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Brownian Motion (I think)

Jason

New member
Feb 5, 2012
28
I need to answer these questions, but I don't have a clue what they mean. Could anybody shed some light?

Find:

(a) $E({B_1^4})$

(b) $E({B_1^6})$

(c) $E(e^{B_1})$

(e) $E({5B_1}^4+{6B_1}^2+{5B_1}^3)$

(e) $E(B_2 B_3)$

(f.) $E(e^{B_2+B_3})$
 

Moo

New member
Feb 12, 2012
26
Hello,

Yes, B indeed stands for a brownian motion, so you have to keep the basic properties in mind.

B0 = 0
Bt-B0 follows a normal distribution with mean 0 and variance (not standard deviation) t.
So in order to solve most of these questions, you need to know the moments of a normal distribution.

For the one involving B2*B3, you have to know that the incremets onf a brownian motion are independent. So in particular, B2*B3=(B3-B2)*B2+B2^2. So computing the expectation knowing the independence is easy.

For question f, you have to know the mgf of a normal distribution, that's all (subtract B0=0 if you have trouble seeing it).



Note for later : the advanced probability is the first thing I look for when I come at MHB, there's no need to send me a PM. If I can answer, I'll answer, if not, I'll try or I leave it :p
 

Jason

New member
Feb 5, 2012
28
I'm absolutely perplexed. Could you work through one of the problems so I get an idea of how to do the other problems? (I promise this isn't a homework assignment; it's an exercise sheet that is... meant to help familiarize me with the "basics.") I'm looking through a zillion stochastic calculus texts, but still don't have a clue.
 

Moo

New member
Feb 12, 2012
26
I'm not scarce on details because it could be an assignment, but trust me you just need the basic properties of a brownian motion to solve these questions. They can be found here : http://en.wikipedia.org/wiki/Brownian_motion#Mathematics (it can be B or W, most of the time, people don't make any difference)

For example the first one, just note that B_0=0, so B_1 = B_1 - B_0, but the third property says that B_1-B_0 follows a normal distribution N(0,1-0), so you have to get the 4th moment of a standard normal distribution. (I think it's 3, you can get it by computing the integral or just looking on the internet).

It's always this trick, you make appear the difference between B_j and B_k and you work on it with the properties you're given. Just give it a try :p I really learned about brownian motion 5 months ago...
 

Jason

New member
Feb 5, 2012
28
Okay, I think I see.

$B_1\sim\textbf{N}(0,1)$

So: $E({B_1})=E(X)=0$, $E({B_1}^2)=E(X^2)=1$, etc.


(a) $E({B_1^4})=E(X^4)=3$

(b) $E({B_1^6})=E(X^6)=15$

(e) $E({5B_1}^4+{6B_1}^2+{5B_1}^3)=5E(X^4)+6E(X^2)+5E(X^3)=5(3)+6(1)+5(0)=21$
 

Moo

New member
Feb 12, 2012
26
Yep, that's a good start ! :)
 

Jason

New member
Feb 5, 2012
28
(e) $ E(B_2 B_3)=E({B_2}^2))=(0)^2+2=2$

(c) $ E(e^{B_1})=e^{1/2}$

(f.) $E(e^{B_1+B_2})=e^{B_1}e^{B_2}=e^1e^{3/2}=e^{5/2}$
 
Last edited:

Moo

New member
Feb 12, 2012
26
That's almost perfect. For the latter one, it's $E[e^{B_1+B_2}]=E[e^{B_1}]E[e^{B_2}]=e^{1/2}e^{2/2}=e^{3/2}$

Maybe you ought to put more details for the step, especially for (e).

See, not that difficult ? :p
 

Jason

New member
Feb 5, 2012
28
Oops, f) should have been: $E[e^{B_2+B_3}]$

Now only ten more questions! Suppose I should start a new thread.