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Stochastic Calculus A Practical

Stochastic Calculus A Practical

作者: jjx323 | 来源:发表于2020-03-05 13:06 被阅读0次

Notes for the following book:
Richard Durrett, Stochastic Calculus A Practical Introduction, CRC Press 2000.

1. Brownian Motion

1.1 Definition and Construction

Exercise 1.2. Show by considering m increments instead of 3 that if \gamma > 1/2+1/m then with probability 1, Brownian paths are not Holder continuous with exponent \gamma at any point of [0,1].
Answer: Let A_n = \{ \omega : \text{there is an }s\in [0,1] \text{ so that }|B_t - B_s|\leq C |t-s|^{\gamma} \text{ when }|t-s|\leq m/n \}. For 1\leq k\leq n-(m-1) let
\begin{align} & Y_{k,n} = \max\left\{ \left| B\left( \frac{k+j}{n}\right) - B\left(\frac{k+j-1}{n}\right) \right| : j=0,1,\cdots, m-1 \right\} \\ & G_n = \left\{ \text{at least one }Y_{k,n} \leq 2C \left( \frac{2m-1}{n} \right)^{\gamma} \right\}. \end{align}
The claim A_n\subset G_n still can be proved when s=1. Specifically, we have
\begin{align} & \left| B( \frac{n-(m-1)}{n} ) - B\left( \frac{n-m}{n} \right)\right| \\ & \leq \left| B\left( \frac{n-(m-1)}{n} \right) - B_1 \right| + \left| B\left( \frac{n-m}{n} \right) - B_1 \right| \\ & \leq C\left\{ \left(\frac{m-1}{n} \right)^{\gamma} + \left( \frac{m}{n} \right)^{\gamma} \right\} \\ & \leq 2C \left( \frac{2m-1}{n} \right)^{\gamma}. \end{align}
Using A_n \subset G_n and the scaling relation as in the book gives
\begin{align} P(A_n) \leq P(G_n) & \leq (n-m) P\left( |B_{1/n}| \leq 2C \left(\frac{2m-1}{n} \right)^{\gamma} \right)^{m} \\ & \leq n\left\{ \frac{2C(2m-1)^{\gamma}}{n^{\gamma-1/2}} \right\}^{m} \end{align}.
To ensure the right hand side goes to zero as n\rightarrow 0, we need
1-m\gamma + m/2 < 0,
which implies \gamma > 1/2+1/m. Obviously, the proof is completed.

2020

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