In mathematics, Itô's lemma or Itô's formula is an identity used in Itô calculus to find the differential of a time-dependent function of a stochastic process. It serves as the stochastic calculus counterpart of the chain rule. It can be heuristically derived by forming the Taylor series expansion of the function up to its second derivatives and retaining terms up to first order in the time increment and second order in the Wiener process increment. The lemma is widely employed in mathematical finance, and its best known application is in the derivation of the Black–Scholes equation for option values. This result was discovered by Japanese mathematician Kiyoshi Itô in 1951.
Motivation Suppose we are given the stochastic differential equation
d X t = μ t d t + σ t d B t , {\displaystyle dX_{t}=\mu _{t}\ dt+\sigma _{t}\ dB_{t},}
where Bt is a Wiener process and the functions μ t , σ t {\displaystyle \mu _{t},\sigma _{t}} are deterministic (not stochastic) functions of time. In general, it's not possible to write a solution X t {\displaystyle X_{t}} directly in terms of B t . {\displaystyle B_{t}.} However, we can formally write an integral solution
X t = ∫ 0 t μ s d s + ∫ 0 t σ s d B s . {\displaystyle X_{t}=\int _{0}^{t}\mu _{s}\ ds+\int _{0}^{t}\sigma _{s}\ dB_{s}.}
This expression lets us easily read off the mean and variance of X t {\displaystyle X_{t}} (which has no higher moments). First, notice that every d B t {\displaystyle \mathrm {d} B_{t}} individually has mean 0, so the expected value of X t {\displaystyle X_{t}} is simply the integral of the drift function:
E [ X t ] = ∫ 0 t μ s d s . {\displaystyle \mathrm {E} [X_{t}]=\int _{0}^{t}\mu _{s}\ ds.}
Similarly, because the d B {\displaystyle dB} terms have variance 1 and no correlation with one another, the variance of X t {\displaystyle X_{t}} is simply the integral of the variance of each infinitesimal step in the random walk:
V a r [ X t ] = ∫ 0 t σ s 2 d s . {\displaystyle \mathrm {Var} [X_{t}]=\int _{0}^{t}\sigma _{s}^{2}\ ds.}
However, sometimes we are faced with a stochastic differential equation for a more complex process Y t , {\displaystyle Y_{t},} in which the process appears on both sides of the differential equation. That is, say
d Y t = a 1 ( Y t , t ) d t + a 2 ( Y t , t ) d B t , {\displaystyle dY_{t}=a_{1}(Y_{t},t)\ dt+a_{2}(Y_{t},t)\ dB_{t},}
for some functions a 1 {\displaystyle a_{1}} and a 2 . {\displaystyle a_{2}.} In this case, we cannot immediately write a formal solution as we did for the simpler case above. Instead, we hope to write the process Y t {\displaystyle Y_{t}} as a function of a simpler process X t {\displaystyle X_{t}} taking the form above. That is, we want to identify three functions f ( t , x ) , μ t , {\displaystyle f(t,x),\mu _{t},} and σ t , {\displaystyle \sigma _{t},} such that Y t = f ( t , X t ) {\displaystyle Y_{t}=f(t,X_{t})} and d X t = μ t d t + σ t d B t . {\displaystyle dX_{t}=\mu _{t}\ dt+\sigma _{t}\ dB_{t}.} In practice, Ito's lemma is used in order to find this transformation. Finally, once we have transformed the problem into the simpler type of problem, we can determine the mean and higher moments of the process.
Derivation We derive Itô's lemma by expanding a Taylor series and applying the rules of stochastic calculus. Suppose X t {\displaystyle X_{t}} is an Itô drift-diffusion process that satisfies the stochastic differential equation
d X t = μ t d t + σ t d B t , {\displaystyle dX_{t}=\mu _{t}\,dt+\sigma _{t}\,dB_{t},}
where Bt is a Wiener process. If f(t,x) is a twice-differentiable scalar function, its expansion in a Taylor series is
Δ f ( t ) d t d t = f ( t + d t , x ) − f ( t , x ) = ∂ f ∂ t d t + 1 2 ∂ 2 f ∂ t 2 ( d t ) 2 + ⋯ Δ f ( x ) d x d x = f ( t , x + d x ) − f ( t , x ) = ∂ f ∂ x d x + 1 2 ∂ 2 f ∂ x 2 ( d x ) 2 + ⋯ {\displaystyle {\begin{aligned}{\frac {\Delta f(t)}{dt}}dt&=f(t+dt,x)-f(t,x)\\&={\frac {\partial f}{\partial t}}\,dt+{\frac {1}{2}}{\frac {\partial ^{2}f}{\partial t^{2}}}\,(dt)^{2}+\cdots \\[1ex]{\frac {\Delta f(x)}{dx}}dx&=f(t,x+dx)-f(t,x)\\&={\frac {\partial f}{\partial x}}\,dx+{\frac {1}{2}}{\frac {\partial ^{2}f}{\partial x^{2}}}\,(dx)^{2}+\cdots \end{aligned}}}
Then use the total derivative and the definition of the partial derivative f y = lim d y → 0 Δ f ( y ) d y {\displaystyle f_{y}=\lim _{dy\to 0}{\frac {\Delta f(y)}{dy}}} :
d f = f t d t + f x d x = lim d x → 0 d t → 0 ∂ f ∂ t d t + ∂ f ∂ x d x + 1 2 ( ∂ 2 f ∂ t 2 ( d t ) 2 + ∂ 2 f ∂ x 2 ( d x ) 2 ) + ⋯ . {\displaystyle {\begin{aligned}df&=f_{t}dt+f_{x}dx\\[1ex]&=\lim _{dx\to 0 \atop dt\to 0}{\frac {\partial f}{\partial t}}\,dt+{\frac {\partial f}{\partial x}}\,dx+{\frac {1}{2}}\left({\frac {\partial ^{2}f}{\partial t^{2}}}\,(dt)^{2}+{\frac {\partial ^{2}f}{\partial x^{2}}}\,(dx)^{2}\right)+\cdots .\end{aligned}}}
Substituting x = X t {\displaystyle x=X_{t}} and therefore d x = d X t = μ t d t + σ t d B t {\displaystyle dx=dX_{t}=\mu _{t}\,dt+\sigma _{t}\,dB_{t}} , we get
d f = lim d B t → 0 d t → 0 ∂ f ∂ t d t + ∂ f ∂ x ( μ t d t + σ t d B t ) + 1 2 [ ∂ 2 f ∂ t 2 ( d t ) 2 + ∂ 2 f ∂ x 2 ( μ t 2 ( d t ) 2 + 2 μ t σ t d t d B t + σ t 2 ( d B t ) 2 ) ] + ⋯ . {\displaystyle {\begin{aligned}df=\lim _{dB_{t}\to 0 \atop dt\to 0}\;&{\frac {\partial f}{\partial t}}\,dt+{\frac {\partial f}{\partial x}}\left(\mu _{t}\,dt+\sigma _{t}\,dB_{t}\right)\\&+{\frac {1}{2}}\left[{\frac {\partial ^{2}f}{\partial t^{2}}}\,{\left(dt\right)}^{2}+{\frac {\partial ^{2}f}{\partial x^{2}}}\left(\mu _{t}^{2}\,{\left(dt\right)}^{2}+2\mu _{t}\sigma _{t}\,dt\,dB_{t}+\sigma _{t}^{2}\,{\left(dB_{t}\right)}^{2}\right)\right]+\cdots .\end{aligned}}}
In the limit d t → 0 {\displaystyle dt\to 0} , the terms ( d t ) 2 {\displaystyle (dt)^{2}} and d t d B t {\displaystyle dt\,dB_{t}} tend to zero faster than d t {\displaystyle dt} . ( d B t ) 2 {\displaystyle (dB_{t})^{2}} is O ( d t ) {\displaystyle O(dt)} (due to the quadratic variation of a Wiener process which says B t 2 = O ( t ) {\displaystyle B_{t}^{2}=O(t)} ), so setting ( d t ) 2 , d t d B t {\displaystyle (dt)^{2},dt\,dB_{t}} and ( d x ) 3 {\displaystyle (dx)^{3}} terms to zero and substituting d t {\displaystyle dt} for ( d B t ) 2 {\displaystyle (dB_{t})^{2}} , and then collecting the d t {\displaystyle dt} terms, we obtain
d f = lim d t → 0 ( ∂ f ∂ t + μ t ∂ f ∂ x + σ t 2 2 ∂ 2 f ∂ x 2 ) d t + σ t ∂ f ∂ x d B t {\displaystyle df=\lim _{dt\to 0}\left({\frac {\partial f}{\partial t}}+\mu _{t}{\frac {\partial f}{\partial x}}+{\frac {\sigma _{t}^{2}}{2}}{\frac {\partial ^{2}f}{\partial x^{2}}}\right)dt+\sigma _{t}{\frac {\partial f}{\partial x}}\,dB_{t}}
as required. Alternatively,
d f = lim d t → 0 ( ∂ f ∂ t + σ t 2 2 ∂ 2 f ∂ x 2 ) d t + ∂ f ∂ x d X t {\displaystyle df=\lim _{dt\to 0}\left({\frac {\partial f}{\partial t}}+{\frac {\sigma _{t}^{2}}{2}}{\frac {\partial ^{2}f}{\partial x^{2}}}\right)dt+{\frac {\partial f}{\partial x}}\,dX_{t}}
Geometric intuition
Suppose we know that X t , X t + d t {\displaystyle X_{t},X_{t+dt}} are two jointly-Gaussian distributed random variables, and f {\displaystyle f} is nonlinear but has a continuous second derivative, then in general, neither of f ( X t ) , f ( X t + d t ) {\displaystyle f(X_{t}),f(X_{t+dt})} is Gaussian, and their joint distribution is also not Gaussian. However, since X t + d t ∣ X t {\displaystyle X_{t+dt}\mid X_{t}} is Gaussian, we might still find f ( X t + d t ) ∣ f ( X t ) {\displaystyle f(X_{t+dt})\mid f(X_{t})} is Gaussian. This is not true when d t {\displaystyle dt} is finite, but when d t {\displaystyle dt} becomes infinitesimal, this becomes true. The key idea is that X t + d t = X t + μ t d t + d W t {\displaystyle X_{t+dt}=X_{t}+\mu _{t}\,dt+dW_{t}} has a deterministic part and a noisy part. When f {\displaystyle f} is nonlinear, the noisy part has a deterministic contribution. If f {\displaystyle f} is convex, then the deterministic contribution is positive (by Jensen's inequality). To find out how large the contribution is, we write X t + d t = X t + μ t d t + σ t d t z {\displaystyle X_{t+dt}=X_{t}+\mu _{t}\,dt+\sigma _{t}{\sqrt {dt}}\,z} , where z {\displaystyle z} is a standard Gaussian, then perform Taylor expansion. f ( X t + d t ) =
f ( X t ) + f ′ ( X t ) μ t d t + f ′ ( X t ) σ t d t z + 1 2 f ″ ( X t ) ( σ t 2 z 2 d t + 2 μ t σ t z d t 3 / 2 + μ t 2 d t 2 ) + o ( d t ) =
[ f ( X t ) + f ′ ( X t ) μ t d t + 1 2 f ″ ( X t ) σ t 2 d t + o ( d t ) ] + [ f ′ ( X t ) σ t