(a)
We can write following two equation for the prices of the put at the two nodes at the end of the first period.
0xp + 14.6(1-p) = 2.6545(1+r) and 14.6p + 35.06(1-p) = 17.09(1+r)
from the first equation 1+r = 5.500094(1-p) , substituting in the second equation we have
14.6p + 35.06(1-p) = 17.09x5.500094(1-p) or 14.6p =58.93661(1-p)
or p = 0.801459 , substituting in the first equation r = 0.09199 or 9.199 %
(b)
The price of the put option at t=0 is
pE(0) = (1+r)-1{2.6545p + 17.09(1-p)}
= 1.09199-1.{2.6545x0.801459 + 17.09x(1- 0.801459)}
= 5.055
The payoff of the put at expiry(after two periods) is K - S2 if S2 < K, or 0 if S2 >or= K
Therefore at the uppermost node at expiry S2> 65 and the option is out of the money
At the remaining two nodes the option is in the money and must have
65 - S2= 14.6 or S2= 50.4 at the middle node and
65 - S2= 35.06 or S2= 29.94 at the bottom node.
In a risk-neutral world the stock price grows at the risk-free rate therefore we must have at t = 1
S1 = (1+r)-1.{50.4p + 29.94(1-p)} at the bottom node
or S1 = 1.09199-1.{50.4x0.801459 + 29.94x(1-0.801459)}
or S1 = 42.4343
Therefore up movement ratio u = 50.4/42.4342 = 1.18772
and down movement ratio d = 29.94/42.43423= 0.705561
Thus S1 = 50.54/d = 71.6309 at the upper node
and S2= 71.609u = 85.076 at the uppermost node
and finally
S0 = 71.6309/u = 63.31
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