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Ito Multiplication Table 📂Stochastic Differential Equations

Ito Multiplication Table

Build-up

s<t<t+us< t < t+u Suppose, meet the following conditions for a stochastic process {Wt}\left\{ W_{t} \right\} to be called a Wiener process.

  • (i): W0=0W_{0} = 0
  • (ii): (Wt+uWt)Ws\left( W_{t+u} - W_{t} \right) \perp W_{s}
  • (iii): (Wt+uWt)N(0,u)\left( W_{t+u} - W_{t} \right) \sim N ( 0, u )
  • (iv): Sample paths of WtW_{t} are almost surely continuous.

The Wiener process has the following properties:

  • [1]: WtN(0,t)\displaystyle W_{t} \sim N ( 0 , t )
  • [2]: E(Wt)=0\displaystyle E ( W_{t} ) = 0
  • [3]: Var(Wt)=t\displaystyle \operatorname{Var} ( W_{t} ) = t
  • [4]: cov(Wt,Ws)=12(t+sts)=min{t,s}\displaystyle \text{cov} ( W_{t} , W_{s} ) = {{1} \over {2}} (|t| + |s| - |t-s|) = \min \left\{ t , s \right\}

Consider a very short infinitesimal interval [t,t+dt]\left[ t , t + d t \right] of the Wiener process {Wt}t0\left\{ W_{t} \right\}_{t \ge 0}. Although it is not an analytically rigorous assumption, let dt>0dt > 0 be (dt)1/2>0\left( dt \right)^{1/2} > 0, and small enough to be considered as (dt)k=0\left( dt \right)^{k} = 0 for any k=2,3,k = 2 , 3, \cdots. In algebraic terms, under this assumption, we treat α+βdt\alpha + \beta dt as an infinitesimal.

Now, suppose dWt:=Wt+dtWtdW_{t} := W_{t + dt} - W_{t}, and we want to consider the multiplications between dtdt and dWtd W_{t}.


Part 1. (dt)2=0\left( dt \right)^{2} = 0

Of course, dt>0dt > 0 holds, but let’s assume dtdt is small enough such that (dt)2=0\left( dt \right)^{2} = 0.


Part 2. dtdWt=0dt d W_{t} = 0

Since WtW_{t} is assumed to follow a Wiener process, it adheres to a normal distribution as dWtN(0,dt2)d W_{t} \sim N \left( 0, \sqrt{dt}^{2} \right).

Properties of mean and variance:

  • [2]: E(aX+b)=aE(X)+bE(aX + b) = a E(X) + b
  • [5]: Var(aX+b)=a2Var(X)\operatorname{Var} (aX + b) = a^2 \operatorname{Var} (X)

The expectation of dtdWtdt d W_{t} brings the constant dtdt outside, E(dtdWt)=dtE(dWt)=dt0=0 E \left( dt d W_{t} \right) = dt E \left( d W_{t} \right) = dt \cdot 0 = 0 Similarly, the variance of dtdWtdt d W_{t} also removes the squared dtdt, Var(dtdWt)=(dt)2Var(dWt)=0Var(dWt)=0 \operatorname{Var} \left( dt d W_{t} \right) = (dt)^{2} \operatorname{Var} \left( d W_{t} \right) = 0 \cdot \operatorname{Var} \left( d W_{t} \right) = 0 Thus, since dtdWtdt d W_{t} has a variance of 00 and an expectation of 00, it must be dtdWt=dWtdt=0 dt d W_{t} = d W_{t} dt = 0


Part 3. (dWt)2=dt\left( d W_{t} \right)^{2} = dt

From Var(dWt)=dt\operatorname{Var} \left( d W_{t} \right) = dt, computing the expectation of dWtdWtd W_{t} \cdot d W_{t}, dt=Var(dWt)=E((dWt)2)[E(dWt)]2=E((dWt)2)02 \begin{align*} dt =& \operatorname{Var} \left( d W_{t} \right) \\ =& E \left( \left( d W_{t} \right)^{2} \right) - \left[ E \left( d W_{t} \right) \right]^{2} \\ =& E \left( \left( d W_{t} \right)^{2} \right) - 0^{2} \end{align*} Thus, it is E((dWt)2)=dtE \left( \left( d W_{t} \right)^{2} \right) = dt.

Expectation of even powers of normally distributed random variables with zero mean: Given a random variable XX following a normal distribution as N(0,σ2)N \left( 0 , \sigma^{2} \right), the expectation of its even power XnX^{n} can be recursively expressed as E(Xn)=(n1)σ2E(Xn2) E \left( X^{n} \right) = (n - 1) \sigma^{2} E \left( X^{n-2} \right) The equation E(Xn)E \left( X^{n} \right) holds nn when it is odd, but for even, it is given by E(X2n)=(2n1)!!σ2n E \left( X^{2n} \right) = \left( 2n - 1 \right)!! \sigma^{2n} Here, the double exclamation mark k!!=k(k2)k!! = k \cdot \left( k - 2 \right) \cdots denotes the double factorial.

Assuming dWtN(0,dt2)d W_{t} \sim N \left( 0, \sqrt{dt}^{2} \right), the variable dWtd W_{t} follows a normal distribution with zero mean, and the variance of (dWt)2\left( d W_{t} \right)^{2} follows from the expectation of the squared random variable E(X2n)=(2n1)!!σ2nE \left( X^{2n} \right) = \left( 2n - 1 \right)!! \sigma^{2n}, Var((dWt)2)=E([(dWt)2]2)[E((dWt)2)]2=E((dWt)22)[dt]2=(221)dt22dt2=3dt2dt2=2dt2=0 \begin{align*} \operatorname{Var} \left( \left( d W_{t} \right)^{2} \right) =& E \left( \left[ \left( d W_{t} \right)^{2} \right]^{2} \right) - \left[ E \left( \left( d W_{t} \right)^{2} \right) \right]^{2} \\ =& E \left( \left( d W_{t} \right)^{2 \cdot 2} \right) - \left[ dt \right]^{2} \\ =& \left( 2 \cdot 2 - 1 \right) \sqrt{dt}^{2 \cdot 2} - dt^{2} \\ =& 3 dt^{2} - dt^{2} \\ =& 2 dt^{2} \\ =& 0 \end{align*} Therefore, (dWt)2\left( d W_{t} \right)^{2} should carry an expected value of dtdt followed by (dWt)2=dt \left( d W_{t} \right)^{2} = dt

Summary 1

Assuming α+βdt\alpha + \beta dt to be an infinitesimal, the product of dtdt and dWtd W_{t} is given by, (dt)2=0dtdWt=0dWtdt=0(dWt)2=dt \begin{align*} \left( dt \right)^{2} =& 0 \\ dt d W_{t} =& 0 \\ d W_{t} dt =& 0 \\ \left( d W_{t} \right)^{2} =& dt \end{align*}


  1. Panik. (2017). Stochastic Differential Equations: An Introduction with Applications in Population Dynamics Modeling: p129. ↩︎