Itô's formula

We want to ''differentiate'' expressions of the form $f\left[ B\left( t\right) \right]$, where $f\left( \cdot \right)$ is a differentiable function. If $B\left( t\right)$ were also differentiable we could just employ the chain rule and have

$$\frac{d}{dt}f\left[ B\left( t\right) \right] = f^{\prime }\left[ B\left( t\right) \right] B^{\prime }\left( t\right)$$

$$df\left[ B\left( t\right) \right] = f^{\prime }\left[ B\left( t\right) %% \right] dB\left( t\right) +o\left( dt\right)$$

Apparently $B$ is not differentiable and we need a way to overcome this problem. Since $f\left( \cdot \right)$ is differentiable we can use a Taylor's expansion and write

$$\Delta _{\delta }f\left[ B\left( t\right) \right] =f^{\prime }\le... ...left( \left[ \Delta _{\delta }B\left( t\right) %% \right] ^{2}\right) \text{.}$$

Now, the fact that $\left[ \Delta _{\delta }B\left( t\right) \right] ^{2}=\delta +o\left( \delta \right)$ implies that

$$\Delta _{\delta }f\left[ B\left( t\right) \right] =f^{\prime }\le... ...me }\left[ B\left( t\right) \right] \delta +o\left( \delta \right) \text{%% ,}$$

which finally gives

$$df\left[ B\left( t\right) \right] =f^{\prime }\left[ B\left( t\ri... ...t( t\right) +\frac{1}{2}f^{\prime \prime }\left[ B\left( t\right) %% \right] dt$$.

The above formula plays the role of the chain rule in stochastic calculus. The extra term $\frac{1}{2}f^{\prime \prime }\left[ B\left( t\right) \right] dt$ appears because $B\left( t\right)$ is not differentiable [technically speaking it does not have zero quadratic variation]. The above formula however does not have any useful meaning itself [one could even argue that it is not mathematically sound] although it can be used in calculations. The meaning will appear when we solve for $f\left[ B\left( t\right) \right]$ which will give

$$f\left[ B\left( t\right) \right] =f\left[ B\left( 0\right) \right... ...t) +\int_{0}^{t}\frac{1}{2}f^{\prime \prime }\left[ B\left( s\right) \right] ds$$.

This is exactly a decomposition which is of paramount use: We achieve a decomposition into an Itô integral $\int_{0}^{t}f^{\prime }\left[ B\left( s\right) \right] dB\left( s\right)$ that we examined before and a Riemann integral $\int_{0}^{t}\frac{1}{2}f^{\prime \prime }\left[ B\left( s\right) \right] ds$ which can be evaluated using elementary calculus.

The above procedure can be extended when the Brownian motion $B\left( t\right)$ is substituted by a general SDE

$$dX\left( t\right) =\mu \left[ X\left( t\right) ,t\right] dt+\sigma \left[ X\left( t\right) ,t\right] dB\left( t\right)$$   ,

and the function $f$ depends explicitly on the time $t$ as well as on the diffusion $X\left( t\right)$, namely $Y\left( t\right) =f\left[ X\left( t\right) ,t\right]$. The resulting formula is then

$$dY=\left[ f_{t}+f_{X}\mu +\frac{1}{2}f_{XX}\sigma ^{2}\right] dt+f_{X}\sigma dB$$,

where the dependences on time have been suppressed for notational convenience.