Let S(u) denote the price of a share of stock at the end of y years. a A populat
ID: 3359741 • Letter: L
Question
Let S(u) denote the price of a share of stock at the end of y years. a A population model for the evolution of these stock prices assumes that)are S(y) S(y-1) independent lognormal random variables with parameters = 0.03 and volatility parameter = 0.15. What is the probability that the price of a share of stock at the end of two years is lower than it is today? (b) Suppose that S(g) follows a Geometric Brownian Motion with drift parameter 0.03 and volatility parameter 0.15. what is the probability that the price of a share of stock at the end of six months is less than 80% of what it is today? (o) Suppose that S(u) follows a Geometric Brownian Motion with drift parameter $80. Find the = 0.01 and volatility parameter = 0.2. Suppose that S(1) probability that the price of a share of stock at the end of ten years is more tharn $72.Explanation / Answer
c)
Since X(t) is a geometric Brownian motion, log(X(t)) is a regular Brownian motion with drift rate m = .01 / yr and s = .2 We want to know the probablity that log(X(10)) ³ log(72) given that log(X(1)) ³ log(80). This means
log(X(10)) - log(X(1)) ³ log(72) - log(80) = log(72/80) » - 0.04576
In this case Z = (log(X(10)) - log(X(1)) – mt)/(s) = (log(X(10)) - log(X(1)) (0.01)(9))/(0.2)(sqrt(9)) = (log(X(10)) - log(X(1)) -0.09)/ .6 is a standard normal random variable. So
Pr{log(X(10)) - log(X(1)) > - 0.04576} = Pr{(log(X(10)) - log(X(1)) -.09)/.6 > (- 0.04576-.09)/.6}
= Pr{ Z > --0.226}
= 1 - Pr{ Z £ -0.226} = 1 – 0.410601 = 0.589399» 59%
b)
Since X(t) is a geometric Brownian motion, log(X(t)) is a regular Brownian motion with 0.03 drift and s = .15. We want to know the probablity that log(X(1/2)) <= log(0.8s(0)) given that log(X(0)) ³ log(s(0)). This means
log(X(1/2)) - log(X(0)) <= log(0.8) » -0.0969
In this case Z = (log(X(1/2)) - log(X(0)) – (.03)(1/2))/(0.15) = (log(X(1/2)) log(X(0))- 0.015)/0.1 is a standard normal random variable. So
Pr{(log(X(1/2)) - log(X(0)) -0.015)/.1 <(-0.0969-0.015)/0.1 }
= Pr{ Z <-1.119 } = 0.13157 » 13%
a)The probability of stock price lower at the end of the two year then the current price can be given by
p(s(0)-s(2)>0)
or evolution function is < 1
=> P(evolution function <1) with mu =0.3 and sigma = .15 => 0.42
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