# 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.

• what is the strike of the call option? – Mark Joshi Feb 1 '15 at 23:55
• strike price is $10$ – and I wondered why nobody responds o_O – Dachser Feb 2 '15 at 8:05

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 $$10 - 10/1.049 = 0.46722.$$

• ok that's what I had in mind too. So the author is wrong with his statement I guess? And can you state an arbitrage opportunity if $V_0>0.46722$? – Dachser Feb 2 '15 at 9:59

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!