CT8 - April 2018 - Q3

[Niall1234]

I'm having trouble understanding the solution to part (ii).

It's only worth 2 marks so should be straight forward.

I have to find E[St] = E[exp^B(5)], where B(5) is standard brownian motion.

From the notes, the expected value of a security price at time t using Geometric brownian motion is :

E[St] = exp((Zo + mu*t) + 0.5 *sigma^2*t)

As B(5) - N(0, 5), I would expect that, E[S(5)] = exp((0+0*5)+0.5*5*5)= exp(12.5),

Where, mu=0, sigma^2=5, t=5, Zo=0

However in the solutions they end up with a much different answer, and E[S(5)] = exp(0.5*5*1^2), so I'm not sure where the 1^2 is coming from.

Any help would be greatly appreciated.

 

[CapitalActuary]

You’re confusing different notation here. The mu and sigma in the notes’ formula for E[St] are the drift and volatility parameters for a Brownian motion.

For standard Brownian motion, drift=0 and volatility=1, so the equation says E[St]=exp(0.5*t). Here we’re at t=5, so the answer is exp(0.5*5).

Equivalently you could note that B(5)~N{0,5}, so exp(B(5)) is lognormal with parameters mu=0 and sigma=sqrt(5). Here mu and sigma are different concepts; they’re the parameters of a lognormal distribution, not the drift and volatility of Brownian motion. Then using the formula for expected value of a lognormal variable, again we get exp(0.5*5).

When you included the “5” term twice in your formula you were double counting by thinking of sigma as a volatility parameter=5, which is not correct.

The 1^2 term comes from sigma=volatility parameter=1, t=5, and E[St]=exp(0.5*t*sigma^2).