本章导读 本章研究扩散过程——漂移与扩散系数为状态的确定性函数的 SDE。§15.1 几何布朗运动（Def 15.1 GBM \(dX_t=X_t(m\,dt+\sigma\,dB_t)\) (15.1)；Prop 15.1 解 \(X_t=X_0 e^{(m-\frac12\sigma^2)t+\sigma B_t}\)，由 Ito 公式 2 验证）。§15.2 扩散过程定义（Def 15.2 \(dX_t=m(t,X_t)dt+\sigma(t,X_t)dB_t\) (15.2)，\(m,\sigma\) 是确定性连续函数；Rmk 15.1 GBM 是其特例）。§15.3 扩散过程的存在性（Thm 15.1：在 \(m,\sigma\) 一致 Lipschitz 条件下，用 Picard 型迭代 \(X_t^{(n+1)}=X_0+\int m(s,X_s^{(n)})ds+\int\sigma(s,X_s^{(n)})dB_s\) 构造，证 \(\mathbb E[|X_t^{(n+1)}-X_t^{(n)}|^2]\le\frac{\lambda^n t^{2n}}{(n+1)!}\to0\)，故极限存在、扩散过程良定义；用到 Hölder 不等式 Def 15.3）。无图。

定义 15.1（几何布朗运动）与命题 15.1 / Definition 15.1 (GBM) and Proposition 15.1 定义 15.1（几何布朗运动）：过程 \(\{X_t\}\) 是参数 \(m,\sigma^2\) 的几何布朗运动，若满足 \(dX_t=X_t(m\,dt+\sigma\,dB_t)\) (15.1)。命题 15.1：定义 15.1 中的几何布朗运动有解 \(X_t=X_0 e^{\left[(m-\frac12\sigma^2)t+\sigma B_t\right]}\)。Definition 15.1 (Geometric Brownian Motion): a process \(\{X_t\}\) is a geometric Brownian motion with parameters \(m,\sigma^2\) if it satisfies \(dX_t=X_t(m\,dt+\sigma\,dB_t)\) (15.1). Proposition 15.1: the geometric Brownian motion in Definition 15.1 has the solution \(X_t=X_0 e^{\left[(m-\frac12\sigma^2)t+\sigma B_t\right]}\).

命题 15.1 证明 / Proof of Proposition 15.1 只需验证 \(X_t=X_0 e^{\left[(m-\frac12\sigma^2)t+\sigma B_t\right]}=f(t,B_t)\) 解 (15.1)。此处 \(\dot f=(m-\frac12\sigma^2)X_t\)、\(f'=\sigma X_t\)、\(f''=\sigma^2 X_t\)。由 Ito 公式 2（定理 14.2）：It suffices to verify that \(X_t=X_0 e^{\left[(m-\frac12\sigma^2)t+\sigma B_t\right]}=f(t,B_t)\) solves (15.1). Here \(\dot f=(m-\frac12\sigma^2)X_t\), \(f'=\sigma X_t\), \(f''=\sigma^2 X_t\). By Ito's Formula 2 (Theorem 14.2):

$$df(t,B_t)=\left(\dot f+\tfrac12 f''\right)dt+f'\,dB_t=\left[(m-\tfrac12\sigma^2)X_t+\tfrac12\sigma^2 X_t\right]dt+\sigma X_t\,dB_t=X_t(m\,dt+\sigma\,dB_t),$$

与 (15.1) 一致，故 \(X_t=X_0 e^{\left[(m-\frac12\sigma^2)t+\sigma B_t\right]}\) 是 (15.1) 的解。\(\blacksquare\)consistent with (15.1), so \(X_t=X_0 e^{\left[(m-\frac12\sigma^2)t+\sigma B_t\right]}\) is the solution to (15.1). \(\blacksquare\)

定义 15.2（扩散过程）与注 15.1 / Definition 15.2 (Diffusion process) and Remark 15.1 定义 15.2（扩散过程）：过程 \(\{X_t\}\) 是（一维）扩散过程，若满足 \(dX_t=m(t,X_t)\,dt+\sigma(t,X_t)\,dB_t\) (15.2)，其中 \(m(t,X_t)\) 与 \(\sigma(t,X_t)\) 是确定性连续函数。注 15.1：几何布朗运动是扩散过程的特例，对应 \(m(t,X_t)=mX_t\)、\(\sigma(t,X_t)=\sigma X_t\)。Definition 15.2 (Diffusion process): a process \(\{X_t\}\) is a (one-dimensional) diffusion process if it satisfies \(dX_t=m(t,X_t)\,dt+\sigma(t,X_t)\,dB_t\) (15.2), where \(m(t,X_t)\) and \(\sigma(t,X_t)\) are deterministic continuous functions. Remark 15.1: the geometric Brownian motion is a special case of the diffusion process with \(m(t,X_t)=mX_t\) and \(\sigma(t,X_t)=\sigma X_t\).

定理 15.1（扩散过程的存在性）/ Theorem 15.1 (Existence) 将证明在 \(\mathcal L^2=\{x\in\mathbb R:\mathbb E[|x|^2]<\infty\}\) 空间中，一维扩散过程 \(\{X_t\}\) 在某些条件下存在且良定义。定理 15.1：设 \(\{X_t\}\) 由扩散过程 \(dX_t=m(t,X_t)\,dt+\sigma(t,X_t)\,dB_t\) (15.3) 定义（\(X_0=x_0\)），\(m(t,X_t)\)、\(\sigma(t,X_t)\) 一致 Lipschitz，即 \(\exists\beta\in\mathbb R\) 使 \(\forall x,y\)，\(|m(t,x)-m(t,y)|+|\sigma(t,x)-\sigma(t,y)|\le\beta|x-y|\)。定义迭代 \(X_t^{(0)}=X_0\)，\(X_t^{(n+1)}=X_0+\int_0^t m(s,X_s^{(n)})\,ds+\int_0^t\sigma(s,X_s^{(n)})\,dB_s\)（\(\forall n\in\mathbb N^+\)）。则极限 \(\lim_{n\to\infty}X_t^{(n)}\) 存在，故扩散过程 \(\{X_t\}\) 由 (15.3) 良定义。We show that in the space \(\mathcal L^2=\{x\in\mathbb R:\mathbb E[|x|^2]<\infty\}\), the one-dimensional diffusion process \(\{X_t\}\) exists and is well-defined under some conditions. Theorem 15.1: suppose \(\{X_t\}\) is defined by the diffusion process \(dX_t=m(t,X_t)\,dt+\sigma(t,X_t)\,dB_t\) (15.3) (\(X_0=x_0\)), with \(m(t,X_t)\), \(\sigma(t,X_t)\) uniformly Lipschitz, i.e. there is \(\beta\in\mathbb R\) such that for all \(x,y\), \(|m(t,x)-m(t,y)|+|\sigma(t,x)-\sigma(t,y)|\le\beta|x-y|\). Define the iteration \(X_t^{(0)}=X_0\), \(X_t^{(n+1)}=X_0+\int_0^t m(s,X_s^{(n)})\,ds+\int_0^t\sigma(s,X_s^{(n)})\,dB_s\) (for all \(n\in\mathbb N^+\)). Then the limit \(\lim_{n\to\infty}X_t^{(n)}\) exists, so the diffusion process \(\{X_t\}\) is well-defined by (15.3).

定义 15.3（Hölder 不等式）/ Definition 15.3 (Hölder's Inequality) 设 \((\Omega,\Sigma,\mu)\) 是测度空间，\(p,q\in[1,\infty]\) 满足 \(\frac1p+\frac1q=1\)。则对 \(\Omega\) 上所有可测函数 \(f,g\)，\(\|fg\|_1\le\|f\|_p\|g\|_q\)，其中 \(\|fg\|_1=\int_\Omega|fg|\,d\mu\)、\(\|f\|_p=\left(\int_\Omega|f|^p\,d\mu\right)^{1/p}\)、\(\|g\|_q=\left(\int_\Omega|g|^q\,d\mu\right)^{1/q}\)。Let \((\Omega,\Sigma,\mu)\) be a measure space and \(p,q\in[1,\infty]\) with \(\frac1p+\frac1q=1\). Then for all measurable functions \(f,g\) on \(\Omega\), \(\|fg\|_1\le\|f\|_p\|g\|_q\), where \(\|fg\|_1=\int_\Omega|fg|\,d\mu\), \(\|f\|_p=\left(\int_\Omega|f|^p\,d\mu\right)^{1/p}\), \(\|g\|_q=\left(\int_\Omega|g|^q\,d\mu\right)^{1/q}\).

定理 15.1 证明（迭代 + Hölder + 归纳）/ Proof of Theorem 15.1 (iteration + Hölder + induction) 要证 \(\lim_{n\to\infty}X_t^{(n)}\) 存在，只需证 \(\lim_{n\to\infty}\mathbb E\!\left[|X_t^{(n+1)}-X_t^{(n)}|^2\right]=0\) (15.4)——因 (15.4) 蕴含 \(\lim\mathbb E[|X_t^{(n+1)}-X_t^{(n)}|]=0\)，故 \(|X_t^{(n+1)}-X_t^{(n)}|\) 同时零均值、零方差（脚注 15.1：由 Markov 不等式蕴含 \(X_t^{(n+1)}\overset{P}{\to}X_t^{(n)}\)，从而 \(X_t^{(n)}\overset{P}{\to}X_t\)），即我们的收敛意义。展开 (15.5)：\(\mathbb E[|X_t^{(n+1)}-X_t^{(n)}|^2]\le2\underbrace{\mathbb E\!\left[\left(\int_0^t(m(s,X_s^{(n)})-m(s,X_s^{(n-1)}))ds\right)^2\right]}_{\text{Part 1}}+2\underbrace{\mathbb E\!\left[\left(\int_0^t(\sigma(s,X_s^{(n)})-\sigma(s,X_s^{(n-1)}))dB_s\right)^2\right]}_{\text{Part 2}}\)。Part 1：用 Lipschitz 与 Hölder 不等式，\(\text{Part 1}\le\mathbb E\!\left[\left(\int_0^t\beta|X_s^{(n)}-X_s^{(n-1)}|ds\right)^2\right]\le\beta^2\mathbb E\!\left[\left(\int_0^t|X_s^{(n)}-X_s^{(n-1)}|^2ds\right)\left(\int_0^t 1\,ds\right)\right]=t\beta^2\int_0^t\mathbb E[|X_s^{(n)}-X_s^{(n-1)}|^2]\,ds\) (15.6)。Part 2 同理（用方差等距）\(=t\beta^2\int_0^t\mathbb E[|X_s^{(n)}-X_s^{(n-1)}|^2]\,ds\) (15.7)。代入 (15.5)：\(\mathbb E[|X_t^{(n+1)}-X_t^{(n)}|^2]\le4t\beta^2\int_0^t\mathbb E[|X_s^{(n)}-X_s^{(n-1)}|^2]\,ds\) (15.8)。归纳得 \(\mathbb E[|X_t^{(n+1)}-X_t^{(n)}|^2]\le\dfrac{\lambda^n t^{2n}}{(n+1)!}\) (15.9)：(1) \(n=0\) 由 \(\{X_t^{(1)}\}\in\mathcal L^2\)，\(\exists\lambda_0\) 使 \(\mathbb E[|X_t^{(1)}-X_0|^2]\le\lambda_0\)；(2) 取 \(\lambda=\max\{\lambda_0,4\beta^2\}\) 使其成立；(3) 设 \(\mathbb E[|X_t^{(k+1)}-X_t^{(k)}|^2]\le\frac{\lambda^k t^{2k}}{(k+1)!}\)，由 (15.8) 验证 \(n=k+1\)：\(\mathbb E[|X_t^{(k+2)}-X_t^{(k+1)}|^2]\le4t\beta^2\int_0^t\frac{\lambda^k s^{2k}}{(k+1)!}ds=4\beta^2\frac{\lambda^k t^{2k+2}}{(k+1)!}\le\frac{\lambda^{k+1}t^{2(k+1)}}{(k+1)!}\)，归纳完成。最后用 (15.9)：\(\lim_{n\to\infty}\mathbb E[|X_t^{(n+1)}-X_t^{(n)}|^2]\le\lim_{n\to\infty}\frac{\lambda^n t^{2n}}{(n+1)!}=0\)，证毕。\(\blacksquare\)To show \(\lim_{n\to\infty}X_t^{(n)}\) exists, we only need \(\lim_{n\to\infty}\mathbb E\!\left[|X_t^{(n+1)}-X_t^{(n)}|^2\right]=0\) (15.4) — since (15.4) implies \(\lim\mathbb E[|X_t^{(n+1)}-X_t^{(n)}|]=0\), so \(|X_t^{(n+1)}-X_t^{(n)}|\) has both zero mean and zero variance (footnote 15.1: by Markov's inequality this implies \(X_t^{(n+1)}\overset{P}{\to}X_t^{(n)}\), hence \(X_t^{(n)}\overset{P}{\to}X_t\)), which is our sense of convergence. Expand (15.5): \(\mathbb E[|X_t^{(n+1)}-X_t^{(n)}|^2]\le2\underbrace{\mathbb E\!\left[\left(\int_0^t(m(s,X_s^{(n)})-m(s,X_s^{(n-1)}))ds\right)^2\right]}_{\text{Part 1}}+2\underbrace{\mathbb E\!\left[\left(\int_0^t(\sigma(s,X_s^{(n)})-\sigma(s,X_s^{(n-1)}))dB_s\right)^2\right]}_{\text{Part 2}}\). Part 1: by Lipschitz and Hölder, \(\text{Part 1}\le\mathbb E\!\left[\left(\int_0^t\beta|X_s^{(n)}-X_s^{(n-1)}|ds\right)^2\right]\le\beta^2\mathbb E\!\left[\left(\int_0^t|X_s^{(n)}-X_s^{(n-1)}|^2ds\right)\left(\int_0^t 1\,ds\right)\right]=t\beta^2\int_0^t\mathbb E[|X_s^{(n)}-X_s^{(n-1)}|^2]\,ds\) (15.6). Part 2 likewise (via the variance isometry) \(=t\beta^2\int_0^t\mathbb E[|X_s^{(n)}-X_s^{(n-1)}|^2]\,ds\) (15.7). Plug into (15.5): \(\mathbb E[|X_t^{(n+1)}-X_t^{(n)}|^2]\le4t\beta^2\int_0^t\mathbb E[|X_s^{(n)}-X_s^{(n-1)}|^2]\,ds\) (15.8). Induction gives \(\mathbb E[|X_t^{(n+1)}-X_t^{(n)}|^2]\le\dfrac{\lambda^n t^{2n}}{(n+1)!}\) (15.9): (1) \(n=0\) from \(\{X_t^{(1)}\}\in\mathcal L^2\), there is \(\lambda_0\) with \(\mathbb E[|X_t^{(1)}-X_0|^2]\le\lambda_0\); (2) take \(\lambda=\max\{\lambda_0,4\beta^2\}\) so it holds; (3) suppose \(\mathbb E[|X_t^{(k+1)}-X_t^{(k)}|^2]\le\frac{\lambda^k t^{2k}}{(k+1)!}\), then by (15.8) for \(n=k+1\): \(\mathbb E[|X_t^{(k+2)}-X_t^{(k+1)}|^2]\le4t\beta^2\int_0^t\frac{\lambda^k s^{2k}}{(k+1)!}ds=4\beta^2\frac{\lambda^k t^{2k+2}}{(k+1)!}\le\frac{\lambda^{k+1}t^{2(k+1)}}{(k+1)!}\), completing the induction. Finally by (15.9): \(\lim_{n\to\infty}\mathbb E[|X_t^{(n+1)}-X_t^{(n)}|^2]\le\lim_{n\to\infty}\frac{\lambda^n t^{2n}}{(n+1)!}=0\), and the proof is complete. \(\blacksquare\)