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Continuous

Challenge

How to describe probability distribution

Brownian Motion/Wiener Process

Basically a continuous version of simple RW: ‘Limit’ of simple RW

Denoted using \(y_t = B(t)\)

Properties

  • Always starts at 0: \(P( y_0 = 0 ) = 1\)
  • Stationary \(\forall s \in [0, t)\)
  • \(y_t - y_s \sim N(0, t-s)\)
    • where \((t-s)\) is the length of the interval
  • Independent increment: If intervals \([s_i, t_i]\) are non-overlapping, then \(y_{t_i} - y_{s_i}\) are independent

Characteristics

  • Cross the independent axis indefinitely-often
  • Does not deviate too much from \(y_t = \sqrt{t}\)
  • Not differentiable
  • Standard calculus cannot be applied
  • Requires Ito calculus
  • Max series
\[ \begin{aligned} & M_t = \max_{s \le t} (y_s) \\ \implies &P(M_t > a) = 2 \cdot P (y_t > a) \\ & \forall \ t, a > 0 \end{aligned} \]
  • Quadratic variation

$$ t = \frac{i}{n} T \ \implies \lim_{n \to \infty} \sum_{i=1}^n (y_{t} - y_{t-1})^2 = T \ \forall T > 0 \

\implies (dB)^2 = dt $$

Implications

\[ \dfrac{d y_t}{y_t} = d B_t \\ d y_t \ne \dfrac{d B_t}{dt} \cdot dt \\ \implies y_t \ne e^{B_t} \]

This is because \(\dfrac{d B_t}{dt}\) is not defined since \(B_t\) is not differentiable

Ito’s Lemma

Consider \(y_t = f(B_t)\), where \(f\) is a smooth function $$ \begin{aligned} y_t &= f(B_t) \ \implies df & \ne f'(B_t) \cdot d B_t \quad [\because (dB)^2 = dt] \ \implies df &= f'(B_t) \cdot d B_t + \dfrac{1}{2} f''(B_t) \cdot dt \end{aligned} $$

IDK

Assuming \(\mu, \sigma\) are constant $$ \begin{aligned} dy_t &= \underbrace{\mu dt}_\text{Drift} + \sigma d B_t \ \implies y_t &= \mu t + \sigma B_t \end{aligned} $$ Using Ito’s Lemma (Basically Taylor’s expansion) $$ d f(t, x) = \dfrac{\partial f}{\partial t} + \mu \dfrac{\partial f}{\partial x} + \dfrac{1}{2} \sigma^2 \dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial f}{\partial x} d B_t $$

Integration

\[ \begin{aligned} F(t, B_t) &= \int f(t, B_t) d B_t + \int g(t, B_t) dt \\ dF &= f dB_t + g dt \end{aligned} \]

Ito integral is the limit of Riemanian sums when we always take leftmost point of each integral

Intuitively, it only uses the data you have seen so far

Adapted Process

A strategy/decision \(D_t\) is said to be adapted to \(y_t\), if \(D_t\) only depends on \(y_s, s \le t, \forall t\)

If \(D_t\) only depends on \(t\) and not on \(B_t\), then \(y_t = \int D_t \cdot d B_t\) is normally-distributed at all times

Ito Isometry

Used to calculate variance of Brownian motion $$ \begin{aligned} D_t &\text{ adapted to } B_t \ \implies V(B_t) &= E \left[ (\int_0^t D_s \cdot dB_s)^2 \right] \ &= E \left[ \int_0^t D^2_t \cdot ds \right] \end{aligned} $$ Due to quadratic variance

Martingale

If \(g(t, B_t)\) is adapted to \(B_t\) then \(\int g(t, B_t) \cdot dB_t\), as long as \(g\) is “reasonable”

\(g\) is reasonable if \(\int \int g^2 \cdot dt \cdot dB_t < \infty\)

If a stochastic differential equation does not have a drift term, then it is a martingale $$ d y_t = \sigma \cdot dB_t \qquad [\mu = 0] $$ Defining stock price as brownian motion, as it is a martingale process $$ \begin{aligned} S_t &= \exp(\frac{-\sigma^2 t}{2} + \sigma B_t) \ \implies \dfrac{d S_t}{S_t} &= \sigma \cdot d B_t \end{aligned} $$

Stochastic Differential Equation

\[ d y_t = \mu \cdot dt + \sigma \cdot dB_t \]

Change of measure

Consider

  • \(B\) is brownian process w/ drift and pdf \(P\)
  • \(\tilde B\) is brownian process w/o drift and pdf \(\tilde P\)
\[ \exists z, \text{ such that } P(t) = z(t) \cdot \tilde P(t) \iff P \equiv \tilde P \]
\[ P \equiv P \text{ if } \\ P (A) > 0 \iff \tilde P (A) > 0, \quad \forall A \subseteq \Omega \]

\(z\) is called the Radon-Nikodym derivative

Girsanov theorem

\[ z(t) = \dfrac{d \tilde P}{dP} (t) = e^{-\mu t T - \frac{\mu^2 T}{2}} \]
\[ E[y_t] = \tilde E[\tilde z_t y_t] \\ \tilde E[y_t] = E[z_t y_t] \]
Last Updated: 2024-05-12 ; Contributors: AhmedThahir

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