X_i$. That is, a player plays the same game until he gets a profit twice in a row. Then the total result of the game, $S_N$, actually equals $S_\tau$, and so it is $\mathcal{F}_\tau$-measurable. However, it is far from $\sigma(\tau)$-measurable, in fact knowing the value of $\tau$ doesn't even tell you the sign of $S_N$, that is, whether the player made a total profit or loss (unless of course $\tau=2$).

• thank you for your response: could you elaborate more on the intuition of the difference of the two $\sigma$-algebras. For example in this case $\mathbb E [S_N \mid \mathcal F_\tau]$ is the expected profit after I stop playing. But what is $\mathbb E[S_N \mid \tau]$? – Cettt Jan 2 '17 at 14:36
• If you're looking for the difference in the conditional expectations, then the extreme example $\tau\equiv N$ may be more illustrative: Then $\mathbb{E}(S_N|\mathcal{F}_\tau)=S_N$ as before, but $\mathbb{E}(S_N|\sigma(\tau))=\mathbb{E}S_N$, which is just a constant. That is of course due the the fact that there isn't any `information that is relevant to the stopping'. – m7e Jan 2 '17 at 18:14

This is just a little more detailed explanation which I derived from m7e's answer. I thought about his/her answer a lot and came up whit this explanation which I personally like a lot:

• $\sigma(\tau)$ contains all the information on when $\tau$ has happened
• $\mathcal F_\tau$ contains all the information on when and how $\tau$ has happened.

To give a little bit more insight consider this slight modification on m7e's example: let $(X_n)_{n \in \{1,\dots,4\}}$ be a an iid sequence such that $\mathbb P(X_1=1)=\mathbb P(X_1=-1)= \frac 12$ and define $S_n= \sum_{k=1}^n X_k$. Intuitively, we can think about this as a game where each round a fair coin is tossed and I either win $1\$$(if X_n=1) or lose 1\$$ (if$X_n=-1$). Also I define a stopping time$\tau$as $$\tau:= \inf \{n \geq 2\mid X_{n-1}=1 \ \text{and} \ X_n=1\} \wedge 5.$$ Following the stopping time$\tau$means that I stop playing after I win two times in a row or at the end of the fifth round at latest. For the following I will be using the following (simple but sloppy) notation, where$W$is short for win and$Lis short for lose: \begin{align} A_{WL}&= \{\omega \in \Omega \mid X_1(\omega)=1 \ \text{and} \ X_2(\omega)=-1 \}, \quad \text{or}\\ A_{LLW}&=\{\omega \in \Omega \mid X_1(\omega)=X_2(\omega)=-1 \ \text{and} \ X_3(\omega)=1 \}, \quad \text{etc} \end{align}. Then the\sigma$-algebra generated by$\tau$is generated by the following sets: $$\sigma(\tau)=\sigma \Bigl\{A_{WW}, \ A_{LWW},\ A_{WLWW}\cup A_{LLWW} \Bigr\}.$$ Here, it is important to note that neither of the sets$A_{WLWW}$and$A_{LLWW}$are contained in$\sigma(\tau)$on there own. Put differently, given the information contained in$\sigma(\tau)$I can distinguish between the events$\{\tau=2\}$(which corresponds to$A_{WW}$),$\{\tau=3\}$(which corresponds to$A_{LWW}$) and$\{\tau=4 \}$(which corresponds to$A_{WLWW}\cup A_{LLWW}$) but I cannot distinguish between$A_{WLWW}$and$A_{LLWW}.$This means that$\sigma(\tau)$tells me when I stopped playing the game, but not the history of the game which has led to the event of me stopping. In particular, if I know that I stopped after four rounds, I do not know what happened in the first two rounds. To sum up$\sigma(\tau)$contains all the information on when$\tau$has happened but not on how$\tau$has happened. On the other hand,$\mathcal F_\tau$contains the following sets: $$\mathcal F_\tau=\sigma \Bigl\{A_{WW}, \ A_{LWW},\ A_{WLWW}, A_{LLWW} \Bigr\}.$$ which allows a differentiation between$A_{WLWW}$and$A_{LLWW}$. So given the information contained in$\mathcal F_\tau\$ and I know when I stopped playing the game and in addition the history up to the stopping of the game.