Delta Hedging A Lookback Option Using the Binomial Asset Pricing Model

[I.K.]

I am studying Stochastic Calculus via Steven Shreve's book 'Stochastic Calculus for Finance I'. I am stuck on exercise 1.5 which involves working out the price of a lookback option at $t=2$ using the Binomial Asset Pricing Model.

The option characteristics are: $S_0 = 4, u = 2, d = \frac{1}{2}, r = \frac{1}{4}$. The payoff of the lookback option is $V_3 = max_{0 \leq n \leq 3} S_n - S_3$.

The question is as follows: An agent sells the look-back option for $V_0 = 1.376$ and buys $\Delta_0 = 0.1733$ shares of stock at time zero. At time one, if the stock goes up, he has a portfolio valued at $V_1(H) = 2.24$. Assume that he now takes a $\Delta_1 = \frac{V_2(HH) - V_2(HT)}{S_2(HH) - S_2(HT)}$ in the stock (ie he re-hedged his short position in the option). Show that, at time two, if the stock goes up again, he will have a portfolio valued at $V_2(HH) = 3.20$, whereas if the stock goes down, his portfolio will be worth $V_2(HT) = 2.40$.

I have worked out that at time two, the value of the portfolio (and so the option value) will be

$X_2(HH) = 6\Delta_1(H) + 2.80$ and $X_2(HT) = -6\Delta_1(T) + 2.80$

But I am not sure how from here how I can show that $V_2(HH) = 3.20$. Any help or hints will be appreciated.

 

[markhobbus]

Just use the risk neutral net present value formula.

p_tilde = q_tilde = 0.5
V_3(HHH) = 0
V_3(HHT) = 8

Therefore, V_2(HH) = (0.5 * 0 + 0.5 * 8) / (1 + 0.25) = 3.2

Hope this helps

 

[I.K.]

Thanks.

I realise I could have used that formula as the book already goes through pricing the option at different periods using the risk neutral formula recursively backwards in time. What I believe the question is asking is how to price option going forwards in time. This is where I am stuck.

At time 1, the question states that it is assumed that the agent has hedged the short position in the option using the standard delta hedging formula. Based on this fact, how do I use this to show/prove that the price is 3.20 at time 2.

 

[markhobbus]

Ok, at t=0 the replicating portfolio is long 0.1733 shares of stock so it invests (1.376 - 0.1733 * 4) in the money market. The stock is assumed to move in the H direction so at t=1, the value of the portfolio is 0.1733 * 8 + (1.376 - 0.1733 * 4) * 1.25 = 2.4. Now the position in the stock is re-hedged so that delta = (3.2 - 2.4) / (16 - 4) = 0.0667. Therefore, the amount of money invested in the money market is (2.4 - 0.0667 * 8). Finally, if at t=2, the stock moves in the H direction, the value of the portfolio is 0.0667 * 16 + (2.4 - 0.0667 * 8) * 1.25 = 3.2 while if it moves in the T direction, the value of the portfolio is 0.0667 * 4 + (2.4 - 0.0667 * 8) * 1.25 = 2.4.