Why Localization Does Not Preserve Expectations Automatically

Localization Controls Paths, Not Automatically Expectations

Localization replaces an integrand that may have excessive energy by stopped integrands with finite expected energy. Each resulting stochastic integral can be a true martingale with mean zero. But even if those stopped integrals converge almost surely to the original local stochastic integral, their expectations need not converge to its expectation. The missing ingredient is control of the tails of the stopped values.

Throughout, work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. An integrand $H$ is in $\mathcal{L}^2$ on $[0,T]$ when it is progressively measurable and $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. For such an integrand, the Ito integral $M_t=\int_0^t H_s\,dB_s$ is a square-integrable martingale, and the Ito isometry gives $\mathbb{E}[M_t^2]=\mathbb{E}\int_0^t H_s^2\,ds$. A locally square-integrable integrand need only satisfy this condition after a localizing sequence of stopping times; its integral is then a continuous local martingale, not automatically a true martingale.

Definition: A family of integrable random variables $\{X_\alpha\}$ is uniformly integrable if $$ \lim_{K\to\infty}\sup_\alpha\mathbb{E}\big[|X_\alpha|\mathbf{1}_{\{|X_\alpha|>K\}}\big]=0. $$ This condition says that the expectations cannot be carried by tails whose probabilities vanish while their magnitudes grow.

When Expectations Pass Through a Limit

Theorem: Suppose $X_n\to X$ almost surely and $\{X_n:n\ge1\}$ is uniformly integrable. Then $X$ is integrable and $\mathbb{E}|X_n-X|\to0$. In particular, $\mathbb{E}[X_n]\to\mathbb{E}[X]$.

Proof. Define the truncation $c_K(x)=\max(-K,\min(x,K))$. For fixed $K$, $c_K(X_n)\to c_K(X)$ almost surely and $|c_K(X_n)-c_K(X)|\le2K$. The bounded convergence theorem therefore gives $\mathbb{E}|c_K(X_n)-c_K(X)|\to0$ as $n\to\infty$. Also, Fatou's lemma applied to $|X_n|\mathbf{1}_{\{|X_n|>K\}}$ yields $$ \mathbb{E}\big[|X|\mathbf{1}_{\{|X|>K\}}\big] \le \liminf_{n\to\infty}\mathbb{E}\big[|X_n|\mathbf{1}_{\{|X_n|>K\}}\big]. $$ In particular, $X$ is integrable: its tail expectation is bounded by the uniformly small tails for sufficiently large $K$, and its truncated part is bounded by $K$. For any real $x$, $|x-c_K(x)|\le |x|\mathbf{1}_{\{|x|>K\}}$. Thus $$ \mathbb{E}|X_n-X| \le \mathbb{E}|c_K(X_n)-c_K(X)| +\mathbb{E}\big[|X_n|\mathbf{1}_{\{|X_n|>K\}}\big] +\mathbb{E}\big[|X|\mathbf{1}_{\{|X|>K\}}\big]. $$ First choose $K$ large so that the last two terms are uniformly small, using uniform integrability and the Fatou bound. Then let $n\to\infty$, so the first term tends to zero. This proves convergence in $L^1$, and $|\mathbb{E}X_n-\mathbb{E}X|\le\mathbb{E}|X_n-X|$ proves convergence of expectations. $\square$

A useful sufficient condition is a uniform second-moment bound. It is especially convenient for stochastic integrals because the Ito isometry identifies their second moments with expected integrand energy.

Corollary: If $X_n\to X$ almost surely and $\sup_n\mathbb{E}[X_n^2]\le C<\infty$, then $\mathbb{E}[X_n]\to\mathbb{E}[X]$.

Proof. For every $K>0$, $$ \mathbb{E}\big[|X_n|\mathbf{1}_{\{|X_n|>K\}}\big] \le \frac{\mathbb{E}[X_n^2]}{K} \le \frac{C}{K}. $$ The first inequality follows pointwise from $|x|\mathbf{1}_{\{|x|>K\}}\le x^2/K$. Sending $K$ to infinity proves uniform integrability. The Theorem now applies. $\square$

A Brownian Stopping Example Where the Limit Is Safe

Worked Example: Stopping Brownian Motion at Increasing Levels

Let $\tau_n=\inf\{t\ge0:|B_t|\ge n\}$, for positive integers $n$, and fix $T>0$. Consider $M_n=B_{T\wedge\tau_n}$. Show that the stopped expectations pass to the limit and find that limit's expectation.

Write the stopped process as a stochastic integral: $$ M_n=\int_0^T\mathbf{1}_{\{s\le\tau_n\}}\,dB_s. $$ The integrand is progressively measurable and its expected energy is at most $T$, so it is in $\mathcal{L}^2$. The Ito integral is a true martingale with expectation zero. By the Ito isometry, $$ \mathbb{E}[M_n^2]=\mathbb{E}\int_0^T\mathbf{1}_{\{s\le\tau_n\}}\,ds =\mathbb{E}[T\wedge\tau_n]\le T. $$

For every continuous Brownian path, $\sup_{0\le s\le T}|B_s|$ is finite. Consequently, for all sufficiently large $n$, the path does not reach level $n$ by time $T$, and $M_n\to B_T$ almost surely. The second moments are uniformly bounded by $T$, so the corollary applies. Therefore $\mathbb{E}[B_T]=\lim_n\mathbb{E}[M_n]=0$.

Check: the limit is $B_T$, whose distribution is $N(0,T)$ and whose mean is zero. The bound $\mathbb{E}[M_n^2]\le T$ is uniform in $n$, exactly the control needed to rule out expectation escaping into rare large values.

Why Almost-Sure Convergence Alone Fails

Worked Example: A Vanishing Tail That Carries the Mean

Let $U$ be uniform on $(0,1)$, and define $X_n=n\mathbf{1}_{\{U<1/n\}}-1$. Compute the expectations and almost-sure limit, and test uniform integrability.

The event $\{U<1/n\}$ has probability $1/n$, so $$ \mathbb{E}[X_n]=n\cdot\frac1n-1=0. $$ For every outcome with $U>0$, once $n\ge 1/U$, the indicator is zero. Hence $X_n\to-1$ almost surely, and $\mathbb{E}[-1]=-1$. The expectations do not pass to the limit.

For $n>K+1$, on $\{U<1/n\}$ the value of $|X_n|$ is $n-1>K$; outside that event it is $1$. Taking $K\ge1$, the tail expectation is $$ \mathbb{E}\big[|X_n|\mathbf{1}_{\{|X_n|>K\}}\big] =(n-1)\frac1n=1-\frac1n. $$ For any fixed $K\ge1$, this approaches $1$ as $n\to\infty$, so the tail expectations cannot tend uniformly to zero as $K\to\infty$. The family is not uniformly integrable.

Check: the rare event has probability $1/n$, but the positive value on it is $n-1$. Its contribution to the mean approaches $1$, balancing the value $-1$ occurring with probability approaching one. The almost-sure limit discards precisely that rare contribution.

This random-variable example isolates the logical error: almost-sure convergence is not enough to interchange limit and expectation. The following stochastic example shows that the same failure can occur for localized stochastic integrals.

A Local Stochastic Integral That Loses Expectation

Worked Example: The Reciprocal of a Three-Dimensional Brownian Radius

Let $\mathbf{B}_t=(B_t^1,B_t^2,B_t^3)$ be a standard three-dimensional Brownian motion on a filtered probability space, and set $\mathbf{Y}_t=(1,0,0)+\mathbf{B}_t$, $R_t=|\mathbf{Y}_t|$, and $X_t=1/R_t$. Three-dimensional Brownian motion started away from the origin almost surely never hits the origin; this standard non-hitting result is used here. For $n\ge2$, define $\tau_n=\inf\{t\ge0:R_t\notin(1/n,n)\}$. These stopping times increase, and $\tau_n\to\infty$ almost surely because on each compact time interval $R$ has a positive minimum and a finite maximum.

Apply multidimensional Ito's formula to $f(y)=|y|^{-1}$, defined for $y\ne0$. Its first partial derivatives are $\partial_i f(y)=-y_i|y|^{-3}$, and its second partial derivatives are $$ \partial_{ij}f(y)=-\delta_{ij}|y|^{-3}+3y_i y_j|y|^{-5}. $$ Thus $\Delta f(y)=\sum_{i=1}^3\partial_{ii}f(y)=-3|y|^{-3}+3|y|^2|y|^{-5}=0$. The Ito second-order correction is $\tfrac12\Delta f(\mathbf{Y}_t)\,dt=0$, so $$ X_t=1-\int_0^t\frac{\mathbf{Y}_s}{R_s^3}\cdot d\mathbf{B}_s. $$ The vector integrand is progressively measurable and is locally square-integrable: before $\tau_n$, its squared norm is $R_s^{-4}\le n^4$, so its stopped expected energy on $[0,T]$ is at most $n^4T$. The displayed integral is therefore a continuous local martingale, but this fact alone does not make $X$ a true martingale.

At each stopping time $\tau_n$, $X_{t\wedge\tau_n}$ is bounded by $n$. Equivalently, its stopped stochastic-integral integrand has norm at most $n^2$ and finite expected energy on $[0,T]$. The stopped integral is square-integrable and has mean zero, so $$ \mathbb{E}[X_{T\wedge\tau_n}]=1. $$ Since $\tau_n\to\infty$ almost surely, $X_{T\wedge\tau_n}\to X_T$ almost surely. But this convergence does not preserve expectations.

To calculate the limiting expectation, the density of $\mathbf{Y}_T$ is $(2\pi T)^{-3/2}\exp(-|y-(1,0,0)|^2/(2T))$. Spherical coordinates, with $r=|y|$ and $u$ the cosine of the angle from the first coordinate axis, give $$ \mathbb{E}[X_T] =\frac{4\pi T}{(2\pi T)^{3/2}}\int_0^\infty \exp\!\left(-\frac{r^2+1}{2T}\right)\sinh\!\left(\frac rT\right)\,dr. $$ Using $$ \exp\!\left(-\frac{r^2+1}{2T}\right)\sinh\!\left(\frac rT\right) =\frac12\left[ \exp\!\left(-\frac{(r-1)^2}{2T}\right) -\exp\!\left(-\frac{(r+1)^2}{2T}\right) \right], $$ the integral equals $\int_0^1\exp(-v^2/(2T))\,dv=\sqrt{\pi T/2}\,\operatorname{erf}(1/\sqrt{2T})$, where $\operatorname{erf}(z)=\frac{2}{\sqrt{\pi}}\int_0^z e^{-v^2}\,dv$. Therefore $$ \mathbb{E}[X_T]=\operatorname{erf}\!\left(\frac{1}{\sqrt{2T}}\right)<1 \qquad(T>0). $$ The strict inequality holds because the integral defining $\operatorname{erf}(z)$ is strictly less than its integral over $[0,\infty)$ for finite $z>0$. The stopped expectations are all $1$, while the limiting expectation is strictly smaller.

Check: as $T$ decreases to zero, $1/\sqrt{2T}$ increases and the expectation approaches $1=X_0$, consistent with continuity at time zero. For every fixed $T>0$, the displayed strict inequality confirms that the limit cannot have the same expectation as the stopped variables.

Using Localization Without Overclaiming

For a local stochastic integral $M_t=\int_0^t H_s\,dB_s$, a localizing sequence makes the stopped processes manageable, but it does not by itself give uniform integrability of their values at a fixed time. A direct route is to establish a uniform bound on the expected stopped energy. For example, if $$ \sup_n\mathbb{E}\int_0^T H_s^2\mathbf{1}_{\{s\le\tau_n\}}\,ds\le C<\infty, $$ then the Ito isometry gives $\sup_n\mathbb{E}[M_{T\wedge\tau_n}^2]\le C$. The second-moment criterion proves uniform integrability. If $M_{T\wedge\tau_n}\to M_T$ almost surely, the theorem then justifies $\mathbb{E}[M_{T\wedge\tau_n}]\to\mathbb{E}[M_T]$.

This is a sufficient condition, not a necessary one: uniform integrability can hold without a uniform second-moment bound. Conversely, the fact that every stopped integral has expectation zero does not establish that the unstopped local martingale has expectation zero. In the reciprocal-radius example, each stopped process is a true martingale, but the family of its fixed-time stopped values is not uniformly integrable.

Common pitfall. The invalid calculation is “$M_{T\wedge\tau_n}\to M_T$ almost surely and $\mathbb{E}[M_{T\wedge\tau_n}]=0$, therefore $\mathbb{E}[M_T]=0$.” The conclusion is valid only after adding a condition that controls the tails, such as uniform integrability or a uniform second-moment bound. Almost-sure convergence controls typical outcomes; expectation also depends on rare outcomes with large magnitude.

Check Your Understanding

For each question, distinguish what follows from localization alone from what requires additional tail control.

State the definition of uniform integrability for a family of integrable random variables.
If $X_n\to X$ almost surely and $\sup_n\mathbb{E}[X_n^2]\le C$, why do the expectations converge?
For $M_n=B_{T\wedge\tau_n}$, with $\tau_n=\inf\{t:|B_t|\ge n\}$, give the uniform second-moment bound and the limiting expectation.
For $X_n=n\mathbf{1}_{\{U<1/n\}}-1$, calculate $\mathbb{E}[X_n]$ and its almost-sure limit. Which hypothesis of the expectation-convergence theorem fails?
In the reciprocal Brownian-radius example, what does the strict inequality $\mathbb{E}[X_T]<X_0$ demonstrate about localization?