How Can I Solve This Complex Problem Involving Brownian Motion Variables?

  • Thread starter Thread starter boliobolo
  • Start date Start date
  • Tags Tags
    Motion
boliobolo
Messages
2
Reaction score
0
can someone show me the solution for E[B(u)B(u+v)B(u+v+w)B(u+v+w+x)], for 0<u<u+v<u+v+w<u+v+w+x .the answer is 3u^2+3uv+uw
 
Physics news on Phys.org
It dosn't look right. There is no x in the answer.
 
it is the answer. with no x
 
I am not familiar with your notation. What is the definition of B(u)?
 
The solution is presumably judicious applications of conditional expectations and properties of Brownian motion. It's certainly worth having a go yourself and if you don't mind me asking how far have you got? As a (rather unhelpful) hint your first move is the reason x doesn't turn up in the solution.
 
Sorry I'm late - finally understood the notation. To refresh, Brownian motion variables are normally distributed with mean 0, variance ~ time, and independent increments.

For the problem stated: B(u) = U, B(u+v) = U + V, B(u+v+w) = U + V + W, and B(u+v+w+x) = U + V + W + X, where U, V, W, X are independent normally distributed random variables with mean 0 and variances u, v, w, x.

The question is then E(U(U+V)(U+V+W)(U+V+W+X)). To evaluate this, note that for any term in the expanded polynomial with any of the variables to the first power, the expectation (E) will be 0.
So we are then left with E(U4 + 3U2V2 + U2W2) = 3u2 + 3uv + uw.
 
Last edited:
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Thread 'Detail of Diagonalization Lemma'
The following is more or less taken from page 6 of C. Smorynski's "Self-Reference and Modal Logic". (Springer, 1985) (I couldn't get raised brackets to indicate codification (Gödel numbering), so I use a box. The overline is assigning a name. The detail I would like clarification on is in the second step in the last line, where we have an m-overlined, and we substitute the expression for m. Are we saying that the name of a coded term is the same as the coded term? Thanks in advance.
Back
Top