hello Please note the following extract from the question: "A binomial model for a non-dividend-paying security with price St at time t is as follows: the price at time (t + 1) is either 1.25St (up-jump) or 0.8St (down-jump). Cash receives interest of 10% per time unit. The value of S0 is 100. A derivative security with price Dt at time t pays the following returns at time 2: D2 = 1 : if S2 = 156.25 D2 = 2 : if S2 = 100 D2 = 0 : if S2 = 64." P(iii) "Derive the corresponding hedging strategy, i.e. the combination of the underlying security and the risk free asset required to hedge an investment in the derivative security." The solution: "to hedge at time 1 if S1 = 125 we let the amount invested in the stock be φ and the amount invested in cash be ψ and solve: 1.25φ + 1.1ψ = 1 0.8φ + 1.1ψ = 2" My question is if we are trying to hedge at time 1 then why do we equate the portfolio value at time 1 to the payoff in time 2? any help would be greatly appreciated Thanks
Hi, We need to carefully read this one. D2 is the payoff of the derivative at maturity. So we're setting up a hedging portfolio that will replicate the payoff of the derivative at time 2. So if we're in the middle state with share price 125 then we need a portfolio that will have value 1 if the share price increases to 156.25 and 2 if the share price decreases to 100. Hence the simultaneous equations above. Hope this helps. Joe
hi Joe, thank you for your reply; I am having difficulty understanding. Please can you confirm if my understanding below is correct? at time 2: then 1.5625φ+ψ=1 as well?
Hi, Might be useful to put this into diagram form. Here's the share price tree with payoffs in brackets at time 2: ----------------156.25 (1) -------125 100 ----------100 (2) -------80 ----------------64 (0) If the share price has risen to 125 then the payoff will either be 1 if share price increases or 2 if it falls. Our hedging portfolio is a portfolio we're setting up at time 1. The value of this portfolio needs to be 1 if share price increases and 2 if it falls, and we earn 10% interest on cash regardless. Hence, with the share price either rising by 25% or falling by 20% over that period: 1.25φ + 1.1ψ = 1 0.8φ + 1.1ψ = 2 I think in your expression below you are looking to set something up at time 0 to match the derivative payoff at time 2. Cash would earn two lots of interest over that period so we'd get the three simultaneous equations: 1.5625φ+1.1^2*ψ=1 φ+1.1^2*ψ=2 0.64φ+1.1^2*ψ=0 However, this can't be solved and besides the value of this portfolio needs to equal the value of the derivative at all future times. Hence, we don't work from time 0 to maturity all in one go but work backwards one time step at a time to ensure that we are holding the appropriate portfolio to hedge values/payoffs one time step into the future. Any clearer? Thanks Joe
hi Joe, Please bear with me; my expression is at time 2, therefore cash would earn nothing and the share price is 156.25 so therefore the share holding plus cash must equal the payoff (1 in this case) at the time 2 when the share is 156.25: 1.5625φ+ψ=1 similary at time 2 when the share price is 100 the payoff is 2 φ+ψ=2 I hope Im making more sense now I look forward to hearing from you DT
Hi DT, I can see where our wires have become crossed in this question. In line with the answer i've got phi in my head as the value of the shares with the share price at 125 whereas I think you may be thinking of phi as the shareholding i.e. the number of shares, which is not unreasonable. If phi is your shareholding then you would need to multiply by 156.25, rather than 1.5625. The latter represents returns on the share whereas the former is of course the share price. So, with phi as the number of shares held: phi*S2+cash at time 2 (cash at time 1 multipled by 1.1) = 1 if S2=156.25; 2 if S2=100. Joe
Dear Joe, Thank you very much! This clears up everything. Please feel free to redeem beer from me should you ever find yourself down in sunny South Africa best wishes Dharshan
