Brownian Motion 1 (birth-death)

  • Thread starter tyler_T
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
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The value of c follows from the varience of the process.( Note that this has to be a 1-dimensional random walk).
 
  • #3
17
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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?
 

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