dstefan
06-10-2008, 02:50 AM
Chapter 3, Problem 3
Over each of three consecutive time intervals of length \tau = 1/12, the price of a stock with spot price S_0 = 40 at time t=0 will either go up by a factor u = 1.05 with probability p=0.6, or down by a factor d = 0.96 with probability 1-p=0.4.
Compute the expected value and the variance of the stock price at time T = 3\tau, i.e., compute E[S_T] and var(S_T).
Solution:
The probability space S is the set of all different paths that the stock could follow three consecutive time intervals, i.e.,
S ~=~ \{ UUU,UUD,UDU,UDD,DUU,DUD,DDU,DDD \}
where U represents an ``up" move and D represents a ``down" move.
The value S_T of the stock at time T is a random variable defined on S, and is given by
S_T(UUU) = S_0 u^3; S_T(DDD) = S_0 d^3;
S_T(UUD) = S_T(UDU) = S_T(DUU) = S_0 u^2 d;
S_T(UDD) = S_T(DUD) = S_T(DDU) = S_0 u d^2
Note that
P(UUU) = p^3; P(DDD) = (1-p)^3;
P(UUD) = P(UDU) = P(DUU) = p^2 (1-p);
P(UDD) = P(DUD) = P(DDU) = p (1-p)^2
We conclude that
E[S_T] = S_0 u^3 \cdot p^3 +~ 3 S_0 u^2 d \cdot p^2 (1-p) +~ 3 S_0 u d^2 \cdot p (1-p)^2 +~ S_0 d^3 \cdot (1-p)^3 = 41.7036
E[(S_T)^2] = (S_0 u^3)^2 \cdot p^3 +~ 3 (S_0 u^2 d)^2 \cdot p^2 (1-p) +~ 3 (S_0 u d^2)^2 \cdot p (1-p)^2 +~ (S_0 d^3)^2 \cdot (1-p)^3 ~=~ 1749.0762
\mbox{var}(S_T) = E[(S_T)^2] - \left( E[S_T] \right)^2 ~=~ 9.8835
Over each of three consecutive time intervals of length \tau = 1/12, the price of a stock with spot price S_0 = 40 at time t=0 will either go up by a factor u = 1.05 with probability p=0.6, or down by a factor d = 0.96 with probability 1-p=0.4.
Compute the expected value and the variance of the stock price at time T = 3\tau, i.e., compute E[S_T] and var(S_T).
Solution:
The probability space S is the set of all different paths that the stock could follow three consecutive time intervals, i.e.,
S ~=~ \{ UUU,UUD,UDU,UDD,DUU,DUD,DDU,DDD \}
where U represents an ``up" move and D represents a ``down" move.
The value S_T of the stock at time T is a random variable defined on S, and is given by
S_T(UUU) = S_0 u^3; S_T(DDD) = S_0 d^3;
S_T(UUD) = S_T(UDU) = S_T(DUU) = S_0 u^2 d;
S_T(UDD) = S_T(DUD) = S_T(DDU) = S_0 u d^2
Note that
P(UUU) = p^3; P(DDD) = (1-p)^3;
P(UUD) = P(UDU) = P(DUU) = p^2 (1-p);
P(UDD) = P(DUD) = P(DDU) = p (1-p)^2
We conclude that
E[S_T] = S_0 u^3 \cdot p^3 +~ 3 S_0 u^2 d \cdot p^2 (1-p) +~ 3 S_0 u d^2 \cdot p (1-p)^2 +~ S_0 d^3 \cdot (1-p)^3 = 41.7036
E[(S_T)^2] = (S_0 u^3)^2 \cdot p^3 +~ 3 (S_0 u^2 d)^2 \cdot p^2 (1-p) +~ 3 (S_0 u d^2)^2 \cdot p (1-p)^2 +~ (S_0 d^3)^2 \cdot (1-p)^3 ~=~ 1749.0762
\mbox{var}(S_T) = E[(S_T)^2] - \left( E[S_T] \right)^2 ~=~ 9.8835