Reading a Filtration as the Information Available at Each Step

A filtration

\mathcal F_t

is the growing collection of events you can distinguish by time

t

. A martingale condition asks for the future process value averaged over outcomes that remain possible given that information. Changing the filtration can change the conditional expectation and therefore the answer to the exercise.

List the information available before taking the expectation

For observed variables

Y_1,\ldots,Y_n

, the natural filtration is

\mathcal F_n=\sigma(Y_1,\ldots,Y_n)

. The inclusion

\mathcal F_n\subseteq\mathcal F_{n+1}

means previously observed information remains available. If

X_n

is adapted, its value is known at time

n

. Provided it is integrable, a one-step martingale proof reduces to

E[X_{n+1}-X_n\mid\mathcal F_n]=0.

To solve a problem, identify the variables you can pull outside the conditional expectation and the variables that still need to be averaged. Independence from $\mathcal F_n$ lets you average a future shock using its original distribution. Observing that shock in advance changes that calculation.

Let $Y_j$ be independent fair increments in $\{-1,1\}$ and $S_n=\sum_{j=1}^nY_j$ with $S_0=0$ . Is $S_n$ a martingale if you observe only completed tosses? What if tomorrow's toss is announced one day early?

With completed tosses,

\mathcal F_n=\sigma(Y_1,\ldots,Y_n)

, so

S_n

is known and

Y_{n+1}

is independent of the information. Its conditional mean is zero. This gives

E[S_{n+1}\mid\mathcal F_n]=S_n

. Adaptedness follows from the sum and integrability follows from

|S_n|\le n

With early announcements, $\mathcal G_n=\sigma(Y_1,\ldots,Y_{n+1})$ . This is still a filtration and $S_n$ is still adapted. But the next value is known, so

E[S_{n+1}\mid\mathcal G_n]=S_n+Y_{n+1}.

The increment is always either plus or minus one, so the conditional mean differs from

S_n

even though the unconditional expectation stays at zero. The process and probability measure were held fixed. The information set alone changed the martingale property.

A useful check is to draw the information tree for two dates. Under completed observations, each node branches into two equally likely outcomes. Under early announcements, the next branch is already determined at that node.

Think of an information set as a partition of the possible histories. After two completed tosses, you know which of four past sequences occurred. Each sequence still allows either next toss. Under the early-announcement filtration, the cell containing the observed history is split by tomorrow's result, so the next increment is already determined. The random walk has the same possible paths in both models. What changes is the conditioning information, so what counts as a surprise changes too.

List the information available before taking the expectation

Turn a fair random walk into a biased one