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Re: Re: Cumulative probability that random walk variable
It sounds like what the OP intended here was to compute the probability
that a standard Brownian motion exceeds the value 2 in the time interval
between 0 and 5. Of course his description does not make it clear that
he is referring to a Brownian motion process - it's not enough simply to
say B[t] is normally distributed with mean o and standard deviation
Sqrt[t], one needs to add some other conditions iid increments and
continuous paths. But that means of course that he is not assuming that
x[t] are independent - only that the increments are independent.
If this is correct than the question refers to a very standard topic
which can be found in almost any introduction to stochastic calculus and
is referred to as the First Pasage Time Distribution. One (of very many)
elementary derivations can be found on page 112 of volume 2 of S.E.
Shreve's "Stochastic Calculus for Finance". I won't give it here as it
has already been done by Roland Franzius in this thread.
Andrzej Kozlowski
On 12 Nov 2009, at 20:05, Leonid Shifrin wrote:
> Hi Kelly,
>
> I am not sure if the notion of "continuous random walk" makes
> sense without refinement (for me at least). I have a feeling that before you
> start using Mathematica to solve your problem, you may want to better
> formulate it and see what is the question you really want to ask.
>
> The probability you ask for is exactly 1 if you assume (as you presumably
> do) that x[t] are independent random variables for different t, and that <t>
> is a continuous variable.
>
> The simplest way to see this is to note that the probability you
> want is
>
> 1 - P{all x[t] are smaller than 2} ==
> 1-Product_{all values of t}(P{x[t]<2})
>
> The product above contains (uncountably) infinitely many terms,
> (one for each value of t), and all of them except for the very first one (at
> t=0) are strictly less than 1 (at least for the normally-distributed x[t],
> where for your numbers, each term in the product is 1/2 (1 +
> Erf[Sqrt[2]/Sqrt[t]])), therefore the product is zero, and the probability
> in question is 1.
>
> "Physically", this happens because, by making x[t] independent
> normally-distributed random variables, you allow for arbitrarily large jumps
> in your trajectories, even for infinitesimally close to each other points
> in time, so a generic trajectory x[t] will be a very messy function of
> <t>, jumping at every point, something like a Dirichlet function (but with
> unbounded jumps). Therefore, the probability that it will not jump over 2
> (or any other finite number for that matter) is just zero.
>
> I can see two ways of making the problem more meaningful. One is to
> discretize the time and fix the number of jumps (steps). In this case, the
> above-mentioned product will be a (possibly very small) finite number.
>
> Another is to somehow restrict your random trajectories to be more "smooth"
> and well-behaved functions. This however will mean that variables x[t] will
> no longer be independently distributed.
> This may eventually lead you to the functional (path) integrals,
> which can be modeled and computed numerically and may provide you with the
> answers. But doing so it sufficiently more complicated and will still
> require a precise formulation of the random system you wish to model, and
> the properties you want to test. You may find that in this approach you will
> be asking different sorts of questions. If you wish to go this way, you may
> find it of interest to look at how this is done for certain simple
> one-dimensional quantum-mechanical or statistical systems. Keywords will
> probably include Quantum Mechanics, Statistical Mechanics, Path Integrals,
> Metropolis algorithm, Monte-Carlo simulations.
>
> Hope this helps. My apologies if I misunderstood the question.
>
> Regards,
> Leonid
>
>
>
>
> On Wed, Nov 11, 2009 at 12:28 PM, Kelly Jones
> <kelly.terry.jones at gmail.com>wrote:
>
>> How can I use Mathematica to solve this problem?
>>
>> Let x[t] be a normally-distributed random variable with mean 0 and
>> standard deviation Sqrt[t].
>>
>> In other words, x[0] is 0, x[1] follows the standard normal
>> distribution, x[2] follows the normal distribution with mean 0 and
>> standard deviation Sqrt[2], etc.
>>
>> It's easy to compute the probability that x[5] > 2 (for example).
>>
>> How do I compute the probability that x[t] > 2 for 0 <= t <= 5.
>>
>> In other words, the probablity that x[t] surpassed 2 at some point
>> between t=0 and t=5, even though x[5] may be less than 2 itself. Notes:
>>
>> % My goal: predicting whether a continuous random walk will exceed a
>> given value in a given period of time.
>>
>> % I realize that saying things like "x[5] may be less than 2" is
>> sloppy, since x[5] is a distribution, not a value. Hopefully, my
>> meaning is clear.
>>
>> % I tried doing this by adding/integrating probabilities like this
>> (psuedo-code):
>>
>> P(x[t] > 2 for 0 <= t <= 5) = Integral[P(x[t] > 2),{t,0,5}]
>>
>> but this overcounts if x[t] > 2 for multiple values of t.
>>
>> --
>> We're just a Bunch Of Regular Guys, a collective group that's trying
>> to understand and assimilate technology. We feel that resistance to
>> new ideas and technology is unwise and ultimately futile.
>>
>>
>
>
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