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解 SDE 并求均值/方差

Solve the SDE 4

专题
Finance / 金融
难度
L4

题目详情

Solve the stochastic differential equation

dXt=t2dt+et2cos⁡WtdWt,dX_{t} = t^{2}dt + e^{\frac{t}{2}}\cos W_{t}dW_{t},

with X0=0X_0 = 0 . Compute E[Xt]\mathbb{E}\left[X_t\right] and var (Xt)(X_t) . 【解答】Let Yt=et2sin⁡(Wt)Y_{t} = e^{\frac{t}{2}}\sin \left(W_{t}\right) . Consider f(t,x)=et2sin⁡xf(t,x) = e^{\frac{t}{2}}\sin x and note that

ft(t,x)=12et2sin⁡x;fx(t,x)=et2cos⁡x;fxx(t,x)=−et2sin⁡x.\begin{array}{l}{f_t(t,x) = \frac{1}{2} e^{\frac{t}{2}}\sin x;\quad f_x(t,x) = e^{\frac{t}{2}}\cos x;}\\ {f_{xx}(t,x) = -e^{\frac{t}{2}}\sin x.} \end{array}

Using Itô's formula to evaluate dYtdY_{t} , we obtain that

dYt=12et2sin⁡Wtdt+et2cos⁡WtdWt−12et2sin⁡Wtdt,=et2cos⁡WtdWt.\begin{array}{l}{dY_{t} = \frac{1}{2} e^{\frac{t}{2}}\sin W_{t}d t + e^{\frac{t}{2}}\cos W_{t}d W_{t}}\\ {\quad \quad -\frac{1}{2} e^{\frac{t}{2}}\sin W_{t}d t,}\\ {\quad \quad = e^{\frac{t}{2}}\cos W_{t}d W_{t}.} \end{array}

From (2.425), it follows that the stochastic differential equation for XtX_{t} can be expressed as

dXt=t2dt+dYt.dX_{t} = t^{2}dt + dY_{t}.
解析

取 Yt=et/2sin⁡WtY_t=e^{t/2}\sin W_t。由 Itô 可得

dYt=et/2cos⁡Wt dWt.dY_t=e^{t/2}\cos W_t\,dW_t.

题目 SDE：dXt=t2dt+et/2cos⁡WtdWt=t2dt+dYtdX_t=t^2dt+e^{t/2}\cos W_t dW_t=t^2dt+dY_t。

且 X0=0,Y0=0X_0=0,Y_0=0，所以

Xt=t33+et/2sin⁡Wt.\boxed{X_t=\frac{t^3}{3}+e^{t/2}\sin W_t}.

均值：E[sin⁡Wt]=0\mathbb{E}[\sin W_t]=0，因此

E[Xt]=t33.\boxed{\mathbb{E}[X_t]=\frac{t^3}{3}}.

方差：常数项不影响方差，

Var⁡(Xt)=etVar⁡(sin⁡Wt)=etE[sin⁡2Wt].\operatorname{Var}(X_t)=e^t\operatorname{Var}(\sin W_t)=e^t\mathbb{E}[\sin^2 W_t].

用 sin⁡2x=(1−cos⁡2x)/2\sin^2x=(1-\cos 2x)/2 且 E[cos⁡(2Wt)]=e−2t\mathbb{E}[\cos(2W_t)]=e^{-2t}，得

Var⁡(Xt)=et⋅1−e−2t2=et−e−t2.\boxed{\operatorname{Var}(X_t)=e^t\cdot\frac{1-e^{-2t}}{2}=\frac{e^t-e^{-t}}{2}}.