Using Put–Call Parity to Build an Arbitrage Portfolio

# PWIQF excercise solution


# PWIQF excercise solution












I am software developer with no previous experience or knowledge in finance and have recently been starting to build my knowledge in this area. I am working through the book: Paul Wilmott Introduces Quantitative Finance. I ran into an exercise question that I haven't been able to fully figure and was hoping someone could enlighten me.

The question:

```
A share currently trades at $60. A European call with exercise price $58 and expiry
in three months trades at $3. The three month default-free discount rate is 5%. A
    put is offered on the market, with exercise price $58 and expiry in three months, for
$1.50. Do any arbitrage opportunities now exist? If there is a possible arbitrage, then
construct a portfolio that will take advantage of it. (This is an application of put-call
parity.)
```

I have been able to figure out that there is in fact arbitrage (I think anyway) in this situation using the formula C - P = S - Ee^-r(T - t) which gives a value of 1.5 on the left side and 2.8 on the right. The part I can't figure out is how to construct a portfolio to take advantage of the arbitrage.

Also, if anyone can clarify what it means when C - P is less than the right side of that equation vs. when it is greater than the right side would be very helpful as well.

## Answer by Jacob M. Morley (score 1, accepted)

https://quant.stackexchange.com/a/10981

The put-call parity equation: $$c-p = S_0 - Ke^{-rT}$$ can be seen as a equality in cash flows--namely, buying a call and selling a put have equivalent cash flows to the underlying stock price less the strike price of the options. Taking this into $t=0$ means the current price of the call less the current price of the put must equal the present value of the stock less the present value of the strike (which is just $S_0 - Ke^{-rT}$).

So if the LHS costs 1.5, the RHS costs 2.8, and both have the same payouts, what do you do?

You buy the cheaper: $c-p$, and sell the more expensive. Buy the call, sell the put, and sell the stock.

$t=0: \\ \text{Call}: -3.00 \\ \text{Put}: +1.50 \\ \text{Stock}: +60.00 \\ $

Reinvest 58.5 at 5%, 3-mo, which at $t=T$ yields: $58.5 e^{.05\times .25} = 59.24$

$t=T: \\ \text{Call}: \max(S_T-58, 0)\\ \text{Put}: -\max(58-S_T, 0) \\ \text{Stock}: -S_T \\$

Thus, if the stock price is above the strike, the call is worth $S_T-58$, the put is worthless. If the stock is below the strike, the put is worth $58-S_T$, the call is worthless. So through your combined position, you just buy at the strike, 58. So, your position is worth $59.24-58=1.24$ at time $t=T$.

Side note: it's probably worthwhile to check out Hull's book for basic derivative questions.