Why Stopping Does Not Automatically Preserve Expectations in the Limit
Localization lets us study a process only while it remains in a bounded region, where integrability is easier to control. A local martingale is, by definition, a process whose suitably stopped versions are martingales. But the martingale identities for those stopped processes do not automatically pass to the unstopped process: convergence almost surely alone does not preserve expectations. Uniform integrability supplies one important sufficient condition for that passage.
Throughout, work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\), with the filtration satisfying the usual conditions. An Ito integral \(\int_0^t H_s\,dB_s\) below has a progressively measurable integrand \(H\) in \(\mathcal{L}^2\) on each finite interval: \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). A martingale is an integrable adapted process satisfying \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) for \(s\le t\); a local martingale satisfies this after stopping along a localizing sequence.
On a fixed finite interval, localization can instead be described by stopping times that eventually exceed each fixed \(t\), almost surely. Stopping an Ito integral replaces its integrand by \(H_s\mathbf{1}_{\{s\le\tau\}}\), which is still progressively measurable; if \(H\in\mathcal{L}^2\), the stopped integrand is also in \(\mathcal{L}^2\). Thus its integral is a genuine martingale, not merely a local martingale.
Two Convergence Facts Behind Localization
Proof. For a simple adapted integrand, each increment of the integral is a finite sum of terms \(Y_i(B_{u_{i+1}}-B_{u_i})\), where \(Y_i\) is measurable with respect to \(\mathcal{F}_{u_i}\). Brownian increments have conditional mean zero given \(\mathcal{F}_{u_i}\); consequently the integral has conditional mean-zero increments. For general \(H\in\mathcal{L}^2\), approximate by simple adapted integrands in the norm \((\mathbb{E}\int_0^T H_s^2\,ds)^{1/2}\). Ito isometry gives convergence of the corresponding integrals in \(L^2(\mathbb{P})\). Conditional expectation is a contraction in \(L^2\), so the conditional mean-zero identity passes to the limit. Ito isometry also gives \(\mathbb{E}[M_t^2]=\mathbb{E}\int_0^t H_s^2\,ds<\infty\), proving integrability. Therefore \(M_t\) is a martingale. \(\square\)
Proof. Fix \(K>0\), and write \(X_n^{(K)}=\max(-K,\min(X_n,K))\), with \(X^{(K)}\) defined in the same way. Almost-sure convergence implies \(X_n^{(K)}\to X^{(K)}\) almost surely, and these truncated variables are bounded in absolute value by \(K\). Bounded convergence gives \(\mathbb{E}[X_n^{(K)}]\to\mathbb{E}[X^{(K)}]\). Also, uniform integrability makes \(\sup_n\mathbb{E}[|X_n-X_n^{(K)}|]\) tend to zero as \(K\to\infty\). Fatou's lemma gives \(\mathbb{E}[|X-X^{(K)}|]\le\liminf_n\mathbb{E}[|X_n-X_n^{(K)}|]\), so the corresponding tail for \(X\) also tends to zero. In particular, \(X\) is integrable. For any \(K\), split \(\mathbb{E}[X_n-X]\) into the difference of the truncated expectations and the two truncation errors. First take \(n\to\infty\), then \(K\to\infty\); all three terms tend to zero. Hence \(\mathbb{E}[X_n]\to\mathbb{E}[X]\). \(\square\)
One practical consequence is that if \(M_{t\wedge\tau_n}\to M_t\) almost surely and the stopped variables \(\{M_{t\wedge\tau_n}:n\ge1\}\) are uniformly integrable, then \(\mathbb{E}[M_t]=\lim_n\mathbb{E}[M_{t\wedge\tau_n}]\). The same reasoning applies to conditional expectations: convergence in \(L^1\) implies convergence of conditional expectations in \(L^1\), since \(\mathbb{E}[|\mathbb{E}[X\mid\mathcal{F}_s]|]\le\mathbb{E}[|X|]\). Thus the stopped martingale identities pass to the limit in \(L^1\).
Worked Stopped-Integral Calculation
Worked Example: Stopping an Ito Integral at a Level
Let \(\tau=\inf\{t\ge0:|B_t|\ge1\}\), and set \(M_t=\int_0^t B_s\mathbf{1}_{\{s\le\tau\}}\,dB_s\). Verify its expectation and compute its second moment at \(T=2\).
The stopped integrand has absolute value at most \(1\), so \(\mathbb{E}\int_0^2 B_s^2\mathbf{1}_{\{s\le\tau\}}\,ds\le2\). It is therefore in \(\mathcal{L}^2\). The proposition shows that \(M_t\) is a square-integrable martingale, and \(\mathbb{E}[M_2]=0\).
By Ito isometry, \(\mathbb{E}[M_2^2]=\mathbb{E}\int_0^2 B_s^2\mathbf{1}_{\{s\le\tau\}}\,ds\). Continuity of Brownian paths and the definition of \(\tau\) give \(|B_s|\le1\) for \(s\le\tau\). Thus \(B_s^2\mathbf{1}_{\{s\le\tau\}}=B_s^2\) on \(\{s\le\tau\}\), and the second moment is \(\mathbb{E}\int_0^{2\wedge\tau}B_s^2\,ds\); unlike the mean, this requires information about the exit time to evaluate exactly. The useful exact bound is \(0\le\mathbb{E}[M_2^2]\le2\).
Check: the bound follows directly from \(B_s^2\mathbf{1}_{\{s\le\tau\}}\le1\), so \(\mathbb{E}\int_0^2 B_s^2\mathbf{1}_{\{s\le\tau\}}\,ds\le\int_0^2 1\,ds=2\). It confirms square integrability and, independently of any exit-time formula, justifies the martingale conclusion.
Exponential Local Martingales: When the Limit Is Safe
For a progressively measurable process \(\theta\), the stochastic exponential is \(Z_t=\exp(\int_0^t\theta_s\,dB_s-\tfrac12\int_0^t\theta_s^2\,ds)\). When \(\int_0^T\theta_s^2\,ds<\infty\) almost surely, Ito's formula gives \(dZ_t=Z_t\theta_t\,dB_t\), so \(Z\) is a positive local martingale. A positive local martingale is a supermartingale, but need not be a martingale; this is why a localizing sequence alone does not justify setting \(\mathbb{E}[Z_t]=Z_0\).
Novikov's condition is the sufficient condition \(\mathbb{E}\exp(\tfrac12\int_0^T\theta_s^2\,ds)<\infty\). The standard Novikov theorem says that under this condition the stochastic exponential is a martingale on \([0,T]\). This named result is used here without proof; its proof belongs with the later treatment of Girsanov's theorem. The condition and conclusion concern a finite horizon \(T\); they do not by themselves assert uniform integrability over an unbounded time interval.
Worked Example: A Bounded-Kernel Exponential
Take \(\theta_s=1\) on \([0,2]\). Then \(Z_t=\exp(B_t-t/2)\). Verify its expectation, a second-moment bound, and uniform integrability of the family \(\{Z_{t\wedge\tau_n}:n\ge1\}\), where \(\tau_n\) is any localizing sequence tending to infinity almost surely.
Here \(\int_0^2\theta_s^2\,ds=2\), so Novikov's expectation is \(\mathbb{E}[e^1]=e<\infty\). The theorem gives \(\mathbb{E}[Z_t]=1\) for \(0\le t\le2\). For the second moment, direct calculation using \(B_t\sim N(0,t)\) gives \(\mathbb{E}[Z_t^2]=\mathbb{E}[e^{2B_t-t}]=e^{2t-t}=e^t\), which is at most \(e^2\) on this interval.
For the stopped family, the stopped integrand is \(\mathbf{1}_{\{s\le\tau_n\}}\). Its squared integral is at most \(2\), so Novikov applies to it and to twice that integrand. Writing the square of its stochastic exponential as the exponential with kernel \(2\mathbf{1}_{\{s\le\tau_n\}}\), multiplied by \(\exp(\int_0^{t\wedge\tau_n}1\,ds)\), gives \(\mathbb{E}[Z_{t\wedge\tau_n}^2]\le e^2\). Hence the stopped family is uniformly integrable: for any \(K>0\), \(\mathbb{E}[|Z_{t\wedge\tau_n}|\mathbf{1}_{\{|Z_{t\wedge\tau_n}|>K\}}]\le e^2/K\), since \(|x|\mathbf{1}_{\{|x|>K\}}\le x^2/K\).
As \(n\to\infty\), \(Z_{t\wedge\tau_n}\to Z_t\) almost surely. Uniform integrability therefore permits passage of expectations to the limit. Check: for \(t=2\), the direct Gaussian calculation gives \(\mathbb{E}[Z_2]=e^{-1}\mathbb{E}[e^{B_2}]=e^{-1}e^{2/2}=1\), agreeing with Novikov's conclusion.
A Martingale Family That Is Not Uniformly Integrable
Worked Example: Exponential Martingale Losing Its Limit Expectation
Let \(Z_n=\exp(B_n-n/2)\) for integers \(n\ge0\). Show that each \(Z_n\) has expectation one, but that the sequence is not uniformly integrable.
For integers \(m\le n\), independent Brownian increments give \(\mathbb{E}[Z_n\mid\mathcal{F}_m]=\exp(B_m-m/2)\mathbb{E}[\exp(B_n-B_m-(n-m)/2)]=Z_m\). The final expectation is one because \(B_n-B_m\sim N(0,n-m)\) and its exponential moment at \(1\) is \(e^{(n-m)/2}\). In particular, \(\mathbb{E}[Z_n]=1\).
To find the almost-sure limit, fix \(\varepsilon>0\). The Gaussian exponential bound gives \(\mathbb{P}(|B_n|>\varepsilon n)\le2e^{-\varepsilon^2n/2}\), and this bound is summable in \(n\). Borel–Cantelli implies \(|B_n|/n\to0\) almost surely, first for each positive rational \(\varepsilon\), hence without restriction on \(\varepsilon\). Therefore \(B_n-n/2\to-\infty\), so \(Z_n\to0\) almost surely.
If \((Z_n)\) were uniformly integrable, the convergence theorem would imply \(\mathbb{E}[Z_n]\to\mathbb{E}[0]=0\). But every expectation equals \(1\). Thus \((Z_n)\) is not uniformly integrable. Check: the contradiction uses both verified facts—almost-sure convergence to zero and the exact expectation \(1\) at every \(n\)—so there is no valid passage of expectations to this limit.
Common Pitfall: Passing an Expectation Through Almost-Sure Convergence
A tempting but invalid argument is: “The stopped martingales converge almost surely to \(M_t\), and each has expectation \(M_0\), so \(\mathbb{E}[M_t]=M_0\).” The second implication does not follow from almost-sure convergence. The sequence \(Z_n\) above is a concrete counterexample: \(Z_n\to0\) almost surely while \(\mathbb{E}[Z_n]=1\) for every \(n\).
Instead, check what convergence information is available. Uniform integrability is sufficient, as proved above. In some problems an independent integrable bound gives dominated convergence; in others an explicit moment estimate gives uniform integrability. If neither is established, the stopped expectation identity cannot simply be transferred to the unstopped limit. This distinction also prevents confusing a local martingale with a true martingale: localization supplies martingales after stopping, while an additional argument is needed to control their limit.
Check Your Understanding
For each question, identify the integrability or convergence fact that justifies the conclusion.
- Why does \(H\in\mathcal{L}^2\) imply that \(\int_0^t H_s\,dB_s\) has a finite second moment?
- State the definition of uniform integrability and explain how it supports passage of expectations under almost-sure convergence.
- For \(\theta_s=1\) on \([0,T]\), compute \(\mathbb{E}[\exp(2B_T-T)]\) and compare it with the first moment of \(\exp(B_T-T/2)\).
- Why does the sequence \(\exp(B_n-n/2)\) fail to be uniformly integrable even though each term has expectation one?
- What additional condition would you seek before passing the expectations of localized martingales to an unstopped limit?