Brownian Motion 1 (birth-death)

tyler_T
Messages
17
Reaction score
0
Problem:

Let X(t), t>0 denote the birth and death process that is allowed to go negative and that has constant birth and death rates Ln = L, un = u (n is integer). Define u and c as functions of L in such a way that cX(t), t>u converges to Brownian motion as L approaches infinity.

Attempt at solution:

Since the expected value of cX(t), must equal 0, it is obvious that u = L.
The answer to the second part is c = 1/sqrt(2L), but I have no idea how to get there.

Can anybody help me make sense of this?

-Tyler
 
Physics news on Phys.org
The value of c follows from the varience of the process.( Note that this has to be a 1-dimensional random walk).
 
That's what I figured, but it just doesn't seem to workout for me.

X(t) ~ binomial(0, number of steps * 1/2 * 1/2)

So if we take L steps per unit of time, var(X(t)) = L*t/4

We want var(cX(t)) = t

So if we take c = 2/sqrt(L), then var(cX(t)) = t.

But this is not the correct answer. The correct answer is 1/sqrt(2*L).

Where am I going wrong?
 
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.

Similar threads

Back
Top