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!
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).
