Ito's Formula

Fix $t$ and a sequence of partitions $0=t_0<\cdots<t_m=t$ with mesh tending to $0$, and write $\Delta_i=W_{t _i}-W_{t_{i-1}}$. Telescoping and the second-order Taylor expansion with Lagrange remainder give, for each interval, a point $\xi_i$ between $W_{t_{i-1}}$ and $W_{t_i}$ with

splitting the second-order term into its value at the left endpoint and a remainder.

The first-order sum. The integrand $f'(W_s)$ is adapted and continuous, so the left-endpoint sums $\sum_i f'(W_{t_{i-1}})\Delta_i$ converge in $L^2$ to the stochastic integral $\int_0^t f'(W_s)\,dW_s$, the simple integrands $f'(W_{t_{i-1}})\mathbf 1_{(t_{i-1},t_i]}$ approximating $f'(W_\cdot)$ in the integral norm.

The second-order sum. Compare $\sum_i f''(W_{t_{i-1}})\Delta_i^2$ with the Riemann sum $\sum_i f''(W_{t_{i -1}})(t_i-t_{i-1})$, which converges to $\int_0^t f''(W_s)\,ds$ because $f''(W_\cdot)$ is continuous. The difference is $\sum_i f''(W_{t_{i-1}})\big(\Delta_i^2-(t_i-t_{i-1})\big)$, and its second moment is, since the increments $\Delta_i^2-(t_i-t_{i-1})$ are mean zero and independent of the past,

the cross terms vanishing by the same mean-zero-and-independence property. So the difference tends to $0$ in $L^2$, and $\sum_i f''(W_{t_{i-1}})\Delta_i^2\to\int_0^t f''(W_s)\,ds$.

The remainder. On the path, $W$ is continuous on $[0,t]$ with compact range, so $f''$ is uniformly continuous there, and $\abs{\xi_i-W_{t_{i-1}}}\le\abs{\Delta_i}$ tends to $0$ uniformly in $i$ as the mesh shrinks. Hence $\max_i\abs{f''(\xi_i)-f''(W_{t_{i-1}})}\to 0$ pathwise, while along the subsequence on which $\sum_i\Delta_i^2\to t$ almost surely the sum $\sum_i\Delta_i^2$ stays bounded, so the remainder is at most $\max_i\abs{f''(\xi_i)-f''(W_{t_{i-1} })}\sum_i\Delta_i^2\to 0$ almost surely along that subsequence.

Each piece converges in probability, so along a subsequence of partitions all converge almost surely, and passing to the limit in Equation (2) yields Equation (1) almost surely at the fixed $t$. Applying this countably often establishes the identity on a fixed countable dense set of $t$ on one null set. The indefinite Ito integral $M_t=\int_0^t f'(W_s)\,dW_s$ has a continuous modification, the standard property of the stochastic integral of an $L^2$ adapted integrand, and $\int_0^t f''(W_s)\,ds$ is continuous in $t$ pathwise, so both sides are continuous in $t$ and the identity extends from the dense set to all $t$ off that null set.

Split each step as $f(t_i,X_{t_i})-f(t_{i-1},X_{t_{i-1}})=[f(t_i,X_{t_i})-f(t_{i-1},X_{t_i})]+[f(t_{i-1},X_{t_i})-f(t_{i-1},X_{t_{i-1}})]$. A first-order expansion in time of the first bracket contributes $\partial_t f\,\Delta t_i$ by the mean value theorem and continuity of $\partial_t f$, and a second-order expansion in space of the second bracket contributes $\partial_x f\,\Delta X_i$ and $\tfrac12\partial_{xx}f\,\Delta X_i^2$. This uses only the assumed $\partial_t f,\partial_x f,\partial_{xx}f$. The increment $\Delta X_i$ splits into its drift part, which contributes $\int b\,\partial_x f\,ds$, and its diffusion part, which contributes $\int\sigma\,\partial_x f\,dW$.

For the second-order space term, write $\Delta X_i=b_{t_{i-1}}\Delta t_i+\sigma_{t_{i-1}}\Delta W_i$ up to the integral errors, so

The first piece converges against $\partial_{xx}f$ to $\int\sigma_s^2\,\partial_{xx}f\,ds$ by replacing $\Delta W_i^2$ with $\Delta t_i$ at $L^2$ cost $2\sum(\sigma^2\partial_{xx}f)^2(\Delta t_i)^2\to0$ as in Equation (3) followed by Riemann convergence; the cross piece is bounded by $C(\max_i\abs{\Delta W_i})\sum_i\Delta t_i\to0$ by continuity of $W$, and the last by $C\,\text{mesh}\sum_i\Delta t_i\to0$. The time term sums to the Riemann integral $\int\partial_t f\,ds$. Collecting the $ds$ and $dW$ terms gives Equation (5). The unbounded-derivative case follows by stopping $X$ when it leaves a large interval $[-N,N]$, on which the continuous functions $f,\partial_t f,\partial_x f,\partial_{xx}f$ are bounded over the compact $[0,t]\times[-N,N]$, so the bounded-derivative estimates apply to the stopped process $X^N$; then letting $N\to\infty$ by continuity of paths removes the boundedness assumption.

Polarise, using only the one-dimensional Theorem 2 applied to $u\mapsto u^2$ (with bounded-derivative localisation) along the single Ito processes $X+Y$, $X$, and $Y$. This gives $d((X+Y)^2)=2(X+Y)\,d(X+Y)+d\qv{X+Y}$, $d(X^2)=2X\,dX+d\qv X$, and $d(Y^2)=2Y\,dY+d\qv Y$, and $\qv{X+Y}=\qv X+2\qv{X,Y}+\qv Y$. Writing $XY=\tfrac12\big((X+Y)^2-X^2-Y^2\big)$ and subtracting,

which is the stated rule.

Apply Theorem 2 with $X=W$ ($b_s=0$, $\sigma_s=1$, $\qv W_t=t$) to $f(t,x)=X_0\exp((\mu-\tfrac12\sigma^2)t+\sigma x)$ evaluated along $W$, so that $X_t=f(t,W_t)$, noting $\partial_t f=(\mu-\tfrac12\sigma^2)f$, $\partial_x f=\sigma f$, and $\partial_{xx}f=\sigma^2 f$. Substituting, the $ds$ coefficient is $(\mu-\tfrac12\sigma^2)f+\tfrac12\sigma^2 f=\mu f$ and the $dW$ coefficient is $\sigma f$, so $dX_t=\mu X_t\,dt+\sigma X_t\,dW_t$. The drift correction $-\tfrac12\sigma^2$ in the exponent is exactly the Ito term, the reason the expected growth rate $\mu$ exceeds the median growth rate.