4.习题1
例1：已知F(t)\mathcal{F}(t)F(t)可测的随机过程X(t)X(t)X(t)的表达式如下
X(t)=exp{θW(t)−12θ2t} X(t) = \exp \{ \theta W(t) - \frac{1}{2} \theta^2 t \}X(t)=exp{θW(t)−21θ2t}
其中，W(t)W(t)W(t)是标准布朗运动，θ\thetaθ是常数。
求X(t)X(t)X(t)的随机微分方程。
解：思路：求dX(t)dX(t)dX(t)，1泰勒公式，2伊藤引理
根据上节原理可得，泰勒公式的求解方法，
dX=Xtdt+XWdW+12XWW(dW)2=∂X∂tdt+∂X∂WdW+12∂2X∂W2(dW)2 \begin{aligned} dX =&X_{t}dt+X_{W}dW+\frac{1}{2}X_{WW}(dW)^{2} \\ =&\frac{\partial X}{\partial t}dt+\frac{\partial X}{\partial W}dW+\frac{1}{2% }\frac{\partial ^{2}X}{\partial W^{2}}(dW)^{2} \end{aligned} dX==Xtdt+XWdW+21XWW(dW)2∂t∂Xdt+∂W∂XdW+21∂W2∂2X(dW)2
dX=exp{θW(t)−12θ2t}−12θ2dt+exp{θW(t)−12θ2t}θdW+12exp{θW(t)−12θ2t}θ2dt=X(t)−12θ2dt+X(t)θdW+12X(t)θ2dt=exp{θW(t)−12θ2t}θdW(t) \begin{aligned} dX =&\exp \{\theta W(t)-\frac{1}{2}\theta ^{2}t\}-\\frac{1}{2}\\theta \^{2}dt+\exp \{\theta W(t)-\frac{1}{2}\theta ^{2}t\}\theta dW+\frac{1}{2}% \exp \{\theta W(t)-\frac{1}{2}\theta ^{2}t\}\theta ^{2}dt \\ =&X(t)-\\frac{1}{2}\\theta \^{2}dt+X(t)\theta dW+\frac{1}{2}X(t)\theta ^{2}dt \\ =&\exp \{\theta W(t)-\frac{1}{2}\theta ^{2}t\}\theta dW(t) \end{aligned} dX===exp{θW(t)−21θ2t}−21θ2dt+exp{θW(t)−21θ2t}θdW+21exp{θW(t)−21θ2t}θ2dtX(t)−21θ2dt+X(t)θdW+21X(t)θ2dtexp{θW(t)−21θ2t}θdW(t)
最后结果
dX(t)=θX(t)dW(t) \begin{equation*} dX(t)=\theta X(t)dW(t) \end{equation*} dX(t)=θX(t)dW(t)
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伊藤引理：将X(t)=exp{Y(t)}X(t)=\exp\{ Y(t) \}X(t)=exp{Y(t)}和Y(t)=θW(t)−12θ2tY(t)=\theta W(t) - \frac{1}{2} \theta^2 tY(t)=θW(t)−21θ2t，然后X(t)→XY(t)X(t) \to XY(t)X(t)→XY(t)。
解2：
dX=Xtdt+XYdY+12XYY(dY)2=XdY+12X(dY)2=XθdW(t)−12θ2dt+12Xθ2dt=XθdW(t) \begin{aligned} dX =&X_{t}dt+X_{Y}dY+\frac{1}{2}X_{YY}(dY)^{2} \\ =&XdY+\frac{1}{2}X(dY)^{2} \\ =&X\left \\theta dW(t)-\\frac{1}{2}\\theta \^{2}dt\\right +\frac{1}{2}X\theta ^{2}dt \\ =&X\theta dW(t) \end{aligned} dX====Xtdt+XYdY+21XYY(dY)2XdY+21X(dY)2XθdW(t)−21θ2dt+21Xθ2dtXθdW(t)
∂X∂t=0,∂X∂Y=X,∂2X∂Y2=XdY=θdW(t)−12θ2dt,(dY)2=θ2dt \begin{aligned} \frac{\partial X}{\partial t} =&0,\quad \frac{\partial X}{\partial Y}% =X,\quad \frac{\partial ^{2}X}{\partial Y^{2}}=X \\ dY =&\theta dW(t)-\frac{1}{2}\theta ^{2}dt,\quad \left( dY\right) ^{2}=\theta ^{2}dt \end{aligned} ∂t∂X=dY=0,∂Y∂X=X,∂Y2∂2X=XθdW(t)−21θ2dt,(dY)2=θ2dt
变量替换的方法（X与t无关，只与Y有关，所以与t求偏导为0），泰勒展开视为一个二元的函数
例2：已知F(t)\mathcal{F}(t)F(t)可测的随机过程X(t)X(t)X(t)的表达式如下
X(t)=W2(t) X(t) = W^{2}(t) X(t)=W2(t)
其中，W(t)W(t)W(t)是标准布朗运动。
求X(t)X(t)X(t)的随机微分方程。
解（泰勒展开）
dX=Xtdt+XWdW+12XWW(dW)2=∂X∂tdt+∂X∂WdW(t)+12∂2X∂W2(dW)2 \begin{aligned} dX =&X_{t}dt+X_{W}dW+\frac{1}{2}X_{WW}(dW)^{2} \\ =&\frac{\partial X}{\partial t}dt+\frac{\partial X}{\partial W}dW(t)+\frac{1% }{2}\frac{\partial ^{2}X}{\partial W^{2}}(dW)^{2} \end{aligned} dX==Xtdt+XWdW+21XWW(dW)2∂t∂Xdt+∂W∂XdW(t)+21∂W2∂2X(dW)2
∂X/∂t=0,∂X/∂W=2W(t),∂2X/∂W2=2 \begin{equation*} \partial X/\partial t=0,\partial X/\partial W=2W(t),\partial ^{2}X/\partial W^{2}=2 \end{equation*} ∂X/∂t=0,∂X/∂W=2W(t),∂2X/∂W2=2
dX=0+2W(t)dW(t)+dt=2W(t)dW(t)+dt \begin{equation*} dX=0+2W(t)dW(t)+dt=2W(t)dW(t)+dt \end{equation*} dX=0+2W(t)dW(t)+dt=2W(t)dW(t)+dt
例3：已知F(t)\mathcal{F}(t)F(t)可测的随机过程f(S(t))=lnS(t)f(S(t))=\ln S(t)f(S(t))=lnS(t)，其中S(t)S(t)S(t)的随机微分方程如下
dS(t)=μS(t)dt+σS(t)dW(t) dS(t) = \mu S(t)dt+\sigma S(t)dW(t)dS(t)=μS(t)dt+σS(t)dW(t)(几何布朗运动)
其中，W(t)W(t)W(t)是标准布朗运动，μ\muμ和σ\sigmaσ是常数。
求f(S(t))f(S(t))f(S(t))的随机微分方程。
解
由于f=lnS(t)f=\ln S(t)f=lnS(t)，又例2类似，可知fff与SSS相关，但是不与ttt相关，所以对ttt求导=0。
assume μ(t)=μS(t),σ(t)=σS(t)\mu (t)=\mu S(t),\sigma (t)=\sigma S(t)μ(t)=μS(t),σ(t)=σS(t).
Then the process S(t)S(t)S(t) satisfies dS(t)=μ(t)dt+σ(t)dW(t),dS(t)=\mu(t)dt+\sigma(t)dW(t),dS(t)=μ(t)dt+σ(t)dW(t),which is an Ito process.
Applying Ito's lemma to f(S)=lnSf(S)=\ln Sf(S)=lnS,
df=ftdt+fSdS+12fSS(dS)2 \begin{equation*} df=f_{t}dt+f_{S}dS+\frac{1}{2}f_{SS}(dS)^{2} \end{equation*} df=ftdt+fSdS+21fSS(dS)2
Since fff does not explicitly depend on ttt,
1 so ft=0f_{t}=0ft=0. 2 fS=1/Sf_{S}=1/SfS=1/S. 3 fSS=−1/S2f_{SS}=-1/S^{2}fSS=−1/S2,
and dS(t)2dS(t)^{2}dS(t)2 contains (dt)2=0(dt)^{2}=0(dt)2=0, dtdW(t)=0dtdW(t)=0dtdW(t)=0, σ2(t)dW(t)2=σ2(t)dt\sigma ^{2}(t)dW(t)^{2}=\sigma ^{2}(t)dtσ2(t)dW(t)2=σ2(t)dt.
df=0+1Sμ(t) dt+σ(t) dW(t)+12−1S2σ2(t)dt=1SμS(t)dt+σS(t)dW(t)−12S2σ2S2(t)dt=μdt+σdW(t)−12σ2dt=μ−12σ2dt+σdW(t) \begin{aligned} df =&0+\frac{1}{S}\left \\mu (t)\\,dt+\\sigma (t)\\,dW(t)\\right +\frac{1}{2}% \frac{-1}{S^{2}}\sigma ^{2}(t)dt \\ =&\frac{1}{S}\left \\mu S(t)dt+\\sigma S(t)dW(t)\\right -\frac{1}{2S^{2}}% \sigma ^{2}S^{2}(t)dt \\ =&\mu dt+\sigma dW(t)-\frac{1}{2}\sigma ^{2}dt \\ =&\left \\mu -\\frac{1}{2}\\sigma \^{2}\\right dt+\sigma dW(t) \end{aligned} df====0+S1μ(t)dt+σ(t)dW(t)+21S2−1σ2(t)dtS1μS(t)dt+σS(t)dW(t)−2S21σ2S2(t)dtμdt+σdW(t)−21σ2dtμ−21σ2dt+σdW(t)
Thus
df=μ−12σ2dt+σdW(t) \begin{equation*} df=\left \\mu -\\frac{1}{2}\\sigma \^{2}\\right dt+\sigma dW(t) \end{equation*} df=μ−21σ2dt+σdW(t)
dlnS=μ−12σ2dt+σdW(t) \begin{equation*} d\ln S=\left \\mu -\\frac{1}{2}\\sigma \^{2}\\right dt+\sigma dW(t) \end{equation*} dlnS=μ−21σ2dt+σdW(t)
Remark: If one writes dS(t)=μS(t) dt+σS(t) dW(t)dS(t)=\mu S(t)\,dt+\sigma S(t)\,dW(t)dS(t)=μS(t)dt+σS(t)dW(t), divides both sides by
S(t)S(t)S(t) to obtain dS(t)/S(t)=μ dt+σ dW(t)dS(t)/S(t)=\mu \,dt+\sigma \,dW(t)dS(t)/S(t)=μdt+σdW(t), and then assumes dlnS(t)=μ dt+σ dW(t)% d\ln S(t)=\mu \,dt+\sigma \,dW(t)dlnS(t)=μdt+σdW(t), this would conflict with the correct
result dlnS=μ−12σ2dt+σ dW(t)d\ln S=\left \\mu -\\frac{1}{2}\\sigma \^{2}\\right dt+\sigma \,dW(t)dlnS=μ−21σ2dt+σdW(t).
Undoubtedly, dlnS=μ−12σ2dt+σ dW(t)d\ln S = \left\\mu - \\frac{1}{2}\\sigma\^2\\rightdt + \sigma\,dW(t)dlnS=μ−21σ2dt+σdW(t) is correct.
why? The result dlnS=μ−12σ2dt+σ dW(t)d\ln S=\left \\mu -\\frac{1}{2}\\sigma \^{2}\\right dt+\sigma \,dW(t)dlnS=μ−21σ2dt+σdW(t) is derived rigorously using Ito's lemma, whereas dlnS(t)=μ dt+σ dW(t)d\ln S(t)=\mu \,dt+\sigma \,dW(t)dlnS(t)=μdt+σdW(t) comes from a naive application of ordinary calculus.
The fundamental source of this discrepancy lies in the nonzero quadratic
variation of Brownian motion (dW(t))2=dt≠0(dW(t))^{2}=dt\neq 0(dW(t))2=dt=0.