Unit 5, lesson 33

Stochastic processes filtrations and stopping times

How a filtration records information, when a process is adapted or predictable, what the predictable sigma algebra and the usual conditions are, which random times are stopping times, and why anticipatory strategies fail.

  • Video 33
  • Information, Brownian motion and quadratic variation

A backtest on a price that moves up or down by one currency unit each period, with equal probabilities, holds one share long before every up move and one share short before every down move. It earns one unit in every period on every path at zero cost, which looks like an arbitrage. The arithmetic is right and the timing is wrong, because each holding uses a price move that has not happened when the position is opened. This lesson makes the timing precise with filtrations, adapted and predictable processes and stopping times, and shows on a four-path model that predictable trading against increments with conditional mean zero earns nothing on average.

Key terms

Adapted process

Definition Fix a probability space (Ω,F,P)(\Omega, \mathcal{F}, \mathbb{P}) and a time set T\mathbb{T}, either the dates {0,1,…,N}\{0, 1, \dots, N\} or the interval [0,∞)[0, \infty). A filtration is a family (Ft)t∈T(\mathcal{F}_t)_{t \in \mathbb{T}} of sigma algebras with

Fs⊆Ft⊆F(s≤t),\mathcal{F}_s \subseteq \mathcal{F}_t \subseteq \mathcal{F} \quad (s \le t),

where an event belongs to Ft\mathcal{F}_t when the information available at time tt decides whether it has occurred. A stochastic process X=(Xt)t∈TX = (X_t)_{t \in \mathbb{T}} is a family of random variables on Ω\Omega, one for each time. It is adapted to the filtration when

Xt is Ft-measurableX_t \text{ is } \mathcal{F}_t\text{-measurable}

for every t∈Tt \in \mathbb{T}. The natural filtration FtX=σ(Xu:u≤t)\mathcal{F}^X_t = \sigma(X_u : u \le t) is the smallest filtration to which XX is adapted. Adaptedness is a statement about information only. It says nothing about continuity or smoothness, or about how well future values can be forecast, and a process adapted to one filtration need not be adapted to a smaller one.

In plain words A filtration is the growing record of what the model lets a trader know, and a process is adapted when its current value can be read off that record. Information is never forgotten as time passes, and the inclusion says nothing about prices going up.

Example In a constructed two-period model, two independent fair coins give price moves ξ1,ξ2∈{−1,1}\xi_1, \xi_2 \in \{-1, 1\} in currency units, each value with probability 12\tfrac12. Let SkS_k be the price at date kk minus the price at date 00, so S0=0S_0 = 0, S1=ξ1S_1 = \xi_1 and S2=ξ1+ξ2S_2 = \xi_1 + \xi_2. The natural filtration has the trivial F0={∅,Ω}\mathcal{F}_0 = \{\varnothing, \Omega\}, the sigma algebra F1\mathcal{F}_1 generated by the two atoms {ξ1=1}\{\xi_1 = 1\} and {ξ1=−1}\{\xi_1 = -1\}, and F2\mathcal{F}_2 containing every subset of the four paths. The process SS is adapted. The process Yk=Sk+1Y_k = S_{k+1} for k=0,1k = 0, 1 is not, because Y0=ξ1Y_0 = \xi_1 is not constant and so is not F0\mathcal{F}_0-measurable.

Predictable process

Definition In discrete time, let HkH_k be the number of shares held from date kk to date k+1k + 1, for k=0,…,N−1k = 0, \dots, N - 1. The strategy is predictable when each HkH_k is Fk\mathcal{F}_k-measurable, so every holding is fixed with the information available when the position is opened. Many texts index the same holding by the right endpoint of its period, and then predictability says that the holding over the period from k−1k - 1 to kk is Fk−1\mathcal{F}_{k-1}-measurable. The gain process, in currency units, is

Gn=∑k=0n−1Hk(Sk+1−Sk),G0=0.\begin{gathered} G_n = \sum_{k=0}^{n-1} H_k (S_{k+1} - S_k), \\[0.4em] G_0 = 0. \end{gathered}

Suppose SS is a martingale under P\mathbb{P} for the filtration, so each SkS_k is integrable and E[ΔSk+1∣Fk]=0\mathbb{E}[\Delta S_{k+1} \mid \mathcal{F}_k] = 0 with ΔSk+1=Sk+1−Sk\Delta S_{k+1} = S_{k+1} - S_k, and suppose each HkH_k is bounded. Then the known factor comes out of the conditional expectation,

E[HkΔSk+1∣Fk]=Hk E[ΔSk+1∣Fk]=0,\begin{aligned} &\mathbb{E}[H_k \Delta S_{k+1} \mid \mathcal{F}_k] \\[0.4em] &\quad = H_k \, \mathbb{E}[\Delta S_{k+1} \mid \mathcal{F}_k] \\[0.4em] &\quad = 0, \end{aligned}

so GG is a martingale and E[Gn]=0\mathbb{E}[G_n] = 0 for every nn. Boundedness of HkH_k can be weakened to integrability of each product HkΔSk+1H_k \Delta S_{k+1}. The argument uses that HkH_k is Fk\mathcal{F}_k-measurable and fails for a holding that depends on ΔSk+1\Delta S_{k+1}.

In plain words A predictable holding is decided before the price move it earns. Against price moves whose conditional mean is zero, such a holding multiplies a zero conditional mean, so it cannot produce an average profit, however cleverly it reacts to the past. In pricing, the same argument is applied to discounted prices under a pricing measure.

Example In the two-coin model, hold H0=1H_0 = 1 share through the first move, then H1=2H_1 = 2 shares if ξ1=1\xi_1 = 1 and H1=0H_1 = 0 if ξ1=−1\xi_1 = -1. Both holdings are predictable, and the four equally likely paths give these gains.

Path (ξ1,ξ2)(\xi_1, \xi_2)Gain G2G_2
(1,1)(1, 1)1+2=31 + 2 = 3
(1,−1)(1, -1)1−2=−11 - 2 = -1
(−1,1)(-1, 1)−1+0=−1-1 + 0 = -1
(−1,−1)(-1, -1)−1+0=−1-1 + 0 = -1

The mean is E[G2]=14(3−1−1−1)=0\mathbb{E}[G_2] = \tfrac14 (3 - 1 - 1 - 1) = 0. After an up move the two remaining gains 33 and −1-1 average to 11, which is exactly G1=1G_1 = 1, and after a down move both paths end at −1=G1-1 = G_1. So E[G2∣F1]=G1\mathbb{E}[G_2 \mid \mathcal{F}_1] = G_1, and doubling the position after a gain adds no conditional average profit.

Predictable sigma algebra

Definition In continuous time with a filtration (Ft)t≥0(\mathcal{F}_t)_{t \ge 0}, the predictable sigma algebra P\mathcal{P} is the sigma algebra on [0,∞)×Ω[0, \infty) \times \Omega generated by all adapted processes with left-continuous paths, each viewed as the map (t,ω)↦Xt(ω)(t, \omega) \mapsto X_t(\omega). Equivalently, it is generated by the sets {0}×A\{0\} \times A with A∈F0A \in \mathcal{F}_0 and (s,t]×A(s, t] \times A with s<ts < t and A∈FsA \in \mathcal{F}_s. A process is predictable when it is measurable with respect to P\mathcal{P}. A simple predictable process has the form

Ht=a01{0}(t)+∑i=0n−1ai1(ti,ti+1](t)\begin{aligned} H_t &= a_0 \mathbf{1}_{\{0\}}(t) \\[0.4em] &\quad + \sum_{i=0}^{n-1} a_i \mathbf{1}_{(t_i, t_{i+1}]}(t) \end{aligned}

with times 0=t0<t1<⋯<tn0 = t_0 < t_1 < \dots < t_n and each aia_i bounded and Fti\mathcal{F}_{t_i}-measurable, and its gain against a price SS up to time tnt_n is ∑iai(Sti+1−Sti)\sum_{i} a_i (S_{t_{i+1}} - S_{t_i}). Every left-continuous adapted process, in particular every continuous adapted process, is predictable, and every predictable process is adapted. An adapted process with jumps need not be predictable.

In plain words The predictable sigma algebra is the continuous-time form of deciding at the left end of each interval. A holding that is chosen at tit_i and kept on the interval from tit_i to ti+1t_{i+1}, left end excluded, is predictable, and so is anything built from such holdings by limits. Such a holding may use everything known just before time tt, but not a jump that happens at time tt itself.

Example Embed the two-coin model in continuous time, with Ft=F⌊t⌋\mathcal{F}_t = \mathcal{F}_{\lfloor t \rfloor} for 0≤t≤20 \le t \le 2 and the price jumping at times 11 and 22. The holding equal to 11 on (0,1](0, 1] and to 2⋅1{ξ1=1}2 \cdot \mathbf{1}_{\{\xi_1 = 1\}} on (1,2](1, 2] is simple predictable, and its gain is the G2G_2 of the table above. The holding equal to ξ2\xi_2 on (1,2](1, 2] is not even adapted, since ξ2\xi_2 is not Ft\mathcal{F}_t-measurable for 1<t<21 < t < 2. With jumps, adaptedness alone is not enough. Let NN be a Poisson process with intensity λ>0\lambda > 0 and St=Nt−λtS_t = N_t - \lambda t, a martingale. The holding Nt−Nt−N_t - N_{t-}, equal to 11 at the jump times and 00 otherwise, is adapted but not predictable. Its pathwise Stieltjes gain on [0,T][0, T] is the number of jumps NTN_T, with mean λT>0\lambda T > 0 at no cost, because the position is opened at the instant of the jump it collects. The left limit Nt−N_{t-} is predictable.

Stopping time

Definition A random time τ\tau with values in T∪{∞}\mathbb{T} \cup \{\infty\} is a stopping time for the filtration when

{τ≤t}∈Ft\{\tau \le t\} \in \mathcal{F}_t

for every t∈Tt \in \mathbb{T}. In discrete time this is equivalent to {τ=k}∈Fk\{\tau = k\} \in \mathcal{F}_k for every date kk. The value ∞\infty means that the event never happens. Constant times are stopping times, and so are the minimum and the maximum of two stopping times. For a continuous adapted price SS and a lower barrier bb, the first hitting time

τb=inf⁡{t≥0:St≤b},\tau_b = \inf\{t \ge 0 : S_t \le b\},

with inf⁡∅=∞\inf \varnothing = \infty, is a stopping time. The set of levels at most bb is closed, so τb≤t\tau_b \le t exactly when the minimum of SS over [0,t][0, t] is at most bb, and by continuity that minimum equals the infimum over the rational times in [0,t][0, t] together with tt, which is Ft\mathcal{F}_t-measurable. For the strict condition St<bS_t < b the first hitting time is a stopping time once the filtration is right-continuous, which is one reason for the usual conditions below.

In plain words A stopping rule is legitimate when at every time you can tell from what you have seen whether you have already stopped. You do not need to know in advance when that will happen. A rule that needs the rest of the path, such as stopping at the highest price of the period, fails this test.

Example In the two-coin model, stop at the first up move, or at date 22 if there is none. Then {τ≤0}=∅\{\tau \le 0\} = \varnothing, {τ≤1}={ξ1=1}∈F1\{\tau \le 1\} = \{\xi_1 = 1\} \in \mathcal{F}_1 and {τ≤2}=Ω\{\tau \le 2\} = \Omega, so τ\tau is a stopping time. Now let σ\sigma be the first date at which SS reaches its maximum over the dates 00, 11 and 22. On the path (1,−1)(1, -1) the values are 0,1,00, 1, 0 and σ=1\sigma = 1, while on (1,1)(1, 1) they are 0,1,20, 1, 2 and σ=2\sigma = 2. The two paths agree up to date 11, yet only one of them has stopped there, so {σ=1}∉F1\{\sigma = 1\} \notin \mathcal{F}_1 and σ\sigma is not a stopping time. The rule that stops at date 11 exactly when ξ2=1\xi_2 = 1 fails for the same reason.

Usual conditions

Definition A filtered probability space (Ω,F,(Ft)t≥0,P)(\Omega, \mathcal{F}, (\mathcal{F}_t)_{t \ge 0}, \mathbb{P}) satisfies the usual conditions when the probability space is complete, F0\mathcal{F}_0 contains every P\mathbb{P}-null set of F\mathcal{F}, and the filtration is right-continuous, which means Ft=Ft+\mathcal{F}_t = \mathcal{F}_{t+} for every t≥0t \ge 0, where

Ft+=⋂u>tFu.\mathcal{F}_{t+} = \bigcap_{u > t} \mathcal{F}_u .

Every filtration can be enlarged to a smallest filtration that satisfies both conditions, its usual augmentation. Completeness ensures that a process equal at each time, almost surely, to an adapted process is itself adapted. Right-continuity ensures that the first time a right-continuous adapted process enters an open set is a stopping time. A standard result, stated here without proof, is that the completed natural filtration of a Brownian motion is already right-continuous and that the process is still a Brownian motion for it. This lesson defines the usual conditions and uses them only as hypotheses.

In plain words Completeness says that events of probability zero never decide anything. Right-continuity says that whatever is known at every moment after tt is already known at tt. They are conventions that make hitting times and modified versions of processes behave well, and they never add information beyond what the model reveals at times arbitrarily close to tt.

Example Let ξ\xi equal 11 or −1-1 with probability 12\tfrac12 each, the direction of a trend that shows immediately, and let Xt=tξX_t = t\xi. The natural filtration has F0X={∅,Ω}\mathcal{F}^X_0 = \{\varnothing, \Omega\}, since X0=0X_0 = 0, and FtX=σ(ξ)\mathcal{F}^X_t = \sigma(\xi) for every t>0t > 0. Hence F0+X=σ(ξ)≠F0X\mathcal{F}^X_{0+} = \sigma(\xi) \ne \mathcal{F}^X_0, and the filtration is not right-continuous. The first time XX enters the open set (0,∞)(0, \infty) is τ=0\tau = 0 when ξ=1\xi = 1 and τ=∞\tau = \infty when ξ=−1\xi = -1. The event {τ≤0}={ξ=1}\{\tau \le 0\} = \{\xi = 1\} is not in F0X\mathcal{F}^X_0, so τ\tau is not a stopping time for the natural filtration, although it is one for the right-continuous filtration (Ft+X)(\mathcal{F}^X_{t+}).

Anticipatory strategy

Definition A trading strategy is anticipatory when some holding is not measurable with respect to the information available when the position is opened, so in discrete time some HkH_k is not Fk\mathcal{F}_k-measurable. Its gains are outside the predictable-gain argument above, and an exercise or stopping rule that is not a stopping time anticipates in the same sense. Revealing a future move early, by replacing Fk\mathcal{F}_k with a larger filtration Gk\mathcal{G}_k for which ΔSk+1\Delta S_{k+1} is Gk\mathcal{G}_k-measurable, changes the model. Then E[ΔSk+1∣Gk]=ΔSk+1\mathbb{E}[\Delta S_{k+1} \mid \mathcal{G}_k] = \Delta S_{k+1}, so the increment no longer has conditional mean zero unless it is zero almost surely, and SS is not a martingale for (Gk)(\mathcal{G}_k).

In plain words An anticipatory strategy peeks at the future. Its backtested profit comes from the timing error, not from the trading idea, and a conclusion such as “predictable gains have mean zero” holds only together with the information assumption it was proved under.

Example In the two-coin model, keep H0=1H_0 = 1 but set H1=ξ2H_1 = \xi_2, the sign of the second move. The second-period gain is ξ2⋅ξ2=1\xi_2 \cdot \xi_2 = 1 on every path, so E[G2]=E[ξ1]+1=1\mathbb{E}[G_2] = \mathbb{E}[\xi_1] + 1 = 1, and the second trade on its own is a costless position that pays one unit in every state. It is not an arbitrage of the model, because H1=ξ2H_1 = \xi_2 is not F1\mathcal{F}_1-measurable. On the atom {ξ1=1}\{\xi_1 = 1\} it equals 11 on one path and −1-1 on the other, while every predictable H1H_1 is constant on each atom and gives

E[H1ξ2∣F1]=H1 E[ξ2∣F1]=0.\begin{aligned} &\mathbb{E}[H_1 \xi_2 \mid \mathcal{F}_1] \\[0.4em] &\quad = H_1 \, \mathbb{E}[\xi_2 \mid \mathcal{F}_1] = 0. \end{aligned}

Check yourself

  1. In the two-coin model with S0=0S_0 = 0, decide whether each random time is a stopping time. The first is the first date at which Sk=−1S_k = -1, with the value ∞\infty if that never happens. The second is the last of the dates 00, 11 and 22 at which Sk=0S_k = 0.
Show the answer

List the paths. On (1,1)(1, 1) the values are 0,1,20, 1, 2, on (1,−1)(1, -1) they are 0,1,00, 1, 0, on (−1,1)(-1, 1) they are 0,−1,00, -1, 0, and on (−1,−1)(-1, -1) they are 0,−1,−20, -1, -2.

The first hitting time of −1-1, call it τ\tau, is ∞\infty on the two paths that start with an up move and 11 on the two that start with a down move. So {τ≤0}=∅\{\tau \le 0\} = \varnothing and {τ≤1}={τ≤2}={ξ1=−1}\{\tau \le 1\} = \{\tau \le 2\} = \{\xi_1 = -1\}, which lies in F1⊆F2\mathcal{F}_1 \subseteq \mathcal{F}_2. It is a stopping time.

The last date with Sk=0S_k = 0, call it σ\sigma, is 00 on (1,1)(1, 1) and (−1,−1)(-1, -1), and 22 on (1,−1)(1, -1) and (−1,1)(-1, 1). So {σ≤0}={ξ1=ξ2}\{\sigma \le 0\} = \{\xi_1 = \xi_2\}, which is not in the trivial F0\mathcal{F}_0. It is not a stopping time. Deciding at date 00 that this is the last visit to 00 requires knowing that the path will not come back, and last exit times typically look into the future in this way.

  1. In the two-coin model let the share price be 10+Sk10 + S_k currency units, take a zero interest rate, and use the fair-coin probabilities as pricing weights, under which this price is a martingale. An American put with strike 1010 pays Zk=(10−(10+Sk))+=(−Sk)+Z_k = (10 - (10 + S_k))^{+} = (-S_k)^{+} if exercised at date k∈{0,1,2}k \in \{0, 1, 2\}. A simulation exercises on each path at the date with the largest payoff and averages the results. Compute that average and the value sup⁡τE[Zτ]\sup_{\tau} \mathbb{E}[Z_\tau] over stopping times τ\tau with values in {0,1,2}\{0, 1, 2\}, and explain the difference.
Show the answer

On the paths (1,1)(1, 1) and (1,−1)(1, -1) the payoffs are 00 at every date. On (−1,1)(-1, 1) they are 0,1,00, 1, 0, and on (−1,−1)(-1, -1) they are 0,1,20, 1, 2. The best payoffs in hindsight are 0,0,1,20, 0, 1, 2, with average

E[max⁡kZk]=34=0.75.\mathbb{E}\Bigl[\max_{k} Z_k\Bigr] = \tfrac{3}{4} = 0.75 .

Backward induction over stopping times, as in American options and optimal stopping, starts with the value Z2Z_2 at date 22. At date 11 after an up move both exercise and continuation are worth 00. After a down move exercise pays 11 and continuation is worth 12(0+2)=1\tfrac12 (0 + 2) = 1, so the value is 11. At date 00 exercise pays 00 and continuation is worth 12(0+1)=0.5\tfrac12 (0 + 1) = 0.5, so the value is 0.50.5.

The hindsight rule exercises at date 11 on (−1,1)(-1, 1) and at date 22 on (−1,−1)(-1, -1). These paths agree up to date 11, so the rule is not a stopping time. Since E[max⁡kZk]≥E[Zτ]\mathbb{E}[\max_k Z_k] \ge \mathbb{E}[Z_\tau] for every stopping time τ\tau, picking the best date path by path never gives less than the value, and here it overstates the value by 0.250.25.

  1. Let SS be a martingale in a discrete model and τ\tau a stopping time. Show that the stopped price is a gain from a predictable strategy,
Sτ∧n−S0=∑k=0n−11{τ>k}ΔSk+1,\begin{aligned} &S_{\tau \wedge n} - S_0 \\[0.4em] &\quad = \sum_{k=0}^{n-1} \mathbf{1}_{\{\tau > k\}} \Delta S_{k+1}, \end{aligned}

and deduce that E[Sτ]=S0\mathbb{E}[S_\tau] = S_0 when τ≤N\tau \le N for a fixed date NN. Check this for the stopping time “first up move, or date 22” in the two-coin model. Then explain why the first time a symmetric random walk started at 00 reaches 11 has E[Sτ]=1\mathbb{E}[S_\tau] = 1 without contradicting the result.

Show the answer

The holding Hk=1{τ>k}H_k = \mathbf{1}_{\{\tau > k\}} is bounded and Fk\mathcal{F}_k-measurable, because {τ>k}\{\tau > k\} is the complement of {τ≤k}∈Fk\{\tau \le k\} \in \mathcal{F}_k. It equals 11 for k<τk < \tau and 00 for k≥τk \ge \tau, so the sum keeps exactly the increments up to date τ∧n\tau \wedge n and telescopes to Sτ∧n−S0S_{\tau \wedge n} - S_0. By the predictable-gain argument, E[Sτ∧n]=S0\mathbb{E}[S_{\tau \wedge n}] = S_0 for every nn. If τ≤N\tau \le N, then Sτ∧N=SτS_{\tau \wedge N} = S_\tau, and so E[Sτ]=S0\mathbb{E}[S_\tau] = S_0.

In the two-coin model, τ=1\tau = 1 with Sτ=1S_\tau = 1 when ξ1=1\xi_1 = 1, and τ=2\tau = 2 with Sτ=−1+ξ2S_\tau = -1 + \xi_2, which is 00 or −2-2, when ξ1=−1\xi_1 = -1. Then E[Sτ]=12⋅1+14⋅0+14⋅(−2)=0=S0\mathbb{E}[S_\tau] = \tfrac12 \cdot 1 + \tfrac14 \cdot 0 + \tfrac14 \cdot (-2) = 0 = S_0.

For the random walk, the first passage time to 11 is a stopping time, and it is finite with probability one, a standard recurrence result not proved here. Since Sτ=1S_\tau = 1 whenever τ\tau is finite, E[Sτ]=1≠0\mathbb{E}[S_\tau] = 1 \ne 0. Each E[Sτ∧n]=0\mathbb{E}[S_{\tau \wedge n}] = 0 still holds, but this τ\tau is unbounded and the limit in nn cannot be taken inside the expectation. Being a stopping time is not enough on its own, and optional sampling results need further conditions such as boundedness.

How this lesson connects