Deriving the Put-Call Parity

• Mar 29th 2010, 08:34 AM
Deriving the Put-Call Parity
An underlying asset with price process $\displaystyle \{S_t\}_{t \in J}$ , where $\displaystyle J = \{0 \Delta t, 2\Delta t, \dots, T\}$ as both a European put and a European call option contract with the same strike price $\displaystyle K$ and maturity $\displaystyle T$ written on it. At time $\displaystyle t = 0$ the price of the call option is given by,

$\displaystyle \Pi_C (0) = e^{-rT} \sum_{j=0}^n {n \choose{j}} (p^*)^j (1-p^*)^{n-j}(S_0 u^j d^{n-j} - K)^+$

where $\displaystyle p^*$ is the risk neutral probability. Write down the pricing formula $\displaystyle \Pi_P (0)$ for the put option.
Derive the Put-Call parity equation,

$\displaystyle \Pi_C (0) + Ke^{-rT} = \Pi_P (0) + S_0$

Solution...

the price $\displaystyle \Pi_C(0)$ of a European call written on a non-dividend-paying stock is given by...

$\displaystyle \Pi_C (0) = e^{-rT} \sum_{j=a_n}^n {n \choose{j}} (p^*)^j (1-p^*)^{n-j}(S_0 u^j d^{n-j} - K)^+$

where $\displaystyle a_n = \min \{j \in \mathbb{N}_0 \textrm{ such that } S_0u^jd^{n-j} \geq K \}$ and the price $\displaystyle \Pi_P (0)$ of a the corresponding European put is given by

$\displaystyle \Pi_C (0) = e^{-rT} \sum_{j=0}^{a_n-1} {n \choose{j}} (p^*)^j (1-p^*)^{n-j}(K-S_0 u^j d^{n-j})^+$

Hence for the put-call parity we get...

$\displaystyle \Pi_C (0) - \Pi_P (0) = e^{-rT} \sum_{j=0}^n {n \choose{j}} (p^*)^j (1-p^*)^{n-j}(S_0 u^j d^{n-j} - K)$
$\displaystyle = S_0 \sum_{j=0}^n {n \choose{j}} (p^*u)^j (1-p^*d)^{n-j} \frac{1}{j!} \frac{1}{(n-j)!} e^{-r \Delta t} + Ke^{-rT} \sum_{j=0}^n {n \choose{j}} (p^*)^j (1-p^*)^{n-j}$

= ...?

I need to simplify the $\displaystyle \frac{1}{j!} \frac{1}{(n-j)!} e^{-r \Delta t}$ part so I can bring it into the $\displaystyle (p^*u)^j (1-p^*d)^{n-j}$ parts and make a substitution but I don't know how to go about doing this...