Brownian Motion
1. Definition
- Continuous sample path but nondifferentiable
- Independent increment
$W_{t_1}-W_0$, $W_{t_2} - W_{t_1}$ … $W_{t_n} - W_{t_{n-1}}$ are i.i.d
- Each of the increments is normally distributed
\[W_{t_m} - W_{t_n} \sim N(0,t_m-t_n)\]
2. Properties
- $E(W_t\vert F_s) = W_s $
- $E(W_t^2-t\vert F_s) = W_s^2-s$
- $E[exp(\lambda W_t - \frac{1}{2}\lambda^2t)\vert F_s] = exp(\lambda W_s-\frac{1}{2}\lambda^2s)$
- $cov(W_t,W_s) = s$
Note: The above proof can be done by replace $W_t = W_t-W_s+W_s$
Quadratic Variation
First-order Variation
Suppose $f(t)$ is a function of time. The first-order of variation of $f$ is defined as below:
\[FV_T(f) = \lim_{||\Pi|| = 0}\sum_{j=0}^{n-1}|f(t_{j+1})-f(t_j)|\]| Where $ | \Pi | = max_{j=0,…,n-1}(t_{j+1}-t_j)$. |
If $f$ has derivative on $t$ at everywhere, the Mean Value Theorem can be applied.
\[f(t_{j+1})-f(t_j) = f'(t^*)(t_{j+1}-t_j)\]Where $t^*$ is a value between $t_{j+1}$ and $t_j$.
Bring this back to first-order Variation, we can get that: \(FV_T(f) = \int_0^T|f'(t)|dt\)
But Brownina motion doesn’t have derivative on t, so this result is not applicable to brownian motion.
Quadratic Variation
Quadratic variation is defined as:
\[\]Ito Integral
1. Definition
The Ito Intergral is defined as below:
\[I \equiv \int_0^TH(t)dW\]$I$ is a random variable. If $H(t)$ is nonrandom function of time, $I$ is normal distributed.
2. Probabilistic Properties
Since $I$ is a random number, so we would care about its expectation and variance.
Expectation
\[E(I_t) = 0\]since $I$ is martingale.
Variance
\[E(I_t^2) = E[\int_0^tH(u)^2du]\]This is also called Ito Isometry.
Linearity
If $I_1(t) = \int_0^TH_1(t)dW$ and $I_2(t) = \int_0^TH_2(t)dW$
\[I_1(t)+I_2(t) = \int_0^T[H_1(t)+H_2(t)]dW\] \[cI_1(t) = c\int_0^TH_1(t)dW \ for \ constant\ c\]Example - $\int_0^TW_tdW_t$ \(Var(I) = E[\int_0^TW_t^2dt]\\=\int_0^TE(W_t^2)dt = \int_0^Ttdt = \frac{1}{2}T^2\)
\[\int_0^TW_tdW_t = \frac{1}{2}W_T^2-\frac{1}{2}T\]
Ito Formula
\[df(t,X_t)= f'_xdX_t+f'_tdt+\frac{1}{2}f''_{xx}dX_tdX_t\]Example - Geometric Brownian Motion Suppose the S follows: $dS_t = \alpha _tS_tdt + \sigma_tS_tdW_t$ Consider a function $f(S_t) = log(S_t)$ \(dlog(S_t) = \frac{1}{S_t}dS_t - \frac{1}{2}*\frac{1}{S_t^2}dS_tdS_t \\ = \alpha_tdt + \sigma_tdW_t - \frac{1}{2S_t^2}*(\sigma_t^2S_t^2dt) \\ = (\alpha_t-\frac{1}{2}\sigma_t^2)dt + \sigma_tdW_t\) As a result \(log(S_T)-log(S_0) = \int_0^T(\alpha_t-\frac{1}{2}\sigma_t^2)dt + \int_0^T\sigma_tdW_t\\ = (\alpha_t-\frac{1}{2}\sigma_t^2)T + \sigma_tW_t\) The $S_T$ can be expressed as below: \(S_T = S_0exp[(\alpha_t-\frac{1}{2}\sigma_t^2)T + \sigma_tW_t]\)
$S_T$ is lognormal distributed.
Example - OU Process Suppose the X follows: $dX_t = (\alpha-\beta X_t)dt + \sigma_tdW_t$ \(d(e^{\beta t}X_t) = \alpha e^{\beta t}dt + e^{\beta t} \sigma dW_t\)
Intergrate it
\[e^{\beta T}X_T - X_0 = \int_0^T\alpha e^{\beta t}dt + \int_0^Te^{\beta t} \sigma dW_t \\ = \frac{\alpha}{\beta}(e^{\beta T} - 1) + \sigma \int_0^Te^{\beta t} dW_t\]
The $X_T$ can be expressed as below:
\[X_T = e^{-\beta T}X_0 + \frac{\alpha}{\beta}(1 - e^{-\beta T}) + \sigma e^{-\beta T}\int_0^Te^{\beta t} dW_t\]
So $X_T$ is normal distributed with mean:
\[e^{-\beta T}X_0 + \frac{\alpha}{\beta}(1 - e^{-\beta T})\]
with variance
\(\sigma^2e^{-2\beta T}*\int_0^Te^{2\beta t} dt = \frac{\sigma^2}{2\beta}(1-e^{-2\beta T})\) As $\lim_{T \to\infty}$,the mean of $X_T$ becomes $\frac{\alpha}{\beta}$
Example - Hard - Not Finished Solve the following equation: \(dS_t = hdt + \sigma S_tdt\)
Assume $X_t$ is the $S_t$ without divident and under the same probability measure \(dX_t = \sigma X_tdw_t\) \(dlogX_t = \frac{1}{X_t}dX_t+\frac{1}{2}*-\frac{1}{X_t^2}dX_tdX_t\) \(dlogX_t = \sigma dw_t - \frac{1}{2}\sigma^2dt\) \(X_T = X_t*exp[\sigma W_{T-t}-\frac{1}{2}\sigma^2(T-t)]\)
\[d(\frac{S_t}{X_t}) = \frac{1}{X_t}dS_t - \frac{1}{2}*\frac{1}{X_t^2}dS_tdX_t -\frac{S_t}{X_t^2}dX_t-\frac{1}{2}*\frac{1}{X_t^2}dX_tdS_t+\frac{S_t}{X_t^3}dX_tdX_t\]
\(=\sigma\frac{S_t}{X_t}dW_t+\frac{h}{X_t}dt-\frac{1}{2}*\frac{\sigma^2S_t}{X_t}dt-\frac{\sigma S_t}{X_t}dW_t-\frac{1}{2}*\frac{\sigma^2S_t}{X_t}dt+\frac{\sigma^2S_t}{X_t}dt\) \(=\frac{h}{X_t}dt\)
\(d(\frac{S_t}{X_t}) = \frac{h}{X_t}dt\) \(\frac{S_T}{X_T} - \frac{S_t}{X_t} = \int_t^T \frac{h}{X_t}dt\)
\(S_T = (\frac{S_t}{X_t} + \int_t^T \frac{h}{X_t}dt)*X_t*exp[\sigma W_{T-t}-\frac{1}{2}\sigma^2(T-t)]\) To make things clear:
\[S_T = (\frac{S_t}{X_t} + \int_t^T \frac{h}{X_u}du)*X_t*exp[\sigma W_{T-t}-\frac{1}{2}\sigma^2(T-t)]\]
\[S_T = (\frac{S_t}{X_t} + \int_t^T \frac{h}{X_t*exp[\sigma W_{u-t}-\frac{1}{2}\sigma^2(u-t)]}du)*X_t*exp[\sigma W_{T-t}-\frac{1}{2}\sigma^2(T-t)]\]
\[S_T = (S_t + h\int_t^T exp[\frac{1}{2}\sigma^2(u-t)-\sigma W_{u-t}]du)*exp[\sigma W_{T-t}-\frac{1}{2}\sigma^2(T-t)]\]
Replace z=u-t
\[S_T = (S_t + h\int_0^{T-t} exp[\frac{1}{2}\sigma^2z-\sigma W_{z}]dz)*exp[\sigma W_{T-t}-\frac{1}{2}\sigma^2(T-t)]\]
Example - Hard Solve the following equation: \(dS_t = (h+\mu S_t)dt + (k + \sigma S_t)dw\)
Create two auxiliary process
\[dX = \mu Xdt + \sigma Xdw\]
\[dY = \alpha(t)dt + \beta(t)dw\]
We wish that $S_t$ can be expressed as $S_t = X_tY_t$
Then $X_0 = 1$ and $Y_0 = S_0$
Since we already know the analystical for $X_t$
\[X_t = X_0exp[(\mu-\frac{1}{2}\sigma^2)T + \sigma W_t] \\ = exp[(\mu-\frac{1}{2}\sigma^2)T + \sigma W_t]\]
Apply Ito’s Lemma to $X_tY_t$
\[d(X_tY_t) = X_tdY_t +Y_tdX_t + dX_tdY_t \\ = \alpha(t)X_tdt + \beta(t)X_tdw + \mu X_tY_tdt + \sigma X_tY_tdw + \sigma X_t\beta(t)dt \\ = (\alpha(t)X_t + \mu X_tY_t + \sigma X_t\beta(t))dt + (\beta(t)X_t + \sigma X_tY_t)dw \\ = (\alpha(t)X_t + \mu S_t + \sigma X_t\beta(t))dt + (\beta(t)X_t + \sigma S_t)dw\]
This term should equal to $dS_t$, to match drift term and diffusion term
\[\alpha(t)X_t + \mu S_t + \sigma X_t\beta(t) = h + \mu S_t\]
\[\beta(t)X_t + \sigma S_t = k + \sigma S_t\]
Wo we can backout $\beta(t)$ and $\alpha(t)$
\[\beta(t) = k * X_t^{-1}\]
\[\alpha(t) = h * X_t^{-1} - \sigma \beta(t)\]
Once $\beta(t)$ and $\alpha(t)$ are known, $Y_t$ can be calculated. Then simply calculate $S_t = X_t*Y_t$
