No-Arbitrage Constraints in a Two-Size Jump Price Model

# Modelling considerations for a jump model


# Modelling considerations for a jump model












The Problem:

Suppose I have a simple jump model for an asset price

$$ dS = S(t-)[\mu dt + YdN(t)] $$

where $N(t)$ is a Poisson process and $Y_i$ are the jump sizes (assume independece of $N(t)$ and $Y_i$). Also for simplicity assume that $Y_i$ can only take two values, lets say $a$ and $b$. What constraints are there for $\mu$, $a$ and $b$?

I should also add that we can assume the existence of a money market account with constant short rate: $$ B(t) = B_0e^{rt} $$

My initial thoughts:

In order to preserve positivity of stock prices I need both $a$ and $b$ to be greater than $-1$. I also need to consider when there might be arbitrage in my model. However I can't work out what additional constraints I need to impose. Can anyone help?

Thanks in advance for any responses.

## Answer by Gordon (score 2)

https://quant.stackexchange.com/a/74861

Note that, \begin{align*} d\big( e^{-\mu t}S_t \big) &= -\mu e^{-\mu t}S_t dt + e^{-\mu t}S_{t-}(\mu dt + Y_t dN_t)\\ &=e^{-\mu t}S_{t-}Y_t dN_t. \end{align*} From the Doleans-Dade exponential formula, \begin{align*} e^{-\mu t}S_t = \prod_{0<s\le t}(1+Y_s)\Delta N_s. \end{align*} That is, \begin{align*} S_t = e^{\mu t} \prod_{0<s\le t}(1+Y_s)\Delta N_s. \end{align*} Therefore, both $a$ and $b$ are greater than $-1$.

Let $\lambda$ be the intensity of the Poisson process $N$. Moreover, for ease of exposition, we assume that $E(Y_s)=L$, for $s\ge 0$, is a constant. Let \begin{align*} M_t = N_t-\lambda t \end{align*} and \begin{align*} X_t = \int_0^t Y_s dN_s -\lambda L t. \end{align*} Then, $\{M_t, t \ge 0 \}$ is a martingale. Moreover, for $u\ge v \ge 0$ and the information set $\mathscr{F}_v$ at time $v$, \begin{align*} E(X_u-X_v\mid \mathscr{F}_v) &= E\bigg(\int_v^u Y_s dN_s \mid \mathscr{F}_v\bigg) - \lambda L (u-v)\\ &= E\bigg(\int_v^u Y_s dM_s \mid \mathscr{F}_v\bigg) + \lambda E\bigg(\int_v^u Y_s ds \mid \mathscr{F}_v\bigg) - \lambda L (u-v)\\ &=0. \end{align*} That is, $\{X_t, t \ge 0\}$ is a martingale.

To rule out any arbitrage opportunity, $e^{-rt}S_t$ has to be a martingale. Note that \begin{align*} d\big( e^{-r t}S_t \big) &= -r e^{-r t}S_t dt + e^{-r t}S_{t-}(\mu dt + Y_t dN_t)\\ &= e^{-r t}S_{t-}\big((\mu-r+\lambda L)dt + dX_t \big). \end{align*} Then, $\mu = r - \lambda L$.