Why Linear Growth Matters for Moments
To control a solution of an SDE, one needs to control the coefficients when the state becomes large. A bound on the coefficients that grows at most proportionally to the state is called a linear growth condition. Its main use here is to bound the drift and diffusion contributions to the change in the squared size of the solution. This gives an energy inequality that can be used in a later step to obtain moment bounds.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying an \(m\)-dimensional Brownian motion \(\mathbf{B}_t\). Consider a \(d\)-dimensional strong solution \(X_t\) of \(dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,d\mathbf{B}_t\), with initial value \(X_0\). Here \(\mu\) takes values in \(\mathbb{R}^d\), and \(\sigma\) takes values in the \(d\)-by-\(m\) matrices. Use Euclidean norm \(|\cdot|\) for vectors and Frobenius norm \(\|\cdot\|_F\) for matrices. Assume the coefficients are measurable, the solution has continuous paths, and \(X_0\) is \(\mathcal{F}_0\)-measurable with \(\mathbb{E}|X_0|^2<\infty\). The drift integral is a time integral; the stochastic integral is the Itô integral. Its integrand is progressively measurable and belongs to \(\mathcal{L}^2\) on \([0,T]\) whenever \(\mathbb{E}\int_0^T\|H_s\|_F^2\,ds<\infty\). If the solution coefficients are only locally integrable, the estimates below are first applied after stopping.
From Linear Growth to Squared Coefficient Bounds
For second-moment calculations, it is convenient to bound the sum of the squared coefficient norms. The following equivalent form packages both coefficients into one estimate.
Proof. First suppose the linear growth condition holds. Since \(a^2+b^2\le(a+b)^2\) for nonnegative \(a,b\), and \((1+|x|)^2\le2(1+|x|^2)\), we have $$ |\mu(t,x)|^2+\|\sigma(t,x)\|_F^2 \le C^2(1+|x|)^2 \le 2C^2(1+|x|^2). $$ Thus the squared bound holds with \(K=\sqrt{2}\,C\). Conversely, suppose the squared bound holds. Then $$ |\mu(t,x)|+\|\sigma(t,x)\|_F \le \sqrt{2}\sqrt{|\mu(t,x)|^2+\|\sigma(t,x)\|_F^2} \le \sqrt{2}K\sqrt{1+|x|^2}. $$ Since \(\sqrt{1+|x|^2}\le1+|x|\), the linear growth condition follows with \(C=\sqrt{2}K\). These implications also cover zero coefficients and \(x=0\); no nonzero-state assumption is needed. \(\square\)
The constants in either version need not be optimal. What matters for moment estimates is that they are finite and independent of \(t\) and \(x\). In particular, the condition gives separate bounds \(|\mu(t,x)|^2\le K^2(1+|x|^2)\) and \(\|\sigma(t,x)\|_F^2\le K^2(1+|x|^2)\). The diffusion bound is just as essential as the drift bound: diffusion contributes to the squared state through quadratic variation.
Worked Example: A Scalar Affine Drift and Diffusion
For \(x\in\mathbb{R}\), let \(\mu(t,x)=2x+1\) and \(\sigma(t,x)=3x\), independent of time. Find one valid constant \(K\) in the squared coefficient bound.
First calculate the sum of squares: $$ |\mu(t,x)|^2+|\sigma(t,x)|^2=(2x+1)^2+(3x)^2=4x^2+4x+1+9x^2=13x^2+4x+1. $$ The inequality \(4x\le2x^2+2\) follows from \(2(x-1)^2\ge0\). Therefore $$ 13x^2+4x+1\le15x^2+3\le15(1+x^2). $$ So \(K=\sqrt{15}\) is valid. To check the bound at \(x=1\), the coefficient sum is \((2+1)^2+3^2=9+9=18\), while \(15(1+1^2)=30\), and \(18\le30\).
This calculation supplies a usable bound; it does not claim that \(\sqrt{15}\) is the smallest possible constant. For example, the estimate was obtained by a convenient square inequality, not by optimizing over all \(x\).
Worked Example: A Two-Dimensional Bound
Let \(X_t\in\mathbb{R}^2\), let \(\mathbf{B}_t\) be two-dimensional, and take \(\mu(t,x_1,x_2)=(x_1+2,-x_2)\) and \(\sigma(t,x_1,x_2)=\begin{pmatrix}1&0\\0&2\end{pmatrix}\). Find a valid \(K\).
The squared drift norm and squared Frobenius norm are $$ |\mu(t,x)|^2=(x_1+2)^2+x_2^2,\qquad \|\sigma(t,x)\|_F^2=1^2+2^2=5. $$ Using \((x_1+2)^2\le2x_1^2+8\), which follows from \((x_1-2)^2\ge0\), gives $$ |\mu(t,x)|^2+\|\sigma(t,x)\|_F^2 \le2x_1^2+x_2^2+13 \le13(1+x_1^2+x_2^2). $$ Thus \(K=\sqrt{13}\) works. At \(x=(1,-1)\), the exact left side is \((1+2)^2+(-(-1))^2+5=9+1+5=15\); the right side is \(13(1+1+1)=39\), so the claimed inequality holds at that state as well. The matrix norm here is the Frobenius norm, not the operator norm.
The Stopped Energy Inequality
The coefficient bounds feed into Itô’s formula for the squared norm. Stopping is important: before an unstopped moment bound has been established, it is not legitimate to assume the stochastic integrand is square-integrable merely because the desired estimate would imply it.
Proof. Apply Itô’s formula to \(f(x)=|x|^2\), whose gradient is \(2x\) and whose Hessian is twice the identity matrix. The quadratic-variation correction is \(\|\sigma(s,X_s)\|_F^2\,ds\). Stopping at \(t\wedge\tau_R\) gives $$ |X_{t\wedge\tau_R}|^2=|X_0|^2+ \int_0^t\mathbf{1}_{\{s<\tau_R\}}\,2\langle X_s,\mu(s,X_s)\rangle\,ds+ \int_0^t\mathbf{1}_{\{s<\tau_R\}}\|\sigma(s,X_s)\|_F^2\,ds+ 2\int_0^t\mathbf{1}_{\{s<\tau_R\}}X_s^\mathsf{T}\sigma(s,X_s)\,d\mathbf{B}_s. $$ The indicator convention at \(s=\tau_R\) does not affect the time integrals. On \(\{s<\tau_R\}\), continuity gives \(|X_s|<R\). Consequently the matrix-valued stochastic integrand \(2\mathbf{1}_{\{s<\tau_R\}}\sigma(s,X_s)^\mathsf{T}X_s\) is progressively measurable and bounded in norm by \(2RK\sqrt{1+R^2}\). Its squared norm has finite time integral in expectation on every finite interval. The Itô integral is therefore a square-integrable martingale: it is integrable, and its conditional expectation at a later time given \(\mathcal{F}_u\) equals its value at time \(u\). In particular, its expectation is zero.
For the drift term, \(2\langle x,\mu\rangle\le |x|^2+|\mu|^2\), because \(0\le|x-\mu|^2\). The squared coefficient bound therefore implies $$ 2\langle x,\mu(s,x)\rangle+\|\sigma(s,x)\|_F^2 \le |x|^2+2K^2(1+|x|^2) =2K^2+(1+2K^2)|x|^2. $$ Take expectations in the stopped Itô identity. The integrand is present only on \(\{s<\tau_R\}\), and on that event \(X_{s\wedge\tau_R}=X_s\). Hence \(\mathbb{E}[\mathbf{1}_{\{s<\tau_R\}}|X_s|^2]\le Y_R(s)\), since the omitted contribution to \(Y_R(s)\) is nonnegative. Also \(\mathbb{P}(s<\tau_R)\le1\). Integrating these inequalities yields the stated bound. \(\square\)
This is an a priori estimate in the sense that its right side depends on the initial second moment and the stopped second moment, not on an explicit formula for the solution. It is not yet a bound independent of \(R\), and it is not yet an unstopped second-moment theorem. To remove the stopping and obtain a closed bound, one must justify the limiting step and control the integral inequality; those are separate steps.
Worked Example: Exact Second Moment for Geometric Brownian Motion
Consider the scalar SDE \(dX_t=X_t\,dt+2X_t\,dB_t\), \(X_0=1\). Its coefficients satisfy the squared growth bound with \(K^2=5\), because \(x^2+4x^2=5x^2\le5(1+x^2)\). Compute and independently check \(\mathbb{E}[X_t^2]\).
Set \(X_t=\exp(-t+2B_t)\). To verify it, use \(f(t,b)=\exp(-t+2b)\). Its derivatives are \(f_t=-f\), \(f_b=2f\), and \(f_{bb}=4f\). Itô’s formula, including its correction term, gives $$ df(t,B_t)=f_t\,dt+f_b\,dB_t+\tfrac12 f_{bb}\,dt =(-f+2f)\,dt+2f\,dB_t=f\,dt+2f\,dB_t. $$ At \(t=0\), \(f(0,B_0)=1\), so this candidate satisfies the SDE and initial condition.
Since \(B_t\) has distribution \(N(0,t)\), its normal exponential moment is \(\mathbb{E}[e^{\lambda B_t}]=e^{\lambda^2t/2}\). Thus $$ \mathbb{E}[X_t^2]=\mathbb{E}[e^{-2t+4B_t}] =e^{-2t}e^{4^2t/2}=e^{6t}. $$ For an independent check, Itô’s formula for \(X_t^2\) gives \(d(X_t^2)=6X_t^2\,dt+4X_t^2\,dB_t\). The stochastic integral has zero expectation: for each finite \(T\), \(\mathbb{E}\int_0^T(4X_s^2)^2\,ds =16\int_0^T e^{28s}\,ds<\infty\), since the same normal moment calculation gives \(\mathbb{E}[X_s^4]=e^{28s}\). Therefore \(\mathbb{E}[X_t^2]=1+6\int_0^t\mathbb{E}[X_s^2]\,ds\), which is satisfied by \(e^{6t}\): \(1+6\int_0^t e^{6s}\,ds=1+(e^{6t}-1)=e^{6t}\). The diffusion contributes \(4X_t^2\,dt\) to the drift of the square; omitting it would give the wrong moment.
Common Pitfall: Bounding Only the Drift
A frequent error is to treat the drift as the only source of change in \(|X_t|^2\). For a scalar SDE, the correct Itô formula is \(d(X_t^2)=2X_t\,dX_t+(dX_t)^2\), where \((dB_t)^2=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\). For \(dX_t=X_t\,dt+2X_t\,dB_t\), the correction is \((2X_t)^2dt=4X_t^2dt\).
The wrong calculation \(d(X_t^2)=2X_t\,dX_t\) would give a drift of \(2X_t^2\) and suggest \(\mathbb{E}[X_t^2]=e^{2t}\). The correction changes that drift to \(6X_t^2\), giving \(e^{6t}\), as verified above. Linear growth controls both coefficient norms, but it does not make the diffusion correction disappear.
Check Your Understanding
Use the squared coefficient bound and the stopped Itô calculation, distinguishing the drift contribution from the diffusion correction.
- If \(|\mu(t,x)|+\|\sigma(t,x)\|_F\le C(1+|x|)\), what value of \(K\) follows from the proof of the equivalence proposition?
- For \(\mu(x)=x\) and \(\sigma(x)=1\), find a finite \(K\) satisfying \(|\mu(x)|^2+|\sigma(x)|^2\le K^2(1+x^2)\).
- In the stopped energy inequality, why is the stopped stochastic integral a true martingale rather than only a local martingale?
- For \(dX_t=3X_t\,dB_t\), identify the quadratic-variation correction in \(d(X_t^2)\).
- Why does the stopped inequality alone not establish an unstopped second-moment bound?