Skip to content
All library documents

Replicating a One-Period Call to Identify Its No-Arbitrage Price

Article Quant Q&A · Author: Dachser
Read the original

Summary

The document analyzes a European call in a one-period, two-outcome stock model and asks how its price can avoid arbitrage. The answer uses replication: because the strike equals the stock’s lower possible price, the call payoff matches the stock price less the strike in both states. A portfolio holding one share and borrowing the discounted strike therefore reproduces the option payoff.

The resulting option value is the current stock price minus the present value of the strike, using the stated risk-free rate. A second portfolio construction illustrates how selling overpriced calls and combining them with shares and bonds can produce a positive payoff in both states. The example shows how replication pins down an option price in a complete market; its conclusion relies on the specified two-state model and does not address transaction costs or other market frictions.

Key ideas

  • When the strike equals the stock’s lower possible terminal price, the call payoff equals the stock price minus the strike in each state.
  • One share combined with borrowing the discounted strike replicates that call payoff.
  • In a complete two-state market, replication determines the no-arbitrage call price.
  • A call priced above its replication cost can be sold against a portfolio of shares and bonds to create an arbitrage in the model.

Tags

OptionsArbitrageDerivatives pricing
Full text
# arbitrage opportunity in a two period model


# arbitrage opportunity in a two period model












I have a little problem evaluating an european call. I Suppose the following:

in $$t=0 : S_0 = 10$$ $$t = 1 : S_1 = \{10,11\}~with ~p=0.5$$

riskless rate : $(1+r)=\beta=1.049$

Strike price: $K=10$

Now the author of my paper states that the value of the call must be $V_0 < 1/\beta (11 - 10) * 0.5 = 0.477$ in order to avoid arbitrage. Can anyone see how this is the case?

I am aware that the stock is not valued with respect to the fair valuation after cox rubinstein which would be $S_0=10.5$. If then $K=10.5$ and for example $V_0=0.5$ i could construct arbitrage by:

in $t=0$: Sell 2 calls and buy a zerobond. If $P$ denotes the portfolio value I have $P_0=2*0.5-1=0$

in $t=1$ when $S_1=10~P_1=1.049-2*0>0$

in $t=1$ when $S_1=11~P_1=1.049-2*(11-10.5)>0$

=> arbitrage

For $S_0=10$: for $V_0 > 1/\beta$ i see that one can short the call and buy a zerobond instead which lets me pay my liability in period 1 in any case just as above.

## Answer by Mark Joshi (score 1)

https://quant.stackexchange.com/a/16427

it's a complete market so the price of the call option is determinable.

There are multiple ways to do it. If the strike is 10 it is particularly easy since the pay-off is $S_1 - 10$ so the value is $S_0 - 10/\beta$ as we can replicate precisely with one unit of $S$ and cash that will be worth $-10$ at time $1$ which $-10/\beta$ units of the cash account today.

The value is therefore $$10 - 10/1.049 = 0.46722.$$

## Answer by Dachser (score 0)

https://quant.stackexchange.com/a/16428

finally:

in $t=0$: sell 19 Bonds, sell 2 calls for $V_0=0.5$, buy 2 shares for 10 => $$P_0=19+2*0.5-20=0$$

in $t=1,S_1=10$ $$P_1 = 2*10-19.9481-2*0=0.069>0$$

in $t=1,S_1=11$ $$P_1=2*11-19.9481-2*1=0.069>0$$

gosh, that took ages.. Thanks anyway!

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.