Categories: Mathematics, Stochastic analysis.

Itō integral

The Itō integral offers a way to integrate a given stochastic process GtG_t with respect to a Wiener process BtB_t, which is also a stochastic process. The Itō integral ItI_t of GtG_t is defined as follows:

ItabGtdBtlimh0t=at=bGt(Bt+hBt)\begin{aligned} \boxed{ I_t \equiv \int_a^b G_t \dd{B_t} \equiv \lim_{h \to 0} \sum_{t = a}^{t = b} G_t \big(B_{t + h} - B_t\big) } \end{aligned}

Where have partitioned the time interval [a,b][a, b] into steps of size hh. The above integral exists if GtG_t and BtB_t are adapted to a common filtration Ft\mathcal{F}_t, and E[Gt2]\mathbf{E}[G_t^2] is integrable for t[a,b]t \in [a, b]. If ItI_t exists, GtG_t is said to be Itō-integrable with respect to BtB_t.


Consider the following simple first-order differential equation for XtX_t, for some function ff:

dXtdt=f(Xt)\begin{aligned} \dv{X_t}{t} = f(X_t) \end{aligned}

This can be solved numerically using the explicit Euler scheme by discretizing it with step size hh, which can be applied recursively, leading to:

Xt+hXt+f(Xt)h    XtX0+s=0s=tf(Xs)h\begin{aligned} X_{t+h} \approx X_{t} + f(X_t) \: h \quad \implies \quad X_t \approx X_0 + \sum_{s = 0}^{s = t} f(X_s) \: h \end{aligned}

In the limit h0h \to 0, this leads to the following unsurprising integral for XtX_t:

0tf(Xs)ds=limh0s=0s=tf(Xs)h\begin{aligned} \int_0^t f(X_s) \dd{s} = \lim_{h \to 0} \sum_{s = 0}^{s = t} f(X_s) \: h \end{aligned}

In contrast, consider the stochastic differential equation below, where ξt\xi_t represents white noise, which is informally the tt-derivative of the Wiener process ξt=dBt/dt\xi_t = \idv{B_t}{t}:

dXtdt=g(Xt)ξt\begin{aligned} \dv{X_t}{t} = g(X_t) \: \xi_t \end{aligned}

Now XtX_t is not deterministic, since ξt\xi_t is derived from a random variable BtB_t. If g=1g = 1, we expect Xt=X0+BtX_t = X_0 + B_t. With this in mind, we introduce the Euler-Maruyama scheme:

Xt+h=Xt+g(Xt)(ξt+hξt)h=Xt+g(Xt)(Bt+hBt)\begin{aligned} X_{t+h} &= X_t + g(X_t) \: (\xi_{t+h} - \xi_t) \: h \\ &= X_t + g(X_t) \: (B_{t+h} - B_t) \end{aligned}

We would like to turn this into an integral for XtX_t, as we did above. Therefore, we state:

Xt=X0+0tg(Xs)dBs\begin{aligned} X_t = X_0 + \int_0^t g(X_s) \dd{B_s} \end{aligned}

This integral is defined as below, analogously to the first, but with hh replaced by the increment Bt+h ⁣ ⁣BtB_{t+h} \!-\! B_t of a Wiener process. This is an Itō integral:

0tg(Xs)dBslimh0s=0s=tg(Xs)(Bs+hBs)\begin{aligned} \int_0^t g(X_s) \dd{B_s} \equiv \lim_{h \to 0} \sum_{s = 0}^{s = t} g(X_s) \big(B_{s + h} - B_s\big) \end{aligned}

For more information about applying the Itō integral in this way, see the Itō calculus.


Since GtG_t and BtB_t must be known (i.e. Ft\mathcal{F}_t-adapted) in order to evaluate the Itō integral ItI_t at any given tt, it logically follows that ItI_t is also Ft\mathcal{F}_t-adapted.

Because the Itō integral is defined as the limit of a sum of linear terms, it inherits this linearity. Consider two Itō-integrable processes GtG_t and HtH_t, and two constants v,wRv, w \in \mathbb{R}:

abvGt+wHtdBt=v ⁣abGtdBt+w ⁣abHtdBt\begin{aligned} \int_a^b v G_t + w H_t \dd{B_t} = v\! \int_a^b G_t \dd{B_t} +\: w\! \int_a^b H_t \dd{B_t} \end{aligned}

By adding multiple summations, the Itō integral clearly satisfies, for a<b<ca < b < c:

acGtdBt=abGtdBt+bcGtdBt\begin{aligned} \int_a^c G_t \dd{B_t} = \int_a^b G_t \dd{B_t} + \int_b^c G_t \dd{B_t} \end{aligned}

A more interesting property is the Itō isometry, which expresses the expectation of the square of an Itō integral of GtG_t as a simpler “ordinary” integral of the expectation of Gt2G_t^2 (which exists by the definition of Itō-integrability):

E(abGtdBt)2=abE[Gt2]dt\begin{aligned} \boxed{ \mathbf{E} \bigg( \int_a^b G_t \dd{B_t} \bigg)^2 = \int_a^b \mathbf{E} \big[ G_t^2 \big] \dd{t} } \end{aligned}

We write out the left-hand side of the Itō isometry, where eventually h0h \to 0:

E[t=at=bGt(Bt+h ⁣ ⁣Bt)]2=t=at=bs=as=bE[GtGs(Bt+h ⁣ ⁣Bt)(Bs+h ⁣ ⁣Bs)]\begin{aligned} \mathbf{E} \bigg[ \sum_{t = a}^{t = b} G_t (B_{t + h} \!-\! B_t) \bigg]^2 &= \sum_{t = a}^{t = b} \sum_{s = a}^{s = b} \mathbf{E} \bigg[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \bigg] \end{aligned}

In the particular case ts ⁣+ ⁣ht \ge s \!+\! h, a given term of this summation can be rewritten as follows using the law of total expectation (see conditional expectation):

E[GtGs(Bt+h ⁣ ⁣Bt)(Bs+h ⁣ ⁣Bs)]=E[E[GtGs(Bt+h ⁣ ⁣Bt)(Bs+h ⁣ ⁣Bs)Ft]]\begin{aligned} \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big] = \mathbf{E} \bigg[ \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big| \mathcal{F}_t \Big] \bigg] \end{aligned}

Recall that GtG_t and BtB_t are adapted to Ft\mathcal{F}_t: at time tt, we have information Ft\mathcal{F}_t, which includes knowledge of the realized values GtG_t and BtB_t. Since ts ⁣+ ⁣ht \ge s \!+\! h by assumption, we can simply factor out the known quantities:

E[GtGs(Bt+h ⁣ ⁣Bt)(Bs+h ⁣ ⁣Bs)]=E[GtGs(Bs+h ⁣ ⁣Bs)E[(Bt+h ⁣ ⁣Bt)Ft]]\begin{aligned} \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big] = \mathbf{E} \bigg[ G_t G_s (B_{s + h} \!-\! B_s) \: \mathbf{E} \Big[ (B_{t + h} \!-\! B_t) \Big| \mathcal{F}_t \Big] \bigg] \end{aligned}

However, Ft\mathcal{F}_t says nothing about the increment (Bt+h ⁣ ⁣Bt)N(0,h)(B_{t + h} \!-\! B_t) \sim \mathcal{N}(0, h), meaning that the conditional expectation is zero:

E[GtGs(Bt+h ⁣ ⁣Bt)(Bs+h ⁣ ⁣Bs)]=0for  ts+h\begin{aligned} \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big] = 0 \qquad \mathrm{for}\; t \ge s + h \end{aligned}

By swapping ss and tt, the exact same result can be obtained for st ⁣+ ⁣hs \ge t \!+\! h:

E[GtGs(Bt+h ⁣ ⁣Bt)(Bs+h ⁣ ⁣Bs)]=0for  st+h\begin{aligned} \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big] = 0 \qquad \mathrm{for}\; s \ge t + h \end{aligned}

This leaves only one case which can be nonzero: [t,t ⁣+ ⁣h]=[s,s ⁣+ ⁣h][t, t\!+\!h] = [s, s\!+\!h]. Applying the law of total expectation again yields:

E[t=at=bGt(Bt+h ⁣ ⁣Bt)]2=t=at=bE[Gt2(Bt+h ⁣ ⁣Bt)2]=t=at=bE[E[Gt2(Bt+h ⁣ ⁣Bt)2Ft]]\begin{aligned} \mathbf{E} \bigg[ \sum_{t = a}^{t = b} G_t (B_{t + h} \!-\! B_t) \bigg]^2 &= \sum_{t = a}^{t = b} \mathbf{E} \Big[ G_t^2 (B_{t + h} \!-\! B_t)^2 \Big] \\ &= \sum_{t = a}^{t = b} \mathbf{E} \bigg[ \mathbf{E} \Big[ G_t^2 (B_{t + h} \!-\! B_t)^2 \Big| \mathcal{F}_t \Big] \bigg] \end{aligned}

We know GtG_t, and the expectation value of (Bt+h ⁣ ⁣Bt)2(B_{t+h} \!-\! B_t)^2, since the increment is normally distributed, is simply the variance hh:

E[t=at=bGt(Bt+h ⁣ ⁣Bt)]2=t=at=bE[Gt2]habE[Gt2]dt\begin{aligned} \mathbf{E} \bigg[ \sum_{t = a}^{t = b} G_t (B_{t + h} \!-\! B_t) \bigg]^2 &= \sum_{t = a}^{t = b} \mathbf{E} \big[ G_t^2 \big] h \longrightarrow \int_a^b \mathbf{E} \big[ G_t^2 \big] \dd{t} \end{aligned}

Furthermore, Itō integrals are martingales, meaning that the average noise contribution is zero, which makes intuitive sense, since true white noise cannot be biased.

We will prove that an arbitrary Itō integral ItI_t is a martingale. Using additivity, we know that the increment It ⁣ ⁣IsI_t \!-\! I_s is as follows, given information Fs\mathcal{F}_s:

E[It ⁣ ⁣IsFs]=E[stGudBuFs]=limh0u=su=tE[Gu(Bu+h ⁣ ⁣Bu)Fs]\begin{aligned} \mathbf{E} \big[ I_t \!-\! I_s | \mathcal{F}_s \big] = \mathbf{E} \bigg[ \int_s^t G_u \dd{B_u} \bigg| \mathcal{F}_s \bigg] = \lim_{h \to 0} \sum_{u = s}^{u = t} \mathbf{E} \Big[ G_u (B_{u + h} \!-\! B_u) \Big| \mathcal{F}_s \Big] \end{aligned}

We rewrite this conditional expectation using the tower property for some FuFs\mathcal{F}_u \supset \mathcal{F}_s, such that GuG_u and BuB_u are known, but Bu+h ⁣ ⁣BuB_{u+h} \!-\! B_u is not:

E[It ⁣ ⁣IsFs]=limh0u=su=tE[E[Gu(Bu+h ⁣ ⁣Bu)Fu]Fs]=0\begin{aligned} \mathbf{E} \big[ I_t \!-\! I_s | \mathcal{F}_s \big] &= \lim_{h \to 0} \sum_{u = s}^{u = t} \mathbf{E} \bigg[ \mathbf{E} \Big[ G_u (B_{u + h} \!-\! B_u) \Big| \mathcal{F}_u \Big] \bigg| \mathcal{F}_s \bigg] = 0 \end{aligned}

We now have everything we need to calculate E[ItFs]\mathbf{E} [ I_t | \mathcal{F_s} ], giving the martingale property:

E[ItFs]=E[IsFs]+E[It ⁣ ⁣IsFs]=Is+E[It ⁣ ⁣IsFs]=Is\begin{aligned} \mathbf{E} \big[ I_t | \mathcal{F}_s \big] = \mathbf{E} \big[ I_s | \mathcal{F}_s \big] + \mathbf{E} \big[ I_t \!-\! I_s | \mathcal{F}_s \big] = I_s + \mathbf{E} \big[ I_t \!-\! I_s | \mathcal{F}_s \big] = I_s \end{aligned}

For the existence of ItI_t, we need E[Gt2]\mathbf{E}[G_t^2] to be integrable over the target interval, so from the Itō isometry we have E[I]2<\mathbf{E}[I]^2 < \infty, and therefore E[I]<\mathbf{E}[I] < \infty, so ItI_t has all the properties of a Martingale, since it is trivially Ft\mathcal{F}_t-adapted.


  1. U.H. Thygesen, Lecture notes on diffusions and stochastic differential equations, 2021, Polyteknisk Kompendie.