Ito’s Lemma

Ito’s lemma, otherwise known as the Ito formula, expresses functions of stochastic processes in terms of stochastic integrals. In standard calculus, the differential of the composition of functions ${f(x), x(t)}$ satisfies ${df(x(t))=f^\prime(x(t))dx(t)}$ . This is just the chain rule for differentiation or, in integral form, it becomes the change of variables formula.

In stochastic calculus, Ito’s lemma should be used instead. For a twice differentiable function ${f}$ applied to a continuous semimartingale ${X}$ , it states the following,

$\displaystyle df(X) = f^\prime(X)\,dX + \frac{1}{2}f^{\prime\prime}(X)\,dX^2.$

This can be understood as a Taylor expansion up to second order in ${dX}$ , where the quadratic term ${dX^2\equiv d[X]}$ is the quadratic variation of the process ${X}$ .

A d-dimensional process ${X=(X^1,\ldots,X^d)}$ is said to be a semimartingale if each of its components, ${X^i}$ , are semimartingales. The first and second order partial derivatives of a function are denoted by ${D_if}$ and ${D_{ij}f}$ , and I make use of the summation convention where indices ${i,j}$ which occur twice in a single term are summed over. Then, the statement of Ito’s lemma is as follows.

Theorem 1 (Ito’s Lemma) Let ${X=(X^1,\ldots,X^d)}$ be a continuous d-dimensional semimartingale taking values in an open subset ${U\subseteq{\mathbb R}^d}$ . Then, for any twice continuously differentiable function ${f\colon U\rightarrow{\mathbb R}}$ , ${f(X)}$ is a semimartingale and,

$\displaystyle df(X) = D_if(X)\,dX^i + \frac{1}{2}D_{ij}f(X)\,d[X^i,X^j].$ (1)

To be clear, written out in integral form and explicitly summing over all indices, this is equivalent to the following,

$\displaystyle f(X) = f(X_0)+\sum_{i=1}^d \int D_if(X)\,dX^i + \frac{1}{2}\sum_{i,j=1}^d\int D_{ij}f(X)\,d[X^i,X^j].$

(2)

The proof of the result is given below. Ito’s lemma has the following consequence for the covariation ${[f(X),Y]}$ with a semimartingale ${Y}$ . The quadratic covariation terms in (2) are FV processes and, by the properties of covariations, do not contribute, giving

$\displaystyle \left[f(X),Y\right]=\sum_{i=1}^d \int D_if(X)\,d[X^i,Y].$

Example: The Doléans exponential

The Doléans exponential of a semimartingale ${X}$ is defined as a process ${U}$ solving the integral equation

$\displaystyle U = 1 + \int U_-\,dX$

(3)

so, in differential notation, ${dU=U_-\,dX}$ . The symbol ${\mathcal{E}(X)}$ is sometimes used for the Doléans exponenial. If ${U,V}$ are the Doléans exponentials of processes ${X,Y}$ then the quadratic covariation is ${d[U,V]=U_-V_-\,d[X,Y]}$ and, applying integration by parts,

$\displaystyle d(UV)=U_-V_-\,dY+U_-V_-\,dX+U_-V_-\,d[X,Y].$

So, the following product formula is satisfied,

$\displaystyle \mathcal{E}\left(X\right)\mathcal{E}\left(Y\right)=\mathcal{E}\left(X+Y+[X,Y]\right).$

For continuous semimartingales, Ito’s lemma can be used to solve for the Doléans exponential. Assuming that the process remains positive, then this gives,

$\displaystyle d\log(U) = U^{-1}\,dU -\frac{1}{2} U^{-2}\,d[U]= dX - \frac{1}{2}\,d[X].$

Therefore, ${\log(U)=X-X_0-[X]/2}$ , and the Doléans exponential of a continuous semimartingale ${X}$ is given by

$\displaystyle \mathcal{E}(X) = \exp\left(X-X_0-\frac{1}{2}[X]\right).$

Example: Geometric Brownian motion

A geometric Brownian motion with volatility ${\sigma}$ and drift ${\mu}$ is defined as the solution to the stochastic differential equation

$\displaystyle dX = \sigma X\,dB + \mu X\,dt$

where ${B}$ is a standard Brownian motion. So, ${X/X_0}$ is the Doléans exponential of ${Z_t\equiv\sigma B_t+\mu t}$ . As Brownian motion has quadratic variation ${[B]_t=t}$ , this gives ${[Z]_t=\sigma^2t}$ and, from the solution above for the Doléans exponential,

$\displaystyle X_t = X_0\exp\left(\sigma B_t + (\mu-\sigma^2/2)t\right).$

Proof of Ito’s Lemma

Writing ${\delta X\equiv X_t-X_s}$ for some times ${s,t}$ , a second order Taylor expansion gives

$\displaystyle f(X_t) = f(X_s) + D_if(X_s)\delta X^i+\frac{1}{2}D_{ij}f(X_s)\delta X^i\delta X^j + R_{s,t}.$

(4)

As ${X}$ is continuous and contained in the open set ${U\subseteq{\mathbb R}^d}$ , the image of its sample path over any bounded interval ${[0,T]}$ is almost surely a compact subset of ${U}$ . It follows that the remainder term ${R_{s,t}}$ is almost surely of the order of ${o(\Vert X_t-X_s\Vert^2)}$ uniformly over the interval ${[0,T]}$ . That is, ${\Vert X_t-X_s\Vert^{-2}R_{s,t}\rightarrow 0}$ whenever ${X_t-X_s\rightarrow 0}$ on ${s,t\in[0,T]}$ .

Now, for a positive integer n, partition the interval ${[0,T]}$ into n equal segments. That is, set ${t^n_k=kT/n}$ for ${k=0,1,\ldots,n}$ . Using the notation ${\delta_{n,k}X\equiv X_{t^n_k}-X_{t^n_{k-1}}}$ , summing (4) over the intervals ${(t^n_{k-1},t^n_k)}$ gives,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle f(X_T) = &\displaystyle f(X_0) + \sum_{k=1}^n D_if(X_{t^n_{k-1}})\delta_{n,k}X^i\smallskip\\ &\displaystyle+ \frac{1}{2}\sum_{k=1}^nD_{ij}f(X_{t^n_{k-1}})\delta_{n,k}X^i\delta_{n,k}X^j + \sum_{k=1}^n R_{t^n_{k-1},t^n_k}. \end{array}$

(5)

Other than the final remainder term, which we will show converges to zero, this expression is a Riemann sum approximation to equation (2). I show separately that each of the terms does indeed converge to the expected limit as ${n\rightarrow\infty}$ . As I will make use of these results again in the extension of Ito’s formula to noncontinuous processes, the more general context of cadlag processes is used where possible. However, for the application to this post, only continuous processes are required, so that ${X_{t-}=X_t}$ .

Lemma 2 If ${Y}$ is a semimartingale and ${U}$ is a cadlag adapted process then

$\displaystyle \sum_{k=1}^nU_{t^n_{k-1}}\delta_{n,k} Y\rightarrow\int_0^T U_-\,dY$

in probability as ${n\rightarrow\infty}$ .

Proof: For any ${0<t\le T}$ the following limit is clear,

$\displaystyle \sum_{k=1}^n U_{t^n_{k-1}}1_{\{t^n_{k-1}<t\le t^n_k\}}\rightarrow U_{t-}.$

Furthermore, the left hand side is dominated by the locally bounded process ${U^*_{t-}\equiv\sup_{s<t}\vert U_s\vert}$ . So, the result follows from dominated convergence

$\displaystyle \sum_{k=1}^n U_{t^n_{k-1}}\delta_{n,k}Y = \int_0^T\sum_{k=1}^n U_{t^n_{k-1}}1_{(t^n_{k-1},t^n_k]}\,dY \rightarrow \int_0^T U_{-}\,dY.$

⬜

Applying this to the first summation term of (5) gives

$\displaystyle \sum_{k=1}^n D_if(X_{t^n_{k-1}})\delta_{n,k}X^i\rightarrow\int_0^T D_if(X_-)\,dX^i$

(6)

in probability as ${n\rightarrow\infty}$ .

Next, the second order terms of the Taylor expansion will converge to the required integral with respect to the quadratic covariation.

Lemma 3 If ${Y,Z}$ are semimartingales and ${U}$ is a cadlag adapted process then

$\displaystyle \sum_{k=1}^nU_{t^n_{k-1}}\delta_{n,k} Y\delta_{n,k}Z\rightarrow\int_0^T U_-\,d[Y,Z]$

in probability as ${n\rightarrow\infty}$ .

Proof: First, rearranging the expression a bit gives

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} \displaystyle\sum_{k=1}^nU_{t^n_{k-1}}\delta_{n,k} Y\delta_{n,k}Z =&\displaystyle\sum_{k=1}^nU_{t^n_{k-1}}\delta_{n,k}(YZ) - \sum_{k=1}^n(UY)_{t^n_{k-1}}\delta_{n,k}Z\smallskip\\ &\displaystyle-\sum_{k=1}^n(UZ)_{t^n_{k-1}}\delta_{n,k}Y. \end{array}$

Lemma 2 can then be applied to each of the terms on the right hand side to give convergence in probability,

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rcl} \displaystyle\sum_{k=1}^nU_{t^n_{k-1}}\delta_{n,k} Y\delta_{n,k}Z &\displaystyle\rightarrow &\displaystyle \int_0^T U_{-}\,d(YZ) - \int_0^T(UY)_{-}\,dZ\smallskip\\ &&\displaystyle -\int_0^T(UZ)_{-}\,dY\smallskip\\ &\displaystyle=&\displaystyle\int_0^TU_-\,d[Y,Z]. \end{array}$

The final equality here is the integration by parts formula. ⬜

Applying this result to the second summation term of (5) gives

$\displaystyle \sum_{k=1}^n D_{ij}f(X_{t^n_{k-1}})\delta_{n,k}X^i\delta_{n,k}X^j\rightarrow\int_0^T D_{ij}f(X_-)\,d[X^i,X^j]$

(7)

in probability as ${n\rightarrow\infty}$ .

Finally, the remainder terms in the Taylor expansion will converge to zero, as required.

Lemma 4 If ${X}$ is a continuous d-dimensional semimartingale and ${R_{s,t}}$ is almost surely of order ${o(\Vert X_t-X_s\Vert^2)}$ on the interval ${[0,T]}$ then,

$\displaystyle \sum_{k=1}^n R_{t^n_{k-1},t^n_k}\rightarrow 0$ (8)

in probability as ${n\rightarrow \infty}$ .

Proof: As the sample paths of ${X}$ are continuous, it follows that ${\max_k\Vert \delta_{n,k}X\Vert}$ goes to zero as ${n\rightarrow\infty}$ . So, as ${R_{s,t}}$ is of order ${o(\Vert X_t-X_s\Vert^2)}$ , the sequence of random variables

$\displaystyle \epsilon_n\equiv\max\left\{\vert R_{t^n_{k-1},t^n_k}\vert/\Vert\delta_{n,k}X\Vert^2\colon k=1,\ldots,n\right\}$

will also go to zero.

$\displaystyle \sum_{k=1}^n \vert R_{t^n_{k-1},t^n_k}\vert\le \epsilon_n \sum_{i=1}^d\sum_{k=1}^n\left(\delta_{n,k} X^i\right)^2.$

The sum over ${k}$ on the right hand side tends to ${[X^i]_T}$ in probability as ${n\rightarrow\infty}$ and, in particular, is bounded in probability. As ${\epsilon_n\rightarrow 0}$ this shows that the right hand side goes to zero in probability as ${n\rightarrow\infty}$ . ⬜

Taking the limit ${n\rightarrow\infty}$ in equation (5) and applying (6,7,8) gives the required integral expression,

$\displaystyle f(X_T) = f(X_0)+\int_0^T D_if(X)\,dX^i+\int_0^TD_{ij}f(X)\,d[X^i,X^j].$

Finally, as ${f(X)}$ is expressed as a sum of stochastic integrals, it is a semimartingale.

12 thoughts on “Ito’s Lemma”

Dave says:

12 February 12 at 2:53 PM

When proving Ito’s lemma, you use Taylor approximation and then show that the terms converge.
The convergence is in probability, but I believe that Ito’s lemma holds almost surely.
Is there no contradiction there?

1. George Lowther says:
  
  12 February 12 at 9:25 PM
  
  Hi.
  
  No, there’s no contradiction. If you have a sequence of random variables Uⁿ which you can show converge in probability to two limits U and V, then it follows that U = V almost surely. That is the situation here, where Uⁿ is equal to f(X_T) for each n but, also, by breaking it into terms corresponding to the Taylor expansion, we show that it converges in probability to the right hand side of Ito’s formula. Hence, Ito’s formula holds almost surely.
  
  1. Anonymous says:
    
    20 December 22 at 3:22 AM
    
    Dear almost sure，
    
    You get the second part of convergence in probability using integration by part. Why you can write it in that way since what I thought was it is derived by the ito formula right?
    
Anonymous says:

27 February 15 at 6:34 PM

Concerning Lemma 2: Why is the running supremum process of a cadlag adapted process locally bounded?

1. Anonymous says:
  
  27 February 15 at 7:17 PM
  
  I think I understand that now (it is easy to find the localizting sequence), but why does the dominated convergence theorem hold for dominating functions that are locally bounded?
  
  1. George Lowther says:
    
    6 June 16 at 3:45 AM
    
    Locally bounded predictable processes are Y-integrable (Lemma 5 of post on properties of the stochastic integral), so dominated convergence holds.
    
Anonymous says:

29 August 19 at 5:00 PM

Hi there. I’m just newbie to the world of semimartingales. I have a question regarding the second term of Ito’s formula applied to the Doleans exponential. It says there -1/2U^{-2}d[U] = -1/2d[X]. Are you saying that -1/2U^{-2}d[U] = -1/2U^{-2}U^{2}d[X] ?

Thank you for even reading this comment.

1. George Lowther says:
  
  20 September 19 at 12:51 AM
  
  I’m not sure what you are asking, but it is true that $U^{-2}d[U]=d[X]$ .
  
Mario says:

20 January 22 at 6:17 AM

What is U_ in your Doleans example?

1. George Lowther says:
  
  20 February 22 at 4:56 PM
  
  Well, U is the Doleans exponential by definition. If you are asking about the ‘-‘ subscript, this refers to the left-limit (I use this notation throughout my notes).
  
  $\displaystyle U_{t-}=\lim_{s\uparrow\uparrow t}U_s$
  
leejulian6702 says:

26 January 22 at 6:35 PM

George, what remainder form are you using for $R_{s,t}$ here to get that the remainder term is of order $o(\Vert X_t - X_s \Vert^2)$ uniformly over [s,t]? I can get this from the Lagrange form if f is of C^3, but I don’t know of a remainder formula that gives this for just C^2 function even if it is compactly supported.

1. George Lowther says:
  
  20 February 22 at 4:26 PM
  
  I just did a quick check online and didn’t see quite the statement that I would like, although I am pretty sure it is standard. Look at the integral form for the remainder.
  $\displaystyle f(y)=f(x)+f^\prime(x)(y-x)+\frac12f^{\prime\prime}(x)(y-x)^2+\int_x^y \left(f^{\prime\prime}(z)-f^{\prime\prime}(x)\right)(z-x)dz$
  For the multidimensional case, the same formula applies if the integral is along a line from x to y, and we use the chain rule to express as partial derivates.
  By uniform continuity of $f^{\prime\prime}$ , the integral is $o(\lvert y-x\rvert^2)$ .

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