From a Local Martingale to a True Martingale
Let \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) be a filtered probability space satisfying the usual conditions, let \(B_t\) be a Brownian motion for this filtration, and fix \(T<\infty\). Let \(\theta\) be progressively measurable with \(\int_0^T\theta_s^2\,ds<\infty\) almost surely. Then \(M_t=\int_0^t\theta_s\,dB_s\) is a continuous local martingale with \(M_0=0\) and quadratic variation \(A_t=[M]_t=\int_0^t\theta_s^2\,ds\). The proof below uses only these two facts, so we write it for a general continuous local martingale \(M\).
The stochastic exponential \(Z_t=\exp(M_t-\tfrac12A_t)\) satisfies \(dZ_t=Z_t\,dM_t\) by Ito's formula, so it is a nonnegative local martingale. A nonnegative local martingale is a supermartingale (Fatou's lemma applied along a localizing sequence), so \(\mathbb{E}[Z_\sigma]\le1\) for every stopping time \(\sigma\le T\). It need not be a martingale: its expectation can fall below one.
The proof has three stages. First, stopping \(A\) at a level makes the stopped exponential a true martingale. Second, we replace \(M\) by \(pM\) with \(0<p<1\); this gives just enough room to prove uniform integrability of the stopped exponentials by Hölder's inequality. Third, a Hölder inequality links \(\mathcal{E}(pM)\) back to \(Z\), and letting \(p\uparrow1\) finishes the argument. Throughout, write \(\mathcal{E}(N)=\exp(N-\tfrac12[N])\) for a continuous local martingale \(N\).
Stage 1: Bounded Quadratic Variation
Proof. Put \(Y=\mathcal{E}(N)\) and let \(\rho_k=\inf\{t:|N_t|\ge k\}\). The stopped process \(Y_{t\wedge\rho_k}\) is a bounded local martingale, hence a martingale. Next, \(Y^2=\exp(2N-[N])=\mathcal{E}(2N)\,e^{[N]}\le e^{c}\,\mathcal{E}(2N)\), and \(\mathcal{E}(2N)\) is a nonnegative local martingale, so \(\mathbb{E}[\mathcal{E}(2N)_{t\wedge\rho_k}]\le1\). Therefore \(\mathbb{E}[Y_{t\wedge\rho_k}^2]\le e^{c}\) for all \(k\) and \(t\le T\). A family bounded in \(L^2\) is uniformly integrable. As \(k\to\infty\), \(\rho_k\to\infty\) by continuity, so \(Y_{t\wedge\rho_k}\to Y_t\) almost surely and hence in \(L^1\). Passing to the limit in \(\mathbb{E}[Y_{t\wedge\rho_k}\mid\mathcal{F}_s]=Y_{s\wedge\rho_k}\) gives \(\mathbb{E}[Y_t\mid\mathcal{F}_s]=Y_s\). \(\square\)
Now set \(\tau_n=\inf\{t\ge0:A_t\ge n\}\). Because \(A\) is continuous and nondecreasing, the stopped process \(M^{\tau_n}\) has \([M^{\tau_n}]_T=A_{T\wedge\tau_n}\le n\). Lemma 1 applies to \(N=pM^{\tau_n}\) for every \(p\in(0,1]\): the process \(\mathcal{E}(pM)_{t\wedge\tau_n}\) is a martingale and \(\mathbb{E}[\mathcal{E}(pM)_{T\wedge\tau_n}]=1\). The constant in Lemma 1 depends on \(n\), so this does not yet show that the family over \(n\) is uniformly integrable. That is the work of Stage 2.
Stage 2: Uniform Integrability for \(p<1\)
Proof. Choose \(s=\tfrac12(1+p^{-2})\), so that \(1<s<p^{-2}\), and let \(s'=s/(s-1)\) be the conjugate exponent. Define \(g(r)=s'p^2r(rs-1)\). Then \(g(1)=s'p^2(s-1)=sp^2<1\), and \(g\) is continuous, so we can fix \(r>1\) with \(g(r)\le1\).
We use the algebraic identity \[ \mathcal{E}(pM)^r=\mathcal{E}(rps\,M)^{1/s}\,\exp\!\Big(\tfrac12\,r p^2(rs-1)\,A\Big). \] To check it, expand the first factor: \(\mathcal{E}(rpsM)^{1/s}=\exp(rpM-\tfrac12r^2p^2s\,A)\). Multiplying by the exponential factor changes the coefficient of \(A\) to \(-\tfrac12r^2p^2s+\tfrac12r^2p^2s-\tfrac12rp^2=-\tfrac12rp^2\), which is the coefficient in \(\mathcal{E}(pM)^r=\exp(rpM-\tfrac12rp^2A)\).
Let \(\sigma\le T\) be any stopping time. Hölder's inequality with exponents \(s\) and \(s'\) gives \[ \mathbb{E}\big[\mathcal{E}(pM)_\sigma^r\big] \le \big(\mathbb{E}[\mathcal{E}(rpsM)_\sigma]\big)^{1/s} \Big(\mathbb{E}\exp\!\big(\tfrac12 g(r)A_\sigma\big)\Big)^{1/s'} \le 1\cdot\big(\mathbb{E}e^{A_T/2}\big)^{1/s'}. \] The first factor is at most one because \(\mathcal{E}(rpsM)\) is a nonnegative local martingale and so a supermartingale. In the second, \(g(r)\le1\) and \(0\le A_\sigma\le A_T\). The bound does not depend on \(\sigma\).
Take \(\sigma=T\wedge\tau_n\) and put \(V_n=\mathcal{E}(pM)_{T\wedge\tau_n}\). Then \(\mathbb{E}[V_n]=1\) by Stage 1, and \(\sup_n\mathbb{E}[V_n^r]<\infty\) with \(r>1\), so \(\{V_n\}\) is uniformly integrable. Since \(\mathbb{E}e^{A_T/2}<\infty\), we have \(A_T<\infty\) almost surely, so \(\tau_n>T\) for all large \(n\) and \(V_n\to\mathcal{E}(pM)_T\) almost surely. Uniform integrability upgrades this to \(L^1\) convergence, hence \(\mathbb{E}[\mathcal{E}(pM)_T]=\lim_n\mathbb{E}[V_n]=1\). \(\square\)
This is the step that almost-sure convergence alone cannot supply. The hypothesis \(p<1\) is what created the room \(g(r)\le1\) with \(r>1\); at \(p=1\) the argument gives only \(r=1\), which yields no uniform integrability.
Stage 3: Letting \(p\uparrow1\)
Proof of Novikov's criterion. For \(0<p<1\) we have \(\mathcal{E}(pM)_T=\exp(pM_T-\tfrac12p^2A_T)=Z_T^{\,p^2}\exp\big(p(1-p)M_T\big)\), because \(Z_T^{p^2}=\exp(p^2M_T-\tfrac12p^2A_T)\). Apply Hölder's inequality with exponents \(1/p^2\) and \(1/(1-p^2)\), and write \(q=p/(1+p)\), so that \(p(1-p)/(1-p^2)=q\): \[ 1=\mathbb{E}[\mathcal{E}(pM)_T]\le\big(\mathbb{E}[Z_T]\big)^{p^2}\big(\mathbb{E}\,e^{qM_T}\big)^{1-p^2}. \] To bound the last factor, write \(e^{qM_T}=\mathcal{E}(2qM)_T^{1/2}\,e^{q^2A_T}\) and use Cauchy–Schwarz: \[ \mathbb{E}\,e^{qM_T}\le\big(\mathbb{E}[\mathcal{E}(2qM)_T]\big)^{1/2}\big(\mathbb{E}\,e^{2q^2A_T}\big)^{1/2}\le\big(\mathbb{E}\,e^{A_T/2}\big)^{1/2}. \] The last inequality uses \(\mathbb{E}[\mathcal{E}(2qM)_T]\le1\) and \(2q^2<\tfrac12\) (because \(q<\tfrac12\)) together with \(A_T\ge0\). Combining, \[ 1\le\big(\mathbb{E}[Z_T]\big)^{p^2}\big(\mathbb{E}\,e^{A_T/2}\big)^{(1-p^2)/2}. \] Since \(\mathbb{E}e^{A_T/2}\) is finite and at least one, the last factor tends to \(1\) as \(p\uparrow1\). Hence \(1\le\mathbb{E}[Z_T]\). The supermartingale property gives \(\mathbb{E}[Z_T]\le1\), so \(\mathbb{E}[Z_T]=1\).
Finally, for \(s\le t\le T\) the supermartingale property gives \(\mathbb{E}[Z_T\mid\mathcal{F}_t]\le Z_t\) and \(\mathbb{E}[Z_t\mid\mathcal{F}_s]\le Z_s\). Both inequalities are between random variables whose expectations are equal (every expectation is one), so each is an equality almost surely. Thus \(Z\) is a martingale on \([0,T]\). \(\square\)
The Exponents in Action
Worked Example: The Hölder Step with \(p=\tfrac12\)
Take \(p=\tfrac12\). Then \(p^2=\tfrac14\), \(1-p^2=\tfrac34\), and \(q=p/(1+p)=\tfrac13\). The Stage 3 inequality reads \[ 1\le\big(\mathbb{E}[Z_T]\big)^{1/4}\big(\mathbb{E}\,e^{M_T/3}\big)^{3/4}, \] and \(2q^2=\tfrac29\le\tfrac12\), so \(\mathbb{E}\,e^{M_T/3}\le(\mathbb{E}e^{A_T/2})^{1/2}\). For Stage 2 with this \(p\), take \(s=\tfrac12(1+4)=\tfrac52\) and \(s'=\tfrac53\); then \(g(r)=\tfrac53\cdot\tfrac14\,r\,(\tfrac52r-1)\). Trying \(r=\tfrac54\) gives \(g=\tfrac{425}{384}>1\), which is too large; \(r=\tfrac98\) gives \(g=\tfrac5{12}\cdot\tfrac98\cdot\tfrac{29}{16}=\tfrac{435}{512}\approx0.85\le1\), so \(r=\tfrac98\) is admissible.
Check: \(g(1)=\tfrac5{12}\cdot\tfrac32=\tfrac58=sp^2=\tfrac52\cdot\tfrac14\), as the formula \(g(1)=sp^2\) predicts, and \(g(\tfrac98)\le1\) as required.
Worked Example: A Deterministic Integrand
Take \(\theta_s=2\) on \([0,1]\). Then \(A_t=4t\), so \(\mathbb{E}e^{A_1/2}=e^2<\infty\), and the criterion says \(Z\) is a martingale. Directly, \(M_t=2B_t\) and \(Z_t=e^{2B_t-2t}\). Since \(B_t\sim N(0,t)\), \(\mathbb{E}Z_t=e^{-2t}\mathbb{E}e^{2B_t}=e^{-2t}e^{(2^2t)/2}=1\).
Check: the Gaussian computation agrees with the criterion, including at \(t=0\), where \(Z_0=1\). The Hölder bound from Stage 3 with \(p=\tfrac12\) gives only \(1\le(\mathbb{E}Z_1)^{1/4}(e^{2})^{3/8}\), that is \(\mathbb{E}Z_1\ge e^{-3}\): a weak bound that becomes sharp as \(p\uparrow1\).
Worked Example: A Bounded Adapted Integrand
Suppose \(\theta\) is progressively measurable and \(|\theta_s|\le3\) for \(s\le2\). Then \(A_2=\int_0^2\theta_s^2\,ds\le18\), so \(\mathbb{E}e^{A_2/2}\le e^9<\infty\). Novikov's criterion shows that \(Z_t=\exp\!\big(\int_0^t\theta_s\,dB_s-\tfrac12\int_0^t\theta_s^2\,ds\big)\) is a martingale on \([0,2]\), even when \(\theta\) depends on the Brownian past.
Check: the hypothesis involves the quadratic variation \(A_2\), not the signed integral \(\int\theta\,dB\); \(A_2\le 3^2\cdot2=18\) and \(18/2=9\).
Worked Example: An Unbounded Integrand
Let \(\theta_s=B_s\) on \([0,T]\) with \(T<1\). Then \(A_T=\int_0^TB_s^2\,ds\). By Jensen's inequality for the uniform average over \([0,T]\), \[ \exp\!\Big(\tfrac12\int_0^TB_s^2\,ds\Big)=\exp\!\Big(\tfrac1T\int_0^T\tfrac T2B_s^2\,ds\Big)\le\frac1T\int_0^T\exp\!\Big(\tfrac T2B_s^2\Big)ds. \] For \(B_s\sim N(0,s)\), \(\mathbb{E}\,e^{\lambda B_s^2}=(1-2\lambda s)^{-1/2}\) when \(2\lambda s<1\); with \(\lambda=T/2\) this is \((1-Ts)^{-1/2}\), and for \(s\le T\) it is at most \((1-T^2)^{-1/2}\). Taking expectations, \(\mathbb{E}e^{A_T/2}\le(1-T^2)^{-1/2}<\infty\), so Novikov's condition holds.
Check: for \(T=\tfrac12\) the bound is \((1-\tfrac14)^{-1/2}=2/\sqrt3\approx1.155\). The exact value is \((\cos T)^{-1/2}=(\cos\tfrac12)^{-1/2}\approx1.067\), which is below the bound, as it must be.
What the Criterion Does—and Does Not—Say
Novikov's condition is sufficient, not necessary. If \(\mathbb{E}e^{A_T/2}=\infty\), the proof is silent: \(Z\) may or may not be a martingale. In particular, failure of the condition does not prove that \(Z\) is a strict local martingale. The criterion also does not construct a new probability measure; it supplies the martingale property \(\mathbb{E}[Z_T]=1\) that a density process needs, and the change of measure is a separate step.
A common mistake is to argue that \(Z^{(n)}_T\to Z_T\) almost surely and \(\mathbb{E}Z^{(n)}_T=1\), therefore \(\mathbb{E}Z_T=1\). This is false without extra control. For \(X_n=n\mathbf{1}_{(0,1/n)}\) on \((0,1)\), we have \(X_n\to0\) almost surely and \(\mathbb{E}X_n=1\), yet the limit has expectation zero. In the proof above, the uniform \(L^r\) bound with \(r>1\) from Lemma 2 supplies exactly the missing uniform integrability, and the passage \(p\uparrow1\) is why the criterion needs an exponential moment rather than just \(\mathbb{E}A_T<\infty\).
Check Your Understanding
Use the three stages and keep track of which hypothesis each one needs.
- Why does \(A_{t\wedge\tau_n}\le n\) imply that \(\mathcal{E}(pM)_{t\wedge\tau_n}\) is a martingale?
- Which step fails if \(p=1\) is used directly in Lemma 2? Where does the choice \(s<p^{-2}\) enter?
- If \(|\theta_s|\le C\) on \([0,T]\), give an explicit upper bound for \(\mathbb{E}e^{A_T/2}\).
- For deterministic \(\theta_s=c\), compute \(A_T\) and verify \(\mathbb{E}Z_T=1\) directly.
- Why does \(\mathbb{E}[Z_T]=1\), together with the supermartingale property, imply the martingale property on all of \([0,T]\)?