Stochastic process: $$X(t), t\in[0,\infty)$$

Brownian motion $$B(t)$$:

• Continuous
• Independent increments
• Normal distribution of increments, i.e.
$Pr[B(s+t)-B(s)\in A]=\int_A \frac{1}{\sqrt{2\pi t}}e^{-x^2/2t} dt \quad\textrm{ for } s,t>0$

Symmetric random walk (SRW):

• $$X_i \in \{-1,1\}$$ with probability $$p$$ and $$q$$ respectively
• SRW is $$M_n = \sum_{i=1}^n X_i$$ and $$M_0 = 0$$
• Scaled SRW: $$B^{(n)}(k/n) = \frac{1}{\sqrt{n}} M_k$$
• Limit: Let $$t=k/n$$, then $$B(t)=\lim_{n\to\infty} B^{(n)}(k/n)$$

Covariance of BM, assumed $$s<t$$,

\begin{aligned} E[B_s B_t] &= E[ (B_s)(B_t-B_s+B_s) ] \\ &= E[B_s(B_t-B_s)] + E[B_s^2] \\ &= 0 + E[B_s^2] \\ &= s \end{aligned}

Therefore, we have

\begin{aligned} E[B_s B_t] &= \min(s,t) \\ \langle B,B\rangle_T &= T \\ dB_t dB_t &= dt \\ dB_t dt &= dt dt = 0 \end{aligned}

Geometric BM:

\begin{aligned} S_t &= S_0 \exp(\sigma B_t + (\alpha-\sigma^2/2)t) \\ \log S_{t_{j+1}} - \log S_{t_j} &= \sigma(B_{t_{j+1}}-B_{t_j}) + (\alpha-\sigma^2/2)(t_{j+1}-t_j) \end{aligned}

First passage time problem:

• Let $$M_t = \max_{0<s<t} X_s$$ be the running maximum of stochastic process $$X_t$$
• Let $$T_a = \inf\{t>0: X_t = a\}$$ be the first passage time
• $$T_a < t$$ is equivalent to $$M_t > a$$

Reflection principle of Brownian motion $$B_t$$:

\begin{aligned} \textrm{Let} && B_s&=a\textrm{ for }s < t \\ \textrm{Then} && B_t-B_s &\perp B_s \\ \textrm{thus} && \Pr[B_t-B_s > 0] &= \Pr[B_t-B_s < 0] \\ && \Pr[B_t > a] &= \Pr[B_s=a, B_t-B_s > 0] \\ && &= \Pr[B_s=a, B_t-B_s < 0] \\ \textrm{so we have} && \Pr[B_s = a] &= \Pr[T_a < t] = 2\Pr[B_t > a]. \end{aligned}

To obtain the p.d.f. of $$T_a$$, we can consider the m.g.f. $$E[\exp(uT_a)]$$.

• Consider the trail of a Brownian motion $$B_t$$ that $$B_0=0$$, $$B_s=a$$, $$B_t=b$$ for $$s<t$$, $$a>b$$, then it has the same probability as $$B'_t$$ that $$B'_0=0$$, $$B'_s=a$$, $$B'_t=2a-b$$, just as the path of the B.M. reflected about $$a$$.
• $$\Pr[T_a < t, u < B_t < v] = \Pr[2a-v < B_t < 2a-u]$$.
• $$\Pr[T_a < t, B_t = u] = \Pr[B_t = 2a-u] = \Pr[M_t > a, B_t = u]$$.
• $$\Pr[M_t = a, B_t = u] = \frac{d}{da}\Pr[M_t > a, B_t = u]$$.

Stochastic integral: $$\int_s^t f(\tau,w)dX_{\tau}(w)$$

Ito integral: $$I(f) = \int_0^t f(s,w)dB_s$$

• $$f$$ is a $$\mathcal{F}_t$$-adapted, square-integrable stochastic process
• It is a martingale
• $$E[I^2(f)] = E[\int_0^t f^2(s,w)ds]$$.

Ito formula:

\begin{aligned} f(t,B_t) &= f(0,B_0) + \int_0^t f_t(\tau,B_{\tau})d\tau + \int_0^t f_x(\tau,B_{\tau})dB_{\tau} + \frac{1}{2}\int_0^t f_{xx}(\tau,B_{\tau})d\tau \\ df(t,B_t) &= f_t(t,B_t)dt + f_x(t,B_t)dB_t + \frac{1}{2}f_{xx}(t,B_t)dt \end{aligned}

Example: Solving $$\int_0^t B_s dB_s$$

\begin{aligned} \textrm{Let }f(t,x) &= \frac{1}{2} x^2 \\ d(\frac{1}{2}B_t^2) &= 0 + B_tdB_t + \frac{1}{2}dt \\ \textrm{thus }\int_0^t d(\frac{1}{2}B_s^2) &= \int_0^t B_sdB_s + \int_0^t \frac{1}{2}ds \\ \frac{1}{2}(B_t^2-B_0^2) &= \int_0^t B_sdB_s + \frac{1}{2}t \\ \int_0^t B_sdB_s &= \frac{1}{2}B_t^2 + \frac{1}{2}t. \end{aligned}

Ito formula in general form:

$df(t,X_t) = f_t dt + f_x dX_t + \frac{1}{2}f_{xx}(dX_t)^2$

where $$dX_t = \mu dt + \nu dB_t$$, and $$(dX_t)^2 = \nu^2 dt$$

Multidimensional version:

\begin{aligned} X_t &= (X_t^1, X_t^2, \cdots, X_t^n) \\ df(t, X_t) &= f_t dt + \sum_{i=0}^n f_{x_i}dX_t^i + \frac{1}{2} \sum_{i,j=1}^n f_{x_ix_j}dX_t^idX_t^j \end{aligned}