How to Audit a Mixed Stochastic Proof
Arguments that combine Itô’s formula, martingale claims, stopping, and a change of measure are especially vulnerable to small errors with large consequences. A missing second-derivative term changes an equation; reversing the Girsanov sign changes the drift; and a limit of stopped martingales need not preserve expectations. The safest approach is to check each operation against its own hypotheses rather than treating a chain of formal differentials as a proof.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Unless explicitly stated otherwise, an Itô integral \(\int_0^t H_s\,dB_s\) has a progressively measurable integrand \(H\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) on each time interval under consideration. An integrand that is only locally square-integrable defines a local stochastic integral; its resulting process is initially a local martingale, not automatically a true martingale.
First Audit: Keep the Itô Correction
For a process \(X_t\) satisfying \(dX_t=\mu_t\,dt+\sigma_t\,dB_t\), Itô’s formula for a twice continuously differentiable function \(f\) is
The last term comes from \(dB_t\,dB_t=dt\); the other multiplication rules are \(dB_t\,dt=0\) and \(dt\,dt=0\). Thus the ordinary chain rule is not the right rule for a process with a Brownian component. In a proof, write down \(f'\), \(f''\), and the correction explicitly before simplifying.
Worked Example: Correcting the Drift of an Exponential
Under \(\mathbb{P}\), let \(X_t=e^{B_t}\). Find its stochastic differential, and test the incorrect claim \(dX_t=X_t\,dB_t\).
Use \(f(x)=e^x\), so \(f'(x)=e^x\) and \(f''(x)=e^x\). Since \(dB_t=0\,dt+1\,dB_t\), the Itô correction is \(\tfrac12 f''(B_t)\,dt=\tfrac12e^{B_t}\,dt\). Therefore
To check independently, \(B_t\) is normal with mean \(0\) and variance \(t\), so \(\mathbb{E}[X_t]=\mathbb{E}[e^{B_t}]=e^{t/2}\). The incorrect equation would make \(X_t-X_0=\int_0^t X_s\,dB_s\). This integral is a true martingale on every finite horizon because \(\mathbb{E}\int_0^T e^{2B_s}\,ds=\int_0^T e^{2s}\,ds<\infty\), and hence it would imply \(\mathbb{E}[X_t]=X_0=1\). That contradicts \(e^{t/2}>1\) for \(t>0\). The missing correction is precisely what resolves the discrepancy.
Second Audit: Fix the Girsanov Sign
Suppose \(\theta_t\) is progressively measurable and define the density process, when it is a true martingale, by
Novikov’s condition on \([0,T]\) is \(\mathbb{E}\exp(\tfrac12\int_0^T\theta_s^2\,ds)<\infty\). It is a sufficient condition for \(Z_t\) to be a martingale with expectation \(1\); it is not a necessary condition. If \(d\mathbb{Q}/d\mathbb{P}|_{\mathcal{F}_T}=Z_T\) and the hypotheses of Girsanov’s theorem hold, then \(B_t^{\mathbb{Q}}=B_t+\int_0^t\theta_s\,ds\) is Brownian under \(\mathbb{Q}\). In particular, \(dB_t=dB_t^{\mathbb{Q}}-\theta_t\,dt\). The minus sign in this substitution determines the drift adjustment.
Worked Example: Checking the Drift Sign
Take \(\theta_t=1\) on \([0,1]\). Verify the change of measure, then rewrite \(dX_t=2\,dt+dB_t\) under \(\mathbb{Q}\).
Here \(Z_t=\exp(-B_t-\tfrac12t)\). The integrand \(H_s=1\) is progressively measurable and \(\mathbb{E}\int_0^1H_s^2\,ds=1\). Novikov’s expectation is \(\mathbb{E}[e^{1/2}]=e^{1/2}<\infty\), so \(Z\) is a true martingale. More directly, for \(s\le t\), independence of \(B_t-B_s\) from \(\mathcal{F}_s\), and its normal moment-generating function, give
Thus \(\mathbb{E}[Z_t]=Z_0=1\), and \(B_t^{\mathbb{Q}}=B_t+t\) is Brownian under \(\mathbb{Q}\). Substitute \(dB_t=dB_t^{\mathbb{Q}}-dt\) into the given equation:
The drift under \(\mathbb{Q}\) is \(1\), not \(3\). As a check, \(X_t=X_0+2t+B_t\), while \(B_t=B_t^{\mathbb{Q}}-t\); hence \(X_t=X_0+t+B_t^{\mathbb{Q}}\), giving the same drift.
Third Audit: A Local Martingale May Lose Expectation
A frequent invalid inference is “the process is a local martingale, therefore its expectation is constant.” A true martingale must be integrable at each time and satisfy the conditional-expectation identity. A local martingale need not have either property globally. Even when every stopped process is a martingale, passing to the limit of their expectations requires additional control, such as uniform integrability.
The following example exhibits the issue on a finite time interval. Let \(0\le t<1\), set \(\theta_t=1/(1-t)\), and define
For every fixed \(t<1\), these integrals are well-defined, and the integrand is square-integrable on \([0,t]\). Define \(Z_1=0\). This endpoint definition is compatible with almost-sure continuity: the quadratic-time parameter \(q(t)=\int_0^t(1-s)^{-2}ds=t/(1-t)\) tends to infinity, and the Brownian term divided by \(q(t)\) tends to zero almost surely, so the exponent tends to minus infinity.
Proof. Fix \(t<1\). The stochastic integral in the exponent is centered normal with variance \(q(t)=t/(1-t)\), and the deterministic term is \(-q(t)/2\). The normal moment-generating function therefore gives \(\mathbb{E}[Z_t]=e^{q(t)/2-q(t)/2}=1\). The same independent-increment calculation between any \(s<t<1\) gives \(\mathbb{E}[Z_t\mid\mathcal{F}_s]=Z_s\). For \(\tau_n=n/(n+1)\), \(q(\tau_n)=n\), so the stopped exponential only uses a finite quadratic-time interval. Its integrand is square-integrable after stopping: by the exponential-normal second moment, \(\mathbb{E}[Z_s^2]=e^{q(s)}\) for \(s\le\tau_n\), and \(\int_0^{\tau_n}(1-s)^{-2}e^{q(s)}ds=\int_0^n e^u du=e^n-1<\infty\). Thus the stopped stochastic integral is a true square-integrable martingale, and \(\mathbb{E}[Z_{1\wedge\tau_n}]=1\). But \(\tau_n\uparrow1\), while the exponent tends to minus infinity almost surely, so \(Z_{\tau_n}\to0\) almost surely. Since \(Z_1=0\), \(\mathbb{E}[Z_1]=0\ne1=\mathbb{E}[Z_0]\); hence \(Z\) is not a martingale on \([0,1]\). The expectation of the stopped limit does not equal the limit of the expectations, so the stopped variables are not uniformly integrable. \(\square\)
Worked Example: When Novikov Cannot Be Invoked
For the preceding \(\theta_t=1/(1-t)\), evaluate Novikov’s condition on \([0,1]\) and identify exactly what conclusion remains valid.
Its quadratic integral diverges at the endpoint:
Consequently \(\mathbb{E}\exp(\tfrac12\int_0^1\theta_s^2ds)=+\infty\), so Novikov’s condition does not apply. On each truncated interval \([0,\tau_n]\), the integral is finite and the stopped density is a martingale, as checked above. But this stopped conclusion does not extend to a true density at time \(1\): \(Z_1=0\) almost surely and \(\mathbb{E}[Z_1]=0\), not \(1\). The example shows both the proper scope of Novikov’s condition and why a sequence of valid truncated changes of measure does not by itself provide a limiting change of measure.
A Practical Proof-Checking Sequence
When auditing a derivation that uses several stochastic tools, record the measure and the status of each process at every step. Under \(\mathbb{P}\), a stochastic integral may be a local martingale; after changing to \(\mathbb{Q}\), the Brownian motion and drift have changed. A martingale assertion under one measure cannot silently be reused under the other. The following checks catch many errors before they propagate.
State the filtration, measure, Brownian motion, and integrability or local-integrability condition for each stochastic integral. Specify the time interval for any exponential-integrability condition.
Write \(f'\), \(f''\), and the term \(\tfrac12 f''\sigma^2dt\) before collecting drift terms. Do not use the ordinary chain rule when the diffusion coefficient is nonzero.
For the density convention above, use \(dB_t=dB_t^{\mathbb{Q}}-\theta_tdt\). Recompute the drift algebraically and name the measure under which the new Brownian motion is Brownian.
For a stochastic integral, check square-integrability or another sufficient integrability condition. For a proposed martingale, verify integrability and the conditional-expectation identity; “local martingale” alone is not enough.
From \(M_{t\wedge\tau_n}\to M_t\) almost surely and \(\mathbb{E}[M_{t\wedge\tau_n}]=\mathbb{E}[M_0]\), do not infer \(\mathbb{E}[M_t]=\mathbb{E}[M_0]\) without a valid limit theorem, for example one supported by uniform integrability.
Common Mistakes and Their Corrections
Wrong calculation: for \(X_t=e^{B_t}\), write \(dX_t=e^{B_t}dB_t\) by applying the ordinary chain rule. Correction: \(f''(x)=e^x\), so Itô’s formula adds \(\tfrac12e^{B_t}dt\). The expectation check distinguishes the two equations.
Wrong calculation: with \(B_t^{\mathbb{Q}}=B_t+\int_0^t\theta_sds\), replace \(dB_t\) by \(dB_t^{\mathbb{Q}}+\theta_tdt\). Correction: rearrange the defining identity first: \(dB_t=dB_t^{\mathbb{Q}}-\theta_tdt\). Then substitute and collect the drift.
Wrong argument: each stopped process has expectation \(1\), and the stopped processes converge almost surely, so the limit has expectation \(1\). Correction: almost-sure convergence does not preserve expectations without additional control. Check uniform integrability or another applicable convergence hypothesis; the stopped exponential example has expectations \(1\) before the limit and \(0\) at the limit.
These checks matter because a formal calculation can produce a plausible-looking equation while failing at the precise point where a theorem’s hypotheses are needed. In particular, Novikov is a sufficient criterion, not a test whose failure proves that an exponential is not a martingale. A failed hypothesis means the theorem has not established the claim; one must use another argument or stop short of the conclusion.
Check Your Understanding
For each question, identify both the correct calculation and the hypothesis or logical step that supports it.
- Apply Itô’s formula to \(f(x)=x^3\) when \(dX_t=\mu_tdt+\sigma_tdB_t\). Write every drift and diffusion term.
- If \(B_t^{\mathbb{Q}}=B_t+\int_0^t\theta_sds\), what substitution for \(dB_t\) gives the drift of \(dX_t=\mu_tdt+\sigma_tdB_t\) under \(\mathbb{Q}\)?
- State Novikov’s condition on a finite horizon and explain why its failure alone does not show that \(Z_t\) is not a martingale.
- Which two properties must be checked to establish that an integrable adapted process is a martingale?
- What additional type of control can justify passing expectations through an almost-sure limit of stopped martingales?