# Lecture 26 (March 30, 2022)

### Jump Diffusion Model

Idea: model jumps by a Poisson process

Def: Given a time $t$, let $x_t$ be the number of events occuring during the time period $[0,t]$, Then, $x_t$ is a Poisson process if $X_t \sim \text{Poisson}(\lambda t)$. i.e.

$P(x_t = k ) = \frac{(\lambda t)^k e^{-\lambda t}}{k!}$

where $\lambda >0$ constant rate parameter.

Assume $(x_t)$ Follows:

1. number of events occuring in non-overlapping intervals are independent.
2. events occur at a constant rate of $\lambda$ per unit time interval
3. events cannot occur simultaneously

Divide the time interval $[0,t]$ by $n$ pieces. Let $x_t$ be the number of events occuring in time $t$ units. Fix $t>0$, Let $Y_i$ be the number of events to occur in the $i$-th bin, $i=1,\dots,n$.

$Y_i \sim \text{Ber}(\lambda \frac{t}{n})$

Then,

$X_t = \sum_{i=1}^n Y_i \sim \text{Binomial}(n, \frac{\lambda t}{n})$

Fix $k$, we can get

\begin{aligned} P(X_t = k) &= {n\choose k}\left(\frac{\lambda t}{n} \right)^k \left(1 - \frac{\lambda t}{n}\right)^{n-k} \\ &=\frac{n!}{k!(n-k)!}\left(\frac{\lambda t}{n}\right)^k \left(1- \frac{\lambda t}{n}\right)^n \left(1 - \frac{\lambda t}{n}\right)^{-k} \\ &\rightarrow \frac{(\lambda t)^k}{k!} e^{-\lambda t} \end{aligned}

This is the logic of Poisson process.

Let $p_t$ be the price of an asset at time $t$, previously , we modeled

$dP_t = \mu P_t dt + \sigma P_t dw_t$

we now have

$\frac{dP_t}{P_t} = \mu dt + \sigma dW_t + d \left[\sum_{i=1}^{n_t} (J_i-1) \right]$

where

\begin{aligned} &W_t \text{ is the standard BM} \\ &n_t \text{ is the Poisson process with rate parameter } \lambda \\ &J_i \text{ is the sequence of iid non-negative r.v.} \end{aligned}

$J_i$ is the random variable that $X_i = \log (J_i)\sim \text{Laplace} (\kappa ,\eta)$

that is

X - \kappa = \left\{\begin{aligned} &\xi \quad \text{ with prob }\frac{1}{2}\\ &-\xi \quad \text{ with prob }\frac{1}{2} \end{aligned}\right.

with $\xi\sim \text{Exp}(\eta)$. with $\mathbb{E}[\xi] = \eta$ and $\text{Var}[\xi] = \eta^2$

Let $t_i$ be the time of the $i$-th jump, $i=1,\dots, n_t$,

$P_{t_1^{-}}= P_0 \exp\left[(\mu-\frac{\sigma^2}{2})t + \sigma w_t \right]$

Then

$P_{t_1} = P_{t_1^{-}} + (J_1-1) P_{t_1^-} = J_1 P_{t_1^-}$

At $t_2$, we have

$P_{t_2^{-}} = P_{t_1} \exp\left[ (\mu-\frac{\sigma^2}{2})(t_2-t_1) + \sigma(w_{t_2}-w_{t_1}) \right]$

and

\begin{aligned} P_{t_2} &= P_{t_1^-} + (J_2-1) P_{t_2}^- = J_2 P_{t_2}^- \\ &= J_2 P_{t_1} \exp\left[(\mu-\frac{\sigma^2}{2})(t_2-t_1) + \sigma (w_{t_2}-w_{t_1}) \right]\\ &= J_2J_1 P_0 \exp\left[ (\mu-\frac{\sigma^2}{2})t_1 + \sigma W_{t_1} \right] \exp \left[(\mu-\frac{\sigma^2}{2})(t_2-t_1) + \sigma (w_{t_2}-w_{t_1}) \right]\\ &= J_2J_1 P_0 \exp \left[(\mu-\frac{\sigma^2}{2})t_2 + \sigma w_{t_2} \right] \end{aligned}

This shows that in general,

$P_t = P_0 \exp\left[ (\mu-\frac{\sigma^2}{2})t + \sigma w_t\right] \prod_{i=1}^{n_t} J_i$

For small $\Delta t$,

\begin{aligned} \frac{P_{t+\Delta t} - P_t}{P_t} &= \frac{P_{t+\Delta t}}{P_t} -1 \\ &=\exp\left[ (\mu-\frac{\sigma^2}{2})\Delta t + \sigma (W_{t+\Delta t} - W_t) \right]\left[\prod_{n_t}^{n_{t+\Delta t}} J_i \right] - 1 \end{aligned}

Note that $X_i = \log J_i$,

\begin{aligned} \frac{P_{t+\Delta t} - P_t}{P_t} &= \exp \left[(\mu-\frac{\sigma^2}{2})\Delta t + \sigma (W_{t+\Delta t} -W_t) + \sum_{n_t}^{n_{t+\Delta t}} X_i \right]-1 \\ &\approx \left[(\mu - \frac{\sigma^2}{2})\Delta t + \sigma \Delta W_t + \sum_{n_t}^{n_{t+\Delta t}} X_i \right] + \frac{1}{2}\sigma^2 \Delta t \\ &= \mu \Delta t + \sigma \sqrt{\Delta t} \epsilon_t + \sum_{n_t}^{n_{t+\Delta t}} X_i \end{aligned}

Updated: