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Calculating a probability if I'm given a conditional density function

  1. Aug 5, 2013 #1

    fluidistic

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    1. The problem statement, all variables and given/known data+Relevant equations+Attempt at a solution
    Hi, I'm stuck on a problem.
    I have that [/itex]P(x,t|y,0)[/itex] represents the probability density that a function has the value x at time t knowing it had the value y at time [/itex]t_0=0[/itex] .
    Where [itex]P(x,t|y,0)=\frac{1}{\sqrt{2\pi t}\sigma} \left[ \exp \{ - \frac{(x-y)^2}{2\sigma ^2 t} \} -\exp \{ - \frac{(x+y-2B)^2}{2\sigma ^2 t} \} \right][/itex]
    Given [itex]B\geq x[/itex], [itex]B>y[/itex] and both x and y are not worth negative infinite, I'm interested in calculating the probability that the event will happen (i.e. that x=B) at some time.

    Worded differently and adding some extra information, the mean time for the event to happen is infinite, nonetheless the probability for it to happen at some time is 1. I don't know how to prove the latter.
    If I'm not wrong and I evaluated well the units, the units of [itex]P(x,t|y,0)[/itex] are [itex]1/\sqrt V[/itex]. While x and y have units of voltage (V).
     
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  3. Aug 5, 2013 #2

    mfb

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    So x is something fluctuating in time? If yes, how? The given probability distribution does not tell that. If I just evaluate it for enough values of t and assume that all of them are independent (I am quite sure they are not!), I would expect that I can get as close to 1 as I like (with 1 as limit).
     
  4. Aug 5, 2013 #3

    fluidistic

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    Yes x is basically the voltage of something. It fluctuates with time due to a white noise (so can't get the explicit x(t)).
    The book I'm dealing with (Stochastic processes in biology) take a path I don't understand and doesn't deal with my exact case either.
     
  5. Aug 5, 2013 #4

    mfb

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    Hmm... assuming this P includes absorption at B (correct?), its integral should go to zero for large t. This should be easy to show.
     
  6. Aug 5, 2013 #5

    fluidistic

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    Yes B is an absorbing point.
    What integral exactly should go to zero exactly?

    Edit: you mean integrating with respect to x and take the limit for when t tends to infinity?
     
  7. Aug 5, 2013 #6

    mfb

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    Right.
    I expect that this is not possible in an analytic way, but an approximation is sufficient.
     
  8. Aug 5, 2013 #7

    Ray Vickson

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    Is the process Markov? Is it "stationary" in the sense that ##P(x,t|y,0) = P(x,t+s|y,s)## for any ##s > 0##? If these things hold you can try to develop a stochastic DE for X(t), then use Ito's Lemma to develop a differential equation for the absorption probability in terms of x---that is, for the probability that starting from x the process is absorbed eventually. If this material reads like gibberish to you, then I suspect you are being asked to solve a problem that needs tools far ahead of the material you have been shown so far.

    As a hint: look up material on reflected Brownian motion; it looks to me like your density function is related to a process of that type. For example, you could start with http://en.wikipedia.org/wiki/Reflected_Brownian_motion . You should do your own search, as you may find articles much more suited to you requirements.
     
  9. Aug 5, 2013 #8

    fluidistic

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    I see. What about the following argument?:
    Probability for a random walker not to be absorbed at x=B after an infinite amount of time is equal to ##\lim _{t\to \infty}\int _{-\infty}^B P(x,y,t)dx=\int_{-\infty}^B \underbrace {\lim _{t\to \infty} P(x,y,t)}_{=0}dx=0##.
    Where the integrand is worth 0 because ##\lim _{t \to \infty} \frac{\exp (-a/t)- \exp (-b/t)}{\sqrt t}=0##.

    And thus the solution to the original problem is 1. The event will happen (the random walker will get absorbed).

    P.S.:I'm worried about the units of that integral. Square root of voltage. It should have no unit if it's meant to represent a probability.
     
  10. Aug 5, 2013 #9

    fluidistic

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    I don't really know if this hold. x represents the voltage of a cell which is under a white noise potential influence (with mean 0 and strength ##\sigma## or ##\sigma ^2##). If the potential of the cell reaches a threshold, the cell "fires". There is an event. In the model, the voltage threshold is "B" and with a finite amount of time, the voltage cannot reach ##-\infty##. I think the book states that -infinity is a reflection point and B is an absorbption point.
    I've seen in some books and on the web the word "Ito", but nothing else. So yes, I know I must catch up a lot to solve what I want.

    Yes, I think the same, thank you.
     
  11. Aug 5, 2013 #10

    mfb

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    You cannot exchange integration and the limit in that way. This is easier to see with a regular diffusion process, which would vanish in your calculation, too.


    σ has to have units of ##\frac{voltage}{\sqrt{time}}##, otherwise your exponential functions do not work. Therefore, P has units 1/voltage, and integrating over voltage gives a dimensionless number.
     
  12. Aug 5, 2013 #11

    Ray Vickson

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    It seems like your process is Markov and stationary: Markov = future depends only on present, and not the past. Stationary: = dynamics is independent of when we decide to start; that is, it does not matter if "time zero" is 2 pm or 7 am or what. These ought to hold if the system is not being filtered by a device with a time constant (so that a high voltage, for example, takes a time to die down) and if the system does not possess "hysteresis".

    So, if your textbook's description is correct, you have a Brownian motion on ##(-\infty,B]## with zero drift and absorption at ##B##. On the web you may be able to find expressions related to Brownian motion on ##[0, \infty)## with absorption at ##0##. In that case your process is just a translation by ##B## and a switching of + and - directions, but is otherwise the same.

    Anyway, if you accept without proof the expression for ##f(x,t) = P(x,t|y,0)##, then you have everything you need. To see this, let ##T## be the time (if ever) that the system is absorbed at level B. The function ##f(x,t)## gives the probability density for being at x at time t, so absorption cannot have happened yet. That is,
    [tex] P\{ T > t \} = \int_{x=-\infty}^B f(x,t) \, dx[/tex]
    because ##T > t## if and only if ##X(t) < B##. (Do you see why?) So, if you can do the integral (and it is doable in terms of the error function or the standard normal cdf) you can ask for the probability that the process is abserbed eventually; this is just ##P\{ T < \infty \}##.
     
    Last edited: Aug 5, 2013
  13. Mar 1, 2014 #12

    fluidistic

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    Ok guys I'm back to this topic for now.
    I am still trying to solve (even with numerical help is welcome) $$\lim _{t\to \infty}\int _{-\infty}^B P(x,y,t)dx$$.
    I cannot put the limit inside the integral as you pointed out. I evaluated numerically the indefinite integral ##\int \exp \left [ - \left ( \frac{1}{c} \right ) (x-a)^2 \right ] dx## to be worth ##\frac{\sqrt \pi }{\left ( \frac{2}{\sqrt c} \right )} \text{erf} \left ( \frac{1}{\sqrt c}x-a \right )##. Using this information I could evaluate the definite integral (by hand) to be worth ##\frac{\sqrt {\pi} \sigma }{\sqrt 2} \left [ 1 +\text{erf} \left ( \frac{B}{\sqrt {2t} \sigma} -y \right ) + 1 +\text{erf} \left ( \frac{B}{\sqrt {2t} \sigma} +2B -y \right ) \right ]##
    Then, when I take the limit t tends to infinity, I get ##1+ \frac{\text{erf}(-y)+\text{erf}(2B-y)}{2}## which, quite unfortunately, differs from 0.

    I do not know whether I made some mistake(s) or whether the value of the limit of the integral really differs from 0. Any comment is appreciated.
     
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