Proving Ito Formula for a Scalar Ito Process

From Partition Increments to the Ito Formula

The scalar Ito formula is often used as a calculation rule, but its extra second-derivative term comes from a precise limit. Along a fine partition, the linear spatial increments converge to a stochastic integral, while the sum of squared spatial increments converges to quadratic variation. The proof below makes those two limits visible and controls the Taylor remainders that connect them.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. Consider the continuous Ito process $X_t=X_0+\int_0^t b_s\,ds+\int_0^t\sigma_s\,dB_s$. Here $b$ and $\sigma$ are progressively measurable, $\int_0^T|b_s|\,ds<\infty$ and $\int_0^T\sigma_s^2\,ds<\infty$ almost surely for every finite $T$. The Ito integral is defined locally by stopping; whenever an integrand is used as an $L^2$ Ito integrand on $[0,T]$, it is progressively measurable and satisfies $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. The drift integral is a pathwise Lebesgue integral.

Theorem: Let $f:[0,T]\times\mathbb{R}\to\mathbb{R}$ have continuous derivatives $f_t$, $f_x$, and $f_{xx}$. Under the assumptions above, for $0\le t\le T$, almost surely, $$ f(t,X_t)=f(0,X_0)+\int_0^t\left(f_t(s,X_s)+b_s f_x(s,X_s)+\tfrac12\sigma_s^2 f_{xx}(s,X_s)\right)ds +\int_0^t f_x(s,X_s)\sigma_s\,dB_s . $$ The last integral is understood locally: after stopping so its squared integrand has finite expected integral, it is an Ito integral with a progressively measurable integrand in $\mathcal{L}^2$. The drift terms are locally integrable.

Proof by Taylor Expansion

Fix a deterministic partition $0=t_0<t_1<\cdots<t_n=T$, write $\Delta t_i=t_i-t_{i-1}$ and $\Delta X_i=X_{t_i}-X_{t_{i-1}}$, and let the mesh tend to zero. First localize. Stop when $|X|$, the accumulated drift variation $\int_0^t|b_s|\,ds$, or the accumulated diffusion energy $\int_0^t\sigma_s^2\,ds$ reaches a bound $N$. On the stopped interval the range of $X$ is bounded, the drift variation is bounded, and the diffusion energy is bounded. Thus the derivatives of $f$ are bounded on the relevant compact time-space set, and the stopped diffusion integrand has finite expected squared integral. Pointwise boundedness of $b_s$ or $\sigma_s$ is neither asserted nor needed. Since the paths of $X$ are continuous and the two accumulated integrals are finite on $[0,T]$, these stopping times eventually exceed $T$ almost surely as $N$ increases. It is therefore enough to prove convergence for the stopped process.

For each partition interval, split the increment into a time change at the right spatial endpoint and a spatial change at the left time endpoint. The ordinary one-variable Taylor formula in space gives

\begin{aligned} f(t_i,X_{t_i})-f(t_{i-1},X_{t_{i-1}}) ={}&f_t(t_{i-1},X_{t_{i-1}})\Delta t_i +f_x(t_{i-1},X_{t_{i-1}})\Delta X_i\\ &+\tfrac12 f_{xx}(t_{i-1},X_{t_{i-1}})(\Delta X_i)^2 +r_i+q_i . \end{aligned}

Here $r_i$ is the spatial Taylor remainder and $q_i$ is the error from replacing the time increment by $f_t(t_{i-1},X_{t_{i-1}})\Delta t_i$. On the stopped compact set, uniform continuity of $f_{xx}$ implies $|r_i|\le \tfrac12\omega(\max_j|\Delta X_j|)(\Delta X_i)^2$, where $\omega(\delta)\to0$ as $\delta\to0$. Uniform continuity of $f_t$ gives $|q_i|\le \eta_n\Delta t_i$, where $\eta_n\to0$ in probability: the time integral defining $q_i$ evaluates $f_t$ at points between $t_{i-1}$ and $t_i$, with spatial coordinate $X_{t_i}$, and both $\max_i\Delta t_i$ and $\max_i|\Delta X_i|$ tend to zero. Consequently, $\sum_i|q_i|\to0$ in probability.

For the spatial remainder, $\sum_i(\Delta X_i)^2$ is bounded in probability: its limit is the quadratic variation $[X]_T=\int_0^T\sigma_s^2\,ds$. This quadratic-variation identity, and the weighted partition-sum version used next, are the standard quadratic-variation results for Ito processes established in the earlier material. Hence $\sum_i|r_i|\to0$ in probability.

Now sum the three displayed Taylor terms. The time sum converges to $\int_0^T f_t(s,X_s)\,ds$ by continuity of $X$ and $f_t$. For the first-order spatial sum, write $\Delta X_i=\int_{t_{i-1}}^{t_i}b_s\,ds+\int_{t_{i-1}}^{t_i}\sigma_s\,dB_s$. Its drift part converges pathwise to $\int_0^T f_x(s,X_s)b_s\,ds$: the left-endpoint values of $f_x$ converge uniformly along each continuous stopped path, and the total drift variation is bounded. Its stochastic part converges to $\int_0^T f_x(s,X_s)\sigma_s\,dB_s$. Indeed, the corresponding left-step integrands converge to $f_x(s,X_s)$; boundedness on the stopped range and the finite expected diffusion energy give convergence by the Ito isometry.

For the second-order sum, put $g_i=f_{xx}(t_{i-1},X_{t_{i-1}})$. The weighted quadratic-variation result gives $\sum_i g_i(\Delta X_i)^2\to\int_0^T f_{xx}(s,X_s)\sigma_s^2\,ds$ in probability. Its applicability here can also be read from the decomposition into drift and martingale increments: the finite-variation squared increments vanish because their maximum tends to zero while their total variation is bounded; their cross terms with martingale increments vanish in probability; the martingale squared increments give the bracket $\int\sigma^2ds$. The continuous weights converge to $f_{xx}(s,X_s)$.

Thus the sums of the Taylor identities converge in probability to the right-hand side of the claimed formula at $T$. Their left-hand sides telescope exactly to $f(T,X_T)-f(0,X_0)$. The two limits must therefore agree almost surely. Applying the same argument on each fixed subinterval $[0,t]$, and removing the localization since the stopping times eventually exceed $T$, proves the stated identity. The correction has coefficient $\tfrac12$ because the spatial Taylor term is $\tfrac12 f_{xx}(\Delta X_i)^2$, and the squared increments converge to $\int\sigma^2ds$. This is the Ito formula, not the ordinary chain rule. A Stratonovich integral uses a different convention; when the integrand is a continuous semimartingale $H$, its conversion is $\int H\circ dB=\int H\,dB+\tfrac12[H,B]$. $\square$

Examples: Reading Each Term in the Formula

Worked Example: A Time Factor and a Squared Process

Let $X_t=2+B_t$ and $f(t,x)=tx^2$. Thus $b_t=0$, $\sigma_t=1$, and $f_t=x^2$, $f_x=2tx$, $f_{xx}=2t$. The process is defined on the filtered Brownian space above; on any finite interval the stochastic integrand $2tX_t$ is in $\mathcal{L}^2$, since $\mathbb{E}[(2tX_t)^2]=4t^2(4+t)$.

Substitution gives $d(tX_t^2)=(X_t^2+t)\,dt+2tX_t\,dB_t$. In integral form, $tX_t^2=\int_0^t(X_s^2+s)\,ds+\int_0^t2sX_s\,dB_s$, because $f(0,X_0)=0$. To check independently, take expectations. Since $\mathbb{E}[X_s^2]=4+s$, the expected drift integral at $t=2$ is $\int_0^2(4+s+s)\,ds=\int_0^2(4+2s)\,ds=8+4=12$. The Ito integral has mean zero: its integrand is in $\mathcal{L}^2$, so the Ito isometry and conditional-mean property apply. Directly, $\mathbb{E}[2X_2^2]=2(4+2)=12$, agreeing with the formula.

Worked Example: Exponentiating a Process with Drift

Let $X_t=t+B_t$ and $f(t,x)=e^x$. Here $b_t=1$, $\sigma_t=1$, $f_t=0$, $f_x=e^x$, and $f_{xx}=e^x$. Therefore $d(e^{X_t})=\tfrac32e^{X_t}\,dt+e^{X_t}\,dB_t$. The stochastic integrand is in $\mathcal{L}^2$ on finite intervals because $\mathbb{E}[e^{2X_s}]=e^{4s}$, so $\int_0^T e^{4s}ds<\infty$.

At $t=1$, the expected drift integral is $\int_0^1\tfrac32\mathbb{E}[e^{X_s}]\,ds =\int_0^1\tfrac32e^{3s/2}ds=e^{3/2}-1$. The stochastic integral has expectation zero by square integrability. Since $X_1$ is normal with mean $1$ and variance $1$, direct calculation gives $\mathbb{E}[e^{X_1}]=e^{1+1/2}=e^{3/2}$. The initial value is $e^{X_0}=1$; thus the expected right-hand side is $1+(e^{3/2}-1)=e^{3/2}$, as required.

Worked Example: The Exponential Martingale

Take $X_t=B_t$ and $f(t,x)=e^{x-t/2}$. Its derivatives are $f_t=-\tfrac12e^{x-t/2}$, $f_x=e^{x-t/2}$, and $f_{xx}=e^{x-t/2}$. The time derivative and Ito correction cancel, yielding $Z_t=e^{B_t-t/2}=1+\int_0^t e^{B_s-s/2}\,dB_s$.

This integral is an $L^2$ Ito integral on every finite interval because $\mathbb{E}\int_0^T e^{2B_s-s}ds=\int_0^T e^sds=e^T-1<\infty$. The martingale claim can also be checked directly: for $s<t$, independent Brownian increments give $\mathbb{E}[Z_t\mid\mathcal{F}_s]=e^{B_s-s/2}\mathbb{E}[e^{B_t-B_s-(t-s)/2}]=Z_s$, since the conditional exponential factor has expectation $1$. Also $\mathbb{E}[Z_t^2]=e^t<\infty$. Thus this is a true martingale, not merely a local martingale.

A Common Error: Omitting the Quadratic-Variation Term

For $X_t=B_t$ and $f(x)=x^2$, the incorrect ordinary-chain-rule calculation is $d(B_t^2)=2B_t\,dB_t$. It omits $\tfrac12 f_{xx}\,dt=\tfrac12(2)\,dt=dt$. The correct formula is $d(B_t^2)=2B_t\,dB_t+dt$. This correction is not optional: the Ito integral has mean zero, so the incorrect formula would imply $\mathbb{E}[B_T^2]=0$, whereas the Brownian variance is $\mathbb{E}[B_T^2]=T$. The missing $dt$ integrates to exactly $T$.

The proof also explains why localization must be stated carefully. Bounding $\int_0^t|b_s|ds$ and $\int_0^t\sigma_s^2ds$ does not make the coefficients pointwise bounded. It does provide the finite variation and diffusion-energy controls used in the partition limits; together with a bounded stopped range, those are the controls the Taylor argument requires.

Check Your Understanding

For each question, identify which partition sum or derivative produces the relevant term.

For $X_t=B_t$ and $f(t,x)=t x$, compute $f_t$, $f_x$, and $f_{xx}$, then write the Ito formula.
In the proof, why does the spatial Taylor remainder vanish when summed over the partition?
Why does bounded accumulated diffusion energy suffice for the localized Ito-isometry argument even if $\sigma_s$ is not pointwise bounded?
For $X_t=t+B_t$ and $f(x)=x^2$, identify the drift and diffusion terms and check the expected drift rate directly.
Explain why $dB_t\,dB_t=dt$ supplies the second-derivative correction, while $dB_t\,dt=0$ and $dt\,dt=0$ supply no additional Taylor terms.

Proving Ito Formula for a Scalar Ito Process

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