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布朗运动的特征函数与偶次矩

Exponential Brownian

专题
Finance / 金融
难度
L4

题目详情

Let BtB_{t} be Brownian motion on R,B0=0\mathbf{R},B_{0} = 0 . Prove that

E[eiuBt]=exp⁡(−12u2t)for allu∈R\mathbb{E}\left[e^{iuB_{t}}\right] = \exp \left(-\frac{1}{2} u^{2}t\right)\quad \text{for all} u\in \mathbf{R}

Use the power series expansion of the exponential function on both sides, compare the terms with the same power of uu and deduce that

E[Bt4]=3t2\mathbb{E}\left[B_{t}^{4}\right] = 3t^{2}

and more generally that

E[Bt2k]=(2k)!2k⋅k!tk;k∈N\mathbb{E}\left[B_{t}^{2k}\right] = \frac{(2k)!}{2^{k}\cdot k!} t^{k};\quad k\in \mathbf{N}
解析

因 Bt∼N(0,t)B_t\sim N(0,t)，其特征函数为

E[eiuBt]=exp⁡(−12u2t).\boxed{\mathbb{E}[e^{iuB_t}]=\exp\left(-\frac12u^2t\right)}.

把两边对 uu 做幂级数展开并对齐系数，可得所有偶次矩：

E[Bt2k]=(2k)!2kk! tk,k∈N,\boxed{\mathbb{E}[B_t^{2k}]=\frac{(2k)!}{2^k k!}\,t^k},\quad k\in\mathbb{N},

奇次矩为 0。特别地

E[Bt4]=3t2.\boxed{\mathbb{E}[B_t^4]=3t^2}.