From a Local Martingale to a Martingale
Let \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) be a filtered probability space carrying a Brownian motion \(B_t\) relative to this filtration. For an adapted, progressively measurable process \(H_t\), its energy on \([0,T]\) is the random variable \(\int_0^T H_s^2\,ds\). The Ito integral is first available locally when this energy is finite almost surely on every finite time interval. That local construction gives a continuous local martingale, but a local martingale need not be a martingale.
A martingale \(M_t\) must be integrable at each time and satisfy \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) for \(s\le t\). By contrast, a local martingale has a sequence of stopping times increasing to infinity almost surely such that the stopped processes are martingales; this definition does not itself guarantee that \(M_t\) is integrable. A useful way to ensure integrability for an Ito integral is to control the expected energy, not merely its value on almost every path.
The Expected-Energy Criterion
Proof. The Ito integral for \(\mathcal{L}^2\) integrands is constructed by approximation in the norm \(\left(\mathbb{E}\int_0^T H_s^2\,ds\right)^{1/2}\) by elementary adapted integrands. The Ito isometry, a standard property of this construction, gives \(\mathbb{E}[M_t^2]=\mathbb{E}\int_0^t H_s^2\,ds\), which is finite by the hypothesis. For an elementary adapted integrand, the integral is a sum of terms \(Y_i(B_{v_i}-B_{u_i})\), where \(Y_i\) is measurable at the left endpoint \(u_i\). Brownian increments after \(u_i\) have conditional mean zero given the past, so these integrals have the martingale conditional-expectation property. For a general \(H\in\mathcal{L}^2\), choose elementary integrands \(H^{(n)}\) converging to \(H\) in that norm. The Ito isometry gives convergence of their integrals to \(M_t\) in \(L^2(\mathbb{P})\), and therefore in \(L^1(\mathbb{P})\). Conditional expectation is a contraction in \(L^1\), so passing to the limit in \(\mathbb{E}[M_t^{(n)}\mid\mathcal{F}_s]=M_s^{(n)}\) yields \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\). Thus both integrability and the martingale property hold. \(\square\)
The expected-energy assumption is sufficient, not necessary in every possible setting. Two immediate ways to check it are to bound the expected energy directly or to have a deterministic bound on the energy. In particular, if \(\int_0^T H_s^2\,ds\le C\) almost surely for a finite constant \(C\), then its expectation is at most \(C\), and the theorem applies. Likewise, if \(|H_s|\le K\) for almost every \((s,\omega)\), where \(K\) is deterministic, then \(\mathbb{E}\int_0^T H_s^2\,ds\le K^2T\).
Worked Example: A Constant Integrand
Let \(T=3\), let \(H_s=2\) on \([0,3]\), and set \(M_t=\int_0^t2\,dB_s\). The integrand is adapted and progressively measurable. Its energy through time \(t\) is
Thus \(H\in\mathcal{L}^2\), and the theorem makes \(M_t\) a square-integrable martingale. Since the integrand is constant, \(M_t=2B_t\). For \(0\le s\le t\le3\), Brownian independent increments give \(\mathbb{E}[2B_t\mid\mathcal{F}_s]=2B_s\). Also, the Ito isometry gives \(\mathbb{E}[M_3^2]=12\), since \(\mathbb{E}[B_3^2]=3\) and \(2^2\cdot3=12\).
Check: the direct expression \(M_3=2B_3\) has mean zero and variance \(2^2\cdot3=12\), agreeing with the integral calculation.
Worked Example: A Brownian-Dependent Integrand
On \([0,2]\), take \(H_s=B_s\) and \(M_t=\int_0^t B_s\,dB_s\). The integrand is adapted and progressively measurable. Since \(\mathbb{E}[B_s^2]=s\), its expected energy is
Hence \(H\in\mathcal{L}^2\), so \(M_t\) is a square-integrable martingale. To check the integral explicitly, apply Ito's formula to \(f(t,x)=\tfrac12x^2\). Its derivatives are \(f_t=0\), \(f_x=x\), and \(f_{xx}=1\); the Ito correction is \(\tfrac12 f_{xx}\,dt=\tfrac12\,dt\). Therefore
Check: for \(s\le t\), the Brownian increment \(B_t-B_s\) is independent of \(\mathcal{F}_s\), has mean zero, and has variance \(t-s\). Thus \(\mathbb{E}[B_t^2\mid\mathcal{F}_s]=B_s^2+(t-s)\), and consequently \(\mathbb{E}[\tfrac12(B_t^2-t)\mid\mathcal{F}_s]=\tfrac12(B_s^2-s)\). This verifies the martingale property directly. At \(t=2\), the Ito isometry gives \(\mathbb{E}[M_2^2]=2\). Independently, \(B_2\) is normal with variance \(2\), so \(\operatorname{Var}(B_2^2)=2(2^2)=8\); therefore \(\operatorname{Var}(\tfrac12(B_2^2-2))=\tfrac14\cdot8=2\), in agreement.
Almost-Surely Finite Energy Is Weaker
Local square integrability asks that energy be finite almost surely, not that its expectation be finite. A nonnegative random variable can be finite on every outcome in an event of probability one while having infinite expectation. This distinction matters for the Ito integral: almost-surely finite energy supports the local construction, whereas the theorem above uses expected energy to guarantee square integrability and the true martingale property.
Worked Example: Finite Energy Almost Surely but No True Martingale
Let \(A\) be independent of \(B\), measurable with respect to \(\mathcal{F}_0\), and distributed on the positive integers by \(\mathbb{P}(A=n)=6/(\pi^2n^2)\) for \(n\ge1\). Use the filtration generated by \(A\) and \(B\), completed in the usual way. Independence ensures that \(B_t\) is still Brownian motion relative to this filtration. The integrand \(H_s=A\) is adapted and progressively measurable. On \([0,1]\), its energy is \(A^2\), which is finite almost surely because \(A\) is a finite positive integer almost surely. But
The locally defined integral is \(M_t=A B_t\). For every \(t>0\), independence and nonnegativity give \(\mathbb{E}|M_t|=\mathbb{E}[A]\mathbb{E}|B_t|=\infty\): indeed \(\mathbb{E}[A]=\frac{6}{\pi^2}\sum_{n=1}^{\infty}1/n=\infty\), while \(\mathbb{E}|B_t|=\sqrt{2t/\pi}>0\). Thus \(M_t\) is not integrable and cannot be a true martingale.
Localization is visible here: for each positive integer \(k\), stop at \(\tau_k=\min(1,k/A^2)\). This is a stopping time because \(A\) is known at time zero. The stopped integrand has expected energy \(\mathbb{E}\min(A^2,k)\le k\), so the theorem makes \(M_{t\wedge\tau_k}\) a square-integrable martingale. The energies \(A^2\) are finite almost surely, and \(\tau_k\) increases to \(1\) almost surely. Nevertheless, the unstopped value at any fixed \(t>0\) is not integrable.
Check: \(\mathbb{E}|A B_t|=\infty\) follows from the actual distribution: each term in the sum for \(\mathbb{E}[A]\) is \(n\cdot6/(\pi^2n^2)=6/(\pi^2n)\), and the harmonic series diverges. So the failure is not a failure of pathwise energy finiteness; it is a failure of integrability.
A Common Mistake and Its Correction
A tempting but invalid calculation is: “Each stopped integral \(M_{t\wedge\tau_k}\) has expectation zero, and \(M_{t\wedge\tau_k}\) converges almost surely to \(M_t\), so \(\mathbb{E}[M_t]=\lim_k\mathbb{E}[M_{t\wedge\tau_k}]=0\).” The last step needs a justification for exchanging limit and expectation, such as an applicable domination or uniform integrability argument. Almost-sure convergence alone does not provide one. In the example above, the stopped integrals do have expectation zero, but \(M_t=A B_t\) is not even integrable, so its expectation cannot be obtained by that limit.
The practical test is to separate two questions. First, is the integrand locally square-integrable, so that the Ito integral is defined as a local martingale? Second, is there enough integrability—most directly, \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\)—to apply the true-martingale theorem? A positive answer to the first question does not settle the second.
Check Your Understanding
Use the expected-energy criterion and the distinction between local and true martingales.
- If \(\mathbb{E}\int_0^T H_s^2\,ds=5\), what can you conclude about \(\int_0^t H_s\,dB_s\) for \(0\le t\le T\)?
- For \(H_s=c\), where \(c\) is a deterministic constant, compute the energy on \([0,T]\) and give a sufficient condition ensuring the integral is square-integrable.
- Why does \(\int_0^T H_s^2\,ds<\infty\) almost surely not imply that \(H\in\mathcal{L}^2\)?
- In the random-amplitude example, why does the stopped process have finite expected energy while the unstopped value is not integrable?
- What additional justification is needed to pass from expectations of stopped integrals to the expectation of their almost-sure limit?