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Joined: March 28th, 2012, 12:10 pm

### Exact simulation of brownian motion and of its integral

Hi everyone,maybe this is a very trivial question but I would like to have some suggestions on this. I would like to jointly simulate a brownian motion $W(t)$ and its integral $Y(t):=\int_0^t{W(s)ds}$.More precisely, what I would like to have is an exact updating rule that allows me to simulate $W(t_2)$ and $Y(t_2)$ given $W(t_1)$ and $Y(t_1)$. Many thanks in advance

Alan
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### Exact simulation of brownian motion and of its integral

just thinking out loud...the first step is developing the joint distribution of $(W(t),Y(t))$.is it a bivariate normal or something else? Left to you to nail down. let's say it *is* bivariate normal. Then, I would draw by(i) making a draw of $W(t)$; (ii) then drawing $Y(t) = A(t) [\rho(t) W(t) + \sqrt{1 -\rho(t)^2} Z(t)]$,choosing $(A(t),\rho(t))$ to make the variance, and covariance of $Y(t)$ come out right. (We know the mean is correctly 0).Here $Z(t)$ is an independent normal draw with mean 0 and variance t. Once that's nailed down, doing it from some t1 instead of t=0 should be clear.Anyway, something like that ...
Last edited by Alan on February 28th, 2016, 11:00 pm, edited 1 time in total.

Topic Author
Posts: 33
Joined: March 28th, 2012, 12:10 pm

### Exact simulation of brownian motion and of its integral

Thanks Alan, that's precisely what I am trying to figure out. Surely $Y(t)$ is normal, however what I miss is the correlation structure.I am trying to write $Y(t)$ in a clever way but I am a bit stuck...

Alan
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### Exact simulation of brownian motion and of its integral

Well, the joint moment generating function will satisfy a 'bond-like' pde problem, which should be easy to solve. Then, from that, get the covariance by differentiating.

Topic Author
Posts: 33
Joined: March 28th, 2012, 12:10 pm

### Exact simulation of brownian motion and of its integral

for the moment I just came out with this,maybe it could be helpful.We rewrite:$Y(t_2) =\int_{t_1}^{t_2}W(s)ds$ by using Ito lemma on the process $W(t)t$. We get$W(t_2)t_2 -W(t_1)t_1 = \int_{t_1}^{t_2} W(s)ds + \int_{t_1}^{t_2}sdW(s)$from this follows $\int_{t_1}^{t_2} W(s)ds = W(t_2)t_2 -W(t_1)t_1 - \int_{t_1}^{t_2}sdW(s)$Now, writing the increment $W(t_2) -W(t_1)$ as $\int_{t_1}^{t_2}dW(s)$ I can compute the 2 out of 3 terms of the covariance:$E[\int_{t_1}^{t_2}dW(s) W(t_1)t_1 ]=0$$E[\int_{t_1}^{t_2}dW(s) \int_{t_1}^{t_2}sdW(s) ]= \int_{t_1}^{t_2}s ds$but I am again stuck with $E[\int_{t_1}^{t_2}dW(s) W(t_2)t_2 ]$....

Alan
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### Exact simulation of brownian motion and of its integral

If you can't get unstuck, try the pde thing.
Last edited by Alan on February 28th, 2016, 11:00 pm, edited 1 time in total.

mj
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Joined: December 20th, 2001, 12:32 pm

### Exact simulation of brownian motion and of its integral

This is well known... The easy way is simply to approximate by a discrete sum and then let the number of steps go infinity. Yes, it is bivariate normal. So all you need to compute are the variances and covariances.

Topic Author
Posts: 33
Joined: March 28th, 2012, 12:10 pm

### Exact simulation of brownian motion and of its integral

QuoteOriginally posted by: CroackingToadfor the moment I just came out with this,maybe it could be helpful.We rewrite:$Y(t_2) =\int_{t_1}^{t_2}W(s)ds$ by using Ito lemma on the process $W(t)t$. We get$W(t_2)t_2 -W(t_1)t_1 = \int_{t_1}^{t_2} W(s)ds + \int_{t_1}^{t_2}sdW(s)$from this follows $\int_{t_1}^{t_2} W(s)ds = W(t_2)t_2 -W(t_1)t_1 - \int_{t_1}^{t_2}sdW(s)$Now, writing the increment $W(t_2) -W(t_1)$ as $\int_{t_1}^{t_2}dW(s)$ I can compute the 2 out of 3 terms of the covariance:$E[\int_{t_1}^{t_2}dW(s) W(t_1)t_1 ]=0$$E[\int_{t_1}^{t_2}dW(s) \int_{t_1}^{t_2}sdW(s) ]= \int_{t_1}^{t_2}s ds$but I am again stuck with $E[\int_{t_1}^{t_2}dW(s) W(t_2)t_2 ]$....sorry also the last passage is easy to compute: just write $E[\int_{t_1}^{t_2}dW(s) W(t_2)t_2 ] = E[\int_{t_1}^{t_2}dW(s) \int_{0}^{t_2}dW(s) t_2 ] = t_2 (t_2-t_1)$So summing up the final covariance should read$t_2 (t_2-t_1) - \frac{1}{2}(t_2^2-t_1^2) = \frac{1}{2}(t_2-t_1)^2$.Do you find this reasonable?

Alan
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### Exact simulation of brownian motion and of its integral

It's dimensionally correct, has the right sign, and vanishes with the interval. I guess that leaves $\alpha (t_2 - t_1)^2$. p.s. I just recalled I have a table in my stoch. vol. book Table 5A.1 (pg. 170), which has the nice formula, among others$<e^{x W_{\tau}} \int_0^{\tau} W_s \, ds> = \exp(\frac{1}{2} x^2 \tau) (\frac{1}{2} x \tau^2)$, so differentiating confirms your answer.
Last edited by Alan on February 28th, 2016, 11:00 pm, edited 1 time in total.

Topic Author
Posts: 33
Joined: March 28th, 2012, 12:10 pm

### Exact simulation of brownian motion and of its integral

many thanks for your help Alan.Regards

Alan
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### Exact simulation of brownian motion and of its integral

You're very welcome.

MaxwellSheffield
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Joined: December 17th, 2013, 11:08 pm

### Exact simulation of brownian motion and of its integral

Hello All,I have a similar question : Let 's take a system of random variables $(X_{k})$ with $X_{k}(t)=\int_{0}^{t}{\beta_{k}(u)^{T}dW(u)}$ where W is a brownian vector of size n, beta is a deterministic vector, bounded, and k varies between 1 and N, where N>>n.Obviously, the system is big, and the distribution is entirely defined by the covariance matrix of size N*N. Let assume that we know $X_{k}(t)$, and we would like to simulate $X_{k}(s)$, where s>t. How do we "square root" the correlation matrix efficiently ? Application : Libor market model on which we already applied PCA , and wish to use a big step scheme .

Alan
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### Exact simulation of brownian motion and of its integral

I think it boils down to drawing from a multivariate normal, with the method explained at wikipediaIf you read my first response, that's essentially what I'm doing.

MaxwellSheffield
Posts: 70
Joined: December 17th, 2013, 11:08 pm

### Exact simulation of brownian motion and of its integral

Won't this imply generating N random variables ?

Alan
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### Exact simulation of brownian motion and of its integral

Sorry, I didn't notice the N vs. n business. In any event, the answer is yes. I'm not sure that is really an obstruction.If you haven't implemented it yet, I would try it in, say C/C++, and then see if the run times are acceptable.

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