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Math Help - an almost random walk SP

  1. #1
    tst
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    an almost random walk SP

    Hi, i hope that i am at the correct category.

    This question is bugging me for too long, so i hope someone will be able to help.

    I have a discrete stochastic process defined as

    Let 0\le k, a_0\le 1,
    a_{n+1} becomes a_n+k(1-a_n) with probability a_n
    and a_{n+1} becomes a_n-k a_n with probability 1-a_n

    we have E(a_{n+1}|a_n)=a_n and V(a_{n+1}|a_n)=k^2 a_n(1-a_n) with V(a_{n+1}|a_n) denoting the variance of a_{n+1} with a_n known.

    Numerically i found that a_t \backsimeq N[a_0,k^2 a_0(1-a_0)t] but this is an aproximation because there should not be any chance of a_t being outside of [0,1].

    So what i am trying to do is to construct the continuous process which is the limit of the above discrete one. Much like random walk and weiner process.

    I would also be content if i was able to construct the corresponding SDE of the process, or the PDE for the evolution of the probability density. In short, any help is more than welcome.

    EDIT:The probability of increase is a_n and not a
    Last edited by tst; July 14th 2010 at 09:45 PM.
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  2. #2
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    Using the fact that E \left[ E (a_{n+1}|a_n) \right] = E [a_{n+1}] you can show E [a_{n}] = a_0 for all n by induction.

    Using the same fact, for the variance you can derive the following recurrence:
    (hopefully I have not made any mistakes in my calculations)

    <br />
Var (a_{n+1}) = (1-k^2) Var (a_n) + k^2 a_0 ( 1- a_0 )<br />

    The solution to the recurrence (using generating functions for example) is

    <br />
Var(a_n) = (1-a_0) a_0 (1-(1-k^2)^n)<br />
.

    Note I didn't actually solve the recurrence myself, I used wolfram alpha.

    so there may be an error there.

    Now that you know the true expectation and variance, maybe try altering your guess to a normal with those values? Tell me if it seems more reasonable. To prove the actual distribution is normal with that expectation and variance shouldn't be too hard if its true ( I just might bite my tongue on that later .)

    Whats interesting to note is that

    <br />
\lim_{n \rightarrow \infty} {Var(a_n)} = a_0 ( 1 - a_0 )<br />
.
    Last edited by gmatt; July 14th 2010 at 06:50 PM.
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  3. #3
    tst
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    hello, gmatt, thank you for you reply. The expectation was pretty straightforward, but i didn't think to find a recursive relation for variance. It may be proved very helpful.


    I think i should say a bit more about this construction. The process is constructed such way, so it is impossible for a_n to leave the interval [0,1]. This means that the density function must be compactly supported in [0,1] and obviously the normal distribution is not. So i doubt that the solution is a normal distribution. An idea that i had was to use a transformation from [-\infty,\infty] to [0,1]. But i cannot do something with that because i don't have any relation for the solution, so i cannot check my idea.

    I tried to find the PDE for the probability density as it was done for random walk.
    (one can see how this is done at page 34 of this document)
    However unlike random walk the increments are a function of the position which make things more complicated that i can handle. Any idea here is useful as any idea on constructing the SDE of the process.

    I will try to find an explicit expression for Var(a_t). I will update with (any) results.

    cheers
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  4. #4
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    This is just a vague idea I had:

    based on the construction of the weiner process found here (using the same notation),

    let a_t^{\Delta} = a_n \Delta x then

    Var(a_t^{\Delta}) = (\Delta x )^2 Var( a_n )<br />
    <br />
 = (\Delta x )^2 (1-a_0) a_0 (1-(1-k^2)^n)
    <br />
= (\Delta x )^2 (1-a_0) a_0 (1-(1-k^2)^{\frac{t}{\Delta t}})<br />

    Now if you let \Delta x = c ( 1 - \Delta t ) we have that

     Var( a_t ) = \lim _ {\Delta t \rightarrow 0} Var( a_t ^ {\Delta}) = c^2 ( 1 - a_0 ) a _0 . This type of stohastic process would have much different properties than a weiner process and I'm not sure it would be useful.
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