Complete Proof of the Time-Dependent Ito Formula

From a Partition to the Time-Dependent Formula

The time-dependent Ito formula has an additional term compared with the formula for a function of Brownian motion alone: change in the time variable contributes an ordinary time integral. Proving the formula requires keeping this contribution separate from the spatial Taylor expansion. The proof below does so on a finite partition, then establishes the limits of the first-order, second-order, and remainder sums.

Fix $T>0$ and a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ with respect to the filtration. In particular, for $s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$. We use the Ito integral $\int_0^t H_s\,dB_s$ for progressively measurable integrands in $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. These conditions will hold for the integrands in the bounded-derivative proof below.

Theorem: Let $f:[0,T]\times\mathbb{R}\to\mathbb{R}$ have continuous derivatives $f_t,f_x,f_{xx}$. If these three derivatives are bounded, then, for every $t\in[0,T]$, almost surely, $$ f(t,B_t)=f(0,B_0)+\int_0^t f_t(s,B_s)\,ds+\int_0^t f_x(s,B_s)\,dB_s+\frac12\int_0^t f_{xx}(s,B_s)\,ds. $$ Here the stochastic integral is the Ito integral with integrand $H_s=f_x(s,B_s)\in\mathcal{L}^2$. For derivatives that are only locally bounded, the same identity holds locally, by stopping before $B$ leaves compact intervals.

The boundedness hypothesis is a convenient way to make the convergence of the stochastic integrals an $L^2$ argument. Continuity, rather than global uniform continuity, suffices for the Taylor remainders: on each compact set the derivatives are uniformly continuous.

The Partition Decomposition and Its Limits

Take a deterministic partition $0=t_0<t_1<\cdots<t_n=T$, write $\Delta t_i=t_i-t_{i-1}$, $\Delta B_i=B_{t_i}-B_{t_{i-1}}$, and let $|\Pi|=\max_i\Delta t_i$. Split each increment by changing the space coordinate first and the time coordinate second:

\begin{aligned} f(t_i,B_{t_i})-f(t_{i-1},B_{t_{i-1}}) ={}&\big[f(t_{i-1},B_{t_i})-f(t_{i-1},B_{t_{i-1}})\big]\\ &+\big[f(t_i,B_{t_i})-f(t_{i-1},B_{t_i})\big]. \end{aligned}

Taylor's formula in the space coordinate, with its integral remainder, gives

f(t_{i-1},B_{t_i})-f(t_{i-1},B_{t_{i-1}}) =f_x(t_{i-1},B_{t_{i-1}})\Delta B_i +\frac12 f_{xx}(t_{i-1},B_{t_{i-1}})(\Delta B_i)^2+R_i.

For example, the remainder can be written as an integral of the difference between $f_{xx}$ at points between $B_{t_{i-1}}$ and $B_{t_i}$ and its value at $B_{t_{i-1}}$, multiplied by the Taylor weight. This form will let us bound it by a modulus of continuity times $(\Delta B_i)^2$. The time-coordinate change is exactly

f(t_i,B_{t_i})-f(t_{i-1},B_{t_i}) =\int_{t_{i-1}}^{t_i} f_t(s,B_{t_i})\,ds.

The next lemma supplies the weighted quadratic-variation limit needed for the second-order term. In its statement, $a:[0,T]\times\Omega\to\mathbb{R}$ is a bounded, adapted process with continuous paths.

Lemma: For such $a$, as $|\Pi|\to0$, $$ \sum_{i=1}^n a(t_{i-1})(\Delta B_i)^2\ \longrightarrow\ \int_0^T a(s)\,ds $$ in probability.

Proof. First subtract the corresponding time sum. Since $a(t_{i-1})$ is $\mathcal{F}_{t_{i-1}}$-measurable, the Brownian increment assumptions give $\mathbb{E}[(\Delta B_i)^2-\Delta t_i\mid\mathcal{F}_{t_{i-1}}]=0$ and $\mathbb{E}[((\Delta B_i)^2-\Delta t_i)^2\mid\mathcal{F}_{t_{i-1}}]=2(\Delta t_i)^2$. The centered summands $a(t_{i-1})((\Delta B_i)^2-\Delta t_i)$ are orthogonal in $L^2$: for $i<j$, the $i$th summand is $\mathcal{F}_{t_{j-1}}$-measurable, while the conditional mean of the $j$th summand given $\mathcal{F}_{t_{j-1}}$ is zero. If $|a|\le C$, it follows that $$ \mathbb{E}\left[\left(\sum_i a(t_{i-1})((\Delta B_i)^2-\Delta t_i)\right)^2\right] =2\sum_i\mathbb{E}[a(t_{i-1})^2](\Delta t_i)^2 \le 2C^2T|\Pi|\longrightarrow0. $$ The remaining sum $\sum_i a(t_{i-1})\Delta t_i$ converges pathwise to $\int_0^T a(s)\,ds$, because each continuous path of $a$ is Riemann integrable. Combining these two limits proves the claim. $\square$

We now prove the theorem. On summing the partition decomposition, the left side telescopes to $f(T,B_T)-f(0,B_0)$. The first-order spatial sum converges to the Ito integral: its integrand is the left-step process $H^{(n)}_s=f_x(t_{i-1},B_{t_{i-1}})$ on $(t_{i-1},t_i]$. It is progressively measurable and bounded. Continuity of $f_x$ and of Brownian paths gives $H^{(n)}_s\to f_x(s,B_s)$ almost surely for each $s$; boundedness and dominated convergence give $\mathbb{E}\int_0^T|H^{(n)}_s-f_x(s,B_s)|^2\,ds\to0$. The Ito isometry therefore yields $$ \sum_i f_x(t_{i-1},B_{t_{i-1}})\Delta B_i =\int_0^T H^{(n)}_s\,dB_s \longrightarrow\int_0^T f_x(s,B_s)\,dB_s $$ in $L^2(\mathbb{P})$.

Apply the lemma with $a(s)=f_{xx}(s,B_s)$. Its hypotheses hold because this process is adapted, continuous, and bounded. Thus the second-order sum converges in probability to $\int_0^T f_{xx}(s,B_s)\,ds$, with the Taylor coefficient $\tfrac12$ retained.

It remains to justify both remainders. For the spatial Taylor remainder, fix $R>0$ and consider the event $\sup_{0\le s\le T}|B_s|\le R$. On $[0,T]\times[-R-1,R+1]$, continuity of $f_{xx}$ gives a modulus of continuity $\omega_R(\delta)$ tending to zero as $\delta\downarrow0$. Once the increments are at most $1$, the sum of the absolute spatial remainders is bounded by $\tfrac12\omega_R(\max_i|\Delta B_i|)\sum_i(\Delta B_i)^2$. Brownian path continuity gives $\max_i|\Delta B_i|\to0$ almost surely. Also $\sum_i(\Delta B_i)^2$ is bounded in probability: its expectation is $T$, since each increment has second moment $\Delta t_i$, so Markov's inequality applies uniformly over the partitions. The spatial remainder sum consequently tends to zero in probability on this event. The probability of the event tends to one as $R\to\infty$, since continuous Brownian paths are bounded on $[0,T]$.

For the time-coordinate term, replace $f_t(s,B_{t_i})$ by $f_t(t_{i-1},B_{t_{i-1}})$ inside each time integral. On the same compact event, the absolute error over all intervals is bounded by $T$ times the modulus of continuity of $f_t$ at $|\Pi|+\max_i|\Delta B_i|$, which tends to zero almost surely. The resulting Riemann sum converges pathwise to $\int_0^T f_t(s,B_s)\,ds$, because $s\mapsto f_t(s,B_s)$ is continuous. We have proved convergence in probability of every term in the telescoping identity to the corresponding term in the asserted formula. Uniqueness of limits in probability proves the formula at $T$. Applying the same argument at each fixed $t$ proves the stated identity. The integral terms have continuous versions; equality at rational times and continuity give a single probability-one event on which the identity holds for all $t\in[0,T]$. $\square$

Three Checks on the Formula

Worked Example: The Product of Time and Brownian Motion

Let $f(t,x)=tx$, and apply the formula on $[0,T]$. Its derivatives are $f_t(t,x)=x$, $f_x(t,x)=t$, and $f_{xx}(t,x)=0$. Since $f(0,B_0)=0$, the result is $$ tB_t=\int_0^t B_s\,ds+\int_0^t s\,dB_s. $$ The stochastic integrand is in $\mathcal{L}^2$, because $\mathbb{E}\int_0^T s^2\,ds=T^3/3<\infty$.

Check: the finite sums for the stochastic term satisfy the discrete summation-by-parts identity $\sum_i t_{i-1}\Delta B_i=tB_t-\sum_i B_{t_i}\Delta t_i$ on a partition of $[0,t]$. The last sum converges pathwise to $\int_0^t B_s\,ds$. Hence the stochastic sum converges to $tB_t-\int_0^t B_s\,ds$, agreeing with the formula.

Worked Example: A Time-Weighted Square

Take $f(t,x)=tx^2$. The derivatives are $f_t(t,x)=x^2$, $f_x(t,x)=2tx$, and $f_{xx}(t,x)=2t$. The formula, with $f(0,B_0)=0$, gives $$ tB_t^2=\int_0^t B_s^2\,ds+\int_0^t 2sB_s\,dB_s+\int_0^t s\,ds. $$ The stochastic integrand belongs to $\mathcal{L}^2$, since $\mathbb{E}\int_0^T(2sB_s)^2\,ds=4\int_0^T s^2\mathbb{E}[B_s^2]\,ds=4\int_0^T s^3\,ds=T^4<\infty$.

Check: take expectations. The Ito integral has mean zero: its integrand is in $\mathcal{L}^2$, and the Ito isometry construction from adapted step integrands gives zero expectation for the integral. Since $\mathbb{E}[B_s^2]=s$, the right-hand expectation is $\int_0^t s\,ds+\int_0^t s\,ds=t^2$. The left-hand expectation is $t\mathbb{E}[B_t^2]=t^2$, as required.

Worked Example: A Purely Time-Dependent Function

Let $f(t,x)=t^2$. Then $f_t(t,x)=2t$, $f_x(t,x)=0$, and $f_{xx}(t,x)=0$. Thus $$ t^2=0+\int_0^t 2s\,ds+0. $$ There is no stochastic integral contribution, because the function does not depend on the space variable.

Check: evaluating the ordinary integral gives $\int_0^t2s\,ds=[s^2]_0^t=t^2$, exactly the left side. In particular, the time derivative is not part of the Ito correction; it is the ordinary first-order time term.

Why the Decomposition Matters

The proof distinguishes three mechanisms that are easy to conflate. The time increment produces $f_t\,dt$ by ordinary calculus. The first spatial increment produces $f_x\,dB_t$ as an Ito integral. The squared spatial increment survives in the limit and produces $\tfrac12 f_{xx}\,dt$. In differential shorthand, this last contribution is encoded by $dB_t\,dB_t=dt$; the mixed and purely time second-order terms vanish, $dB_t\,dt=0$ and $dt\,dt=0$.

A common error is to apply the ordinary chain rule and write only $df(t,B_t)=f_t(t,B_t)\,dt+f_x(t,B_t)\,dB_t$. For $f(t,x)=tx^2$, this omits $\int_0^t s\,ds=t^2/2$, because $\tfrac12 f_{xx}(s,B_s)=s$. The check by expectations exposes the error: without that term the proposed right side has expectation $t^2/2$, whereas $\mathbb{E}[tB_t^2]=t^2$. The missing term is precisely the limit of the weighted squared increments in the proof.

Takeaway: The time-dependent Ito formula follows by splitting each partition increment into a spatial change and a time change. The time sum converges by ordinary Riemann integration; the spatial first-order sum converges by the Ito isometry; the weighted squared increments converge to the second spatial derivative integral; and continuity controls the Taylor remainders.

Check Your Understanding

Use the partition decomposition and the hypotheses of the theorem to answer these questions.

For $f(t,x)=t^2x$, compute $f_t$, $f_x$, and $f_{xx}$, then write the formula from time $0$ to $t$.
Why does the time-coordinate increment use $f_t(s,B_{t_i})$ before taking a limit, rather than $f_t(s,B_{t_{i-1}})$?
In the weighted quadratic-variation lemma, where is independence of Brownian increments from $\mathcal{F}_{t_{i-1}}$ used?
Explain why boundedness in probability of $\sum_i(\Delta B_i)^2$, together with a vanishing modulus of continuity, makes the spatial Taylor remainder vanish in probability.
For $f(t,x)=tx$, identify which of the time, first spatial, and second spatial terms are nonzero.

Complete Proof of the Time-Dependent Ito Formula

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