Brownian Motion

MaxPower89

Hi,

i need to evaluate E[exp(sigma.W(t))], sigma>0, W(t) is standard brownian motion.

also can i take the expected value inside the brackets and exaluate as
exp(sigma.E[W(t)])???

Also i need to prove that standard brownian motion is a martingale..i can do it for brownian motion but not for standard..

any help much appreciated..

Max

Fremen

i need to evaluate E[exp(sigma.W(t))], sigma>0, W(t) is standard brownian motion.

E[exp(sigma.W(t))] = exp(sigma*sigma*t/2)

You can see this by solving Integral(((1/Sqrt(2*PI)*t))*exp(sigma.Z*(sqrt t)) * exp(-Z*Z/2)dz)
from minus infinity to infinity.

If you don't like that method, you can use ito's lemma to show that

exp(sigma.W(t))*exp(-sigma*sigma*t/2) = integral(exp(sigma.W(t))dW)

which is a martingale with initial value one.

also can i take the expected value inside the brackets and exaluate as
exp(sigma.E[W(t)])???

No.
You can think of the E[] operator as being an integral from minus infinity to infinity. There's no rule which says

integral(exp(x)dx) = exp(integral(x dx))

Also i need to prove that standard brownian motion is a martingale..i can do it for brownian motion but not for standard..

I think you might be a bit confused about what a standard brownian motion is here. It's a brownian motion with mean 0 and variance t.

You can show it's a martingale by

E[w(t) | Fs] = E[w(t) - w(s) + w(s) |Fs] = E[w(t) - w(s)] + E[ w(s) |Fs]

Now, w(t) - w(s) is independant of Fs, so the expectation is zero. w(s) is Fs-measurable, so the expectation is W(s)

so the above simplifies to E[w(t) | Fs] = w(s) so w is a martingale.

MaxPower89

thanks for that.

and how do i prove that the expected value is less than infinity to show its a martingale? i have a solution for brownian motion and it works out to sqrt((2.sigma.sigma.t)/pie)....so can i just remove the sigma^2 here to get sqrt(2.t/pie)..and this is less than infinity?

thanks again

Fremen

MaxPower89 wrote: »

thanks for that.

and how do i prove that the expected value is less than infinity to show its a martingale? i have a solution for brownian motion and it works out to sqrt((2.sigma.sigma.t)/pie)....so can i just remove the sigma^2 here to get sqrt(2.t/pie)..and this is less than infinity?

thanks again

Sorry, not sure that I understand the question. What do you mean by a solution for brownian motion?

MaxPower89

im just stuck on trying to show that the expected value of the absolute value of standard brownian motion is less than infinity..ie...E[|W(t)|] < infinity. I have this worked out for brownian motion and the expectation worked out to sqrt((2.sigma.sigma.t)/pie), which is less that infinity.

I think i should be "getting" this a lot easier but its just not clicking with me.

Fremen

MaxPower89 wrote: »

im just stuck on trying to show that the expected value of the absolute value of standard brownian motion is less than infinity..ie...E[|W(t)|] < infinity. I have this worked out for brownian motion and the expectation worked out to sqrt((2.sigma.sigma.t)/pie), which is less that infinity.

I think i should be "getting" this a lot easier but its just not clicking with me.

E[|W(t)|] <= (E[W(t)^2])^1/2 = t^(1/2)

since if p < q, then the p-norm of a random variable is less than or equal to the q-norm.