103 april 2001 Q7 ii)

[april2105]

Hi,
I am trying to see how we match the expectation to the MGF of the normal distr on page 11 of the tables.

We have E[exp(sigma B_t)].

equating the MGF formula and the above would this not mean that:
exp(sigma B_t) = exp(0.5 sigma^2 t^2)?
Assuimng mu = 0?

How do we then go to the solution of
E[exp(sigma B_t)] = exp(0.5 sigma^2 t)

This puzzles me!

Thanks for your help!

 

[Mike Lewry]

We need to be very careful about parameters being used to mean different things in the Tables from in the question.

In the Tables, we're given a formula for M(t), which means E[exp(tX)]
In the question, we want E[exp(sigma.Bt)] where Bt~N(0,t)

So, looking at the coefficient of the random variable, we can see that "t" in the Tables maps onto "sigma" in the question.
And X in the Tables maps onto Bt in the question, so "mu"=0 and "sigma^2"=t

So, making these substitutions:

exp(0.5 sigma^2 t^2) = exp(0.5 t sigma^2).