Brownian motion
- Thread starter amir
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If the variables are normally distributed then it has already been assumed that they should follow some distribution. In this case the model fitted is the normal distribution. While analyzing the data, what do you do first??? Check if they are iid to fit some model of distribution. If they are not iid and for example you concluded they are time series (like stock prices) then what model are you applying???
That wouldn't be a statistical distribution but try to find some dependence on time. In your particular case the normal distribution is already gives so they are identically independently distributed random variables.
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first I tried the integral method by conditioning on ... but it gets nasty by having the CDF of normal distribution as the integrand of a double integral. so we need approximation and ...
It's a simple integral that needs to be computed provided that we already have been given that integral and care not with the background. When you say that it's not a real application I completely agree since in real application you have to define whether the data you are analyzing is iid or time series or neither in order to fit the model. I got your point you are true.
Do you have trouble in computing the double integral or need some kind of approximation algorithms?
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I think I have to compute this integral
it needs some approx., but my profs. said that you should get a specific number ....
P(B3<B2 | B2>B1, B1>B3) =1 since you already know that condition already B3<B1<B2.
So if you write P(B3<B2) * P(B2>B1, B1>B3) which is your outcome in line 3 of code, that's gonna break since the assumption is false. So you are not gonna get any frictional number in the probability. N(0) = 0.2667 is not reachable from the first part of the statement. Not sure though...
So if you write P(B3<B2) * P(B2>B1, B1>B3) which is your outcome in line 3 of code, that's gonna break since the assumption is false. So you are not gonna get any frictional number in the probability. N(0) = 0.2667 is not reachable from the first part of the statement. Not sure though...
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P(B3<B2 | B2>B1, B1>B3) =1 since you already know that condition already B3<B1<B2.
So if you write P(B3<B2) * P(B2>B1, B1>B3) which is your outcome in line 3 of code, that's gonna break since the assumption is false. So you are not gonna get any frictional number in the probability. N(0) = 0.2667 is not reachable from the first part of the statement. Not sure though...
you are right. I have to change the order and try again
why N(0) = 0.2667?? it is 0.5
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I dont know about the method!!!
If we took my version of path than that'd be 0.2667 but as I said I'm not sure if it's true...
As for Monte Carlo method for integration, Ill upload the PDF here. In Brownian cases it's very common to use multiple integrals and so it's useful (especially its fast when coding) but now in your case since you are evaluating on one dimension than any approximation algorithms are possible.
As for unbounded integrals the way is to rescale the integrand into manageable scale of [-1,1] as Gauss-Legendre algorithm does and integrate over that integral.
BTW, Im currently programming the MC integration in C++/C#. Made quite a big code and seems hard to maintain. Ill do the checking tomorrow and I'll upload it here if needed since that'll give much flexibility in numerical integration over multiple dimensions.I dont know about the method!!!
bob
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Okay, as it turns out the original post had the right idea, but after a crazy day my execution was a little off.
As I understand the problem, (B_1) is a bit of a red herring. We're looking essentially for the probability that the second Brownian increment (x=B_2-B_1 > 0) and then the third Brownian increment (y = B_3-B_2 < -x). Given (x), the probability (P(y < -x) = N(-x)); all we have to do is integrate over all (x) that satisfy the initial condition, weighted by the probability of that outcome, to get the answer.
In other words, we want to evaluate the integral:
(\int_{0}^{+\infty} N'(x)N(-x) dx)
By symmetry, we see that...
(\int_{0}^{+\infty} N'(x)N(-x) dx = \int_{0}^{+\infty} N'(x)[1-N(x)] dx)
(=\int_{0}^{+\infty}N'(x)dx - \int_{0}^{+\infty} N'(x)N(x)dx = \frac{1}{2} - \int_{0}^{+\infty} N'(x)N(x)dx)
That second integral can be done pretty easily (try it by parts) and is equal to (\frac{3}{8}), giving the eventual answer of (\frac{1}{8}).
That final answer actually makes intuitive sense. After all, the simple probability that (x > 0) and then (y < 0) is (\frac{1}{4}). Now, take some path (i) satisfying this condition, and suppose arbitrarily that (x_i > -y_i); then in this case our condition is not satisfied. However, for each such path we can identify an equally probable path by just swapping the values (x_{i'} = -y_i, y_{i'} = -x_i) that does satisfy the condition.
So, among our universe of (\frac{1}{4}) of all possible paths, there is a one-to-one correspondence between paths that satisfy the desired conditions and ones that don't. Since the set of paths ending exactly at zero is of zero measure, we conclude that the probability of satisfying the condition we want is (\frac{1}{8}).
As I understand the problem, (B_1) is a bit of a red herring. We're looking essentially for the probability that the second Brownian increment (x=B_2-B_1 > 0) and then the third Brownian increment (y = B_3-B_2 < -x). Given (x), the probability (P(y < -x) = N(-x)); all we have to do is integrate over all (x) that satisfy the initial condition, weighted by the probability of that outcome, to get the answer.
In other words, we want to evaluate the integral:
(\int_{0}^{+\infty} N'(x)N(-x) dx)
By symmetry, we see that...
(\int_{0}^{+\infty} N'(x)N(-x) dx = \int_{0}^{+\infty} N'(x)[1-N(x)] dx)
(=\int_{0}^{+\infty}N'(x)dx - \int_{0}^{+\infty} N'(x)N(x)dx = \frac{1}{2} - \int_{0}^{+\infty} N'(x)N(x)dx)
That second integral can be done pretty easily (try it by parts) and is equal to (\frac{3}{8}), giving the eventual answer of (\frac{1}{8}).
That final answer actually makes intuitive sense. After all, the simple probability that (x > 0) and then (y < 0) is (\frac{1}{4}). Now, take some path (i) satisfying this condition, and suppose arbitrarily that (x_i > -y_i); then in this case our condition is not satisfied. However, for each such path we can identify an equally probable path by just swapping the values (x_{i'} = -y_i, y_{i'} = -x_i) that does satisfy the condition.
So, among our universe of (\frac{1}{4}) of all possible paths, there is a one-to-one correspondence between paths that satisfy the desired conditions and ones that don't. Since the set of paths ending exactly at zero is of zero measure, we conclude that the probability of satisfying the condition we want is (\frac{1}{8}).
