Milstein stochastic taylor and iterated integrals — Examples

Define $G(t,w)=x_0{e}^{(\mu -{\nu }^2\mathrm{/}2)t+\nu w}$. Its derivatives are $G_t=(\mu -{\nu }^2\mathrm{/}2)G$, $G_w=\nu G$, and $G_{ww}={\nu }^{2}G$. It^o's formula [F2] therefore gives $dG(t,W_t)=(\mu -{\nu }^2\mathrm{/}2)Gdt+\frac{1}{2}{\nu }^{2}Gdt+\nu Gd{W}_t=\mu Gdt+\nu Gd{W}_t$, with $G(0,W_0)=x_0$. Thus the displayed process solves the SDE, including $x_0=0$.

For $b(x)=\mu x$ and $\sigma (x)=\nu x$, ${\sigma }^{\mathrm{\prime }}(x)=\nu$ and $\sigma (x){\sigma }^{\mathrm{\prime }}(x)={\nu }^{2}x$. Substituting these coefficients into [F1] and factoring $Y_n$ gives the displayed Milstein update. If $\nu =0$, its correction vanishes and it reduces to the deterministic Euler step.

Here $b(x)=0$, $\sigma (x)=\sin x$, and ${\sigma }^{\mathrm{\prime }}(x)=\cos x$. At $Y_0=\pi \mathrm{/}4$, both $\sigma (Y_0)$ and ${\sigma }^{\mathrm{\prime }}(Y_0)$ equal $1\mathrm{/}\sqrt{2}$, so [F1] gives the stated update with correction coefficient $\frac{1}{2}(1\mathrm{/}\sqrt{2}{)}^2=1\mathrm{/}4$. The function $\sin x$ and its derivatives through order three are bounded and continuous, so it meets the regularity premise of [L1].

By [F2], the correction has conditional mean $\frac{1}{4}(h-h)=0$ and variance $\frac{1}{16}(3h^2-2h^2+h^2)=h^2\mathrm{/}8$. Its covariance with the linear noise term is $\frac{1}{4\sqrt{2}}\mathbb{E}[(\mathrm{\Delta }{W}_0{)}^3-h\mathrm{\Delta }{W}_0]=0$. Hence the one-step increment variance is $h\mathrm{/}2+h^2\mathrm{/}8$, and the derivative correction is a separate nonzero fluctuation.

Applying [L1] to the SDE with $b=0$ and $\sigma (x)=\sin x$ yields ${\max }_{0\le n\le N}\mathbb{E}\mathrm{\mid }{Y}_n-X_{t_n}{\mathrm{\mid }}^2\le C_T{h}^2$, with $C_T$ independent of $N$, on the supplied uniform grid. Taking square roots yields grid root-mean-square error $O(h)$ on every fixed finite horizon. This is the theorem's bounded-coefficient multiplicative-noise case under the inherited AC convention [A1]; it supplies no weak-order-two statement for Heun.

The Jacobian of ${\sigma }_2$ sends $(1,0)$ to $(0,1)$, so $L_1{\sigma }_2=(0,1)$; the Jacobian of the constant field ${\sigma }_1$ is zero, so $L_2{\sigma }_1=(0,0)$. Thus commutativity fails. Starting from $X_0=(0,0)$, the first coordinate is $W_t^1$ and the second is ${\int }_0^t{W}_s^{1}d{W}_s^2$. The general Milstein Taylor update [F2] therefore has second-coordinate increment $I_{12}$ on the first step.

By [L1], the endpoints determine $I_{12}+I_{21}=UV$, where $U=\mathrm{\Delta }{W}^1$ and $V=\mathrm{\Delta }{W}^2$. Let ${\pi }_n=\{0=s_0<\cdots <s_{q_n}=h\}$ be deterministic partitions with mesh $\mathrm{\mid }{\pi }_n\mathrm{\mid }\to 0$, and put $S_n={\sum }_{k=0}^{q_n-1}{W}_{s_k}^1(W_{s_{k+1}}^2-W_{s_k}^2)$. The It^o isometry [F3] gives $\mathbb{E}\mathrm{\mid }{S}_n-I_{12}{\mathrm{\mid }}^2={\sum }_k{\int }_{s_k}^{s_{k+1}}(s-s_k)ds\le h\mathrm{\mid }{\pi }_n\mathrm{\mid }\mathrm{/}2\to 0$.

For $0<s<h$, write $A=W_s^1$ and $U-A=W_h^1-W_s^1$; their joint density is $p_s(a)p_{h-s}(u-a)$, where $p_r$ is the $N(0,r)$ density. Dividing by the marginal density $p_h(u)$ and completing the square gives $\mathbb{E}[A\mid U=u]=(s\mathrm{/}h)u$. At $s=0$ and $s=h$ the same formula follows from $A=0$ and $A=U$, respectively. For ${\mathrm{\ell }}_k=s_{k+1}-s_k$, the pair $B_k=W_{s_{k+1}}^2-W_{s_k}^2$ and $V-B_k$ has density $p_{{\mathrm{\ell }}_k}(b)p_{h-{\mathrm{\ell }}_k}(v-b)$ when ${\mathrm{\ell }}_k<h$; completing the square gives $\mathbb{E}[B_k\mid V=v]=({\mathrm{\ell }}_k\mathrm{/}h)v$, and if ${\mathrm{\ell }}_k=h$ then $B_k=V$ gives the same result. Independence of the two Brownian coordinate processes [F1] makes the joint density of $(A,U,B_k,V)$ the product of these two pair densities. Dividing by $p_h(u)p_h(v)$ shows that the conditional density of $(A,B_k)$ given $(U,V)=(u,v)$ factors, so $\mathbb{E}[A{B}_k\mid U,V]=(s\mathrm{/}h)U({\mathrm{\ell }}_k\mathrm{/}h)V$. Therefore $\mathbb{E}[S_n\mid \sigma (U,V)]=\frac{UV}{h^2}{\sum }_k{s}_k(s_{k+1}-s_k)⟶\frac{UV}{2},$ because the sum is a left Riemann sum for ${\int }_0^{h}sds=h^2\mathrm{/}2$. By [F7], step 1.2 implies $S_n\to I_{12}$ in $L^1$; conditional expectation is $L^1$-contractive by [F6], so [F5] gives $\mathbb{E}[I_{12}\mid \sigma (U,V)]=UV\mathrm{/}2$.

The isometry [F3] gives $\mathbb{E}{I}_{12}^2={\int }_0^h\mathbb{E}(W_s^1{)}^{2}ds=h^2\mathrm{/}2$. Also $U,V$ are independent with second moments $h$, so step 2.1 gives $J:=\mathbb{E}[I_{12}\mid \sigma (U,V)]=UV\mathrm{/}2$ and $\mathbb{E}{J}^2=\mathbb{E}[U^2{V}^2]\mathrm{/}4=h^2\mathrm{/}4$. By [F8], $\mathbb{E}[(I_{12}-J)J]=0$; expanding the square therefore gives $\mathbb{E}\mathrm{\mid }{I}_{12}-J{\mathrm{\mid }}^2=\mathbb{E}{I}_{12}^2-\mathbb{E}{J}^2=h^2\mathrm{/}4>0$. If $I_{12}$ were a measurable function of $(U,V)$, the conditional-expectation identity [F5] would give $I_{12}=J$ almost surely, contradicting this positive residual. Thus the endpoint increments do not determine the required area sample, and any approximation has a separate error and cost.