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I noticed that if you use two correlated geometric brownian motions, the correlation structure decays in time pretty fast even for really high correlation values. I think that is not replicating reality, is it? I wondered how people solve this problem in practice? Specially for pricing of spread options.It overprices long term options.

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Would you mind writing down some formulas? You have 2 correlated geometric BM. Then you look at the correlation of the returns, $ r_i = \mu_i dt + \sigma_i dB_t^i$ for $i =1,2$ with $B^1$ and $B^2$ correlated, I guess. I have to check myself but I don't see why this correlation should vanish. Am I missing an important point? –  Richard Mar 14 '13 at 19:43
    
I looked at the correlation of the stock prices themselves. For returns I believe the correlation stays constant and there is no problem. I just calculated correlation for different lengths of time using the covariance formula of the multivariate lognormal distribution. en.wikipedia.org/wiki/… –  Amir Yousefi Mar 14 '13 at 20:34
    
Ok, now I understand. The two prices have correlated log-returns and the correlation of the prices themselves decays. Interesting ... I have to check. –  Richard Mar 14 '13 at 20:46
    
Isn't it true that the mean and the variance of the price process is not stationary. Therefore although returns are correlated the prices don't share this property in general. The mean (of prices) wanders around and the variance grows. –  Richard Mar 15 '13 at 14:27
    
I understand the mean of the process changes but why variance? I think below answer is complete and the way I think about it is that assume the prices are equal at the beginning they have different volatilities and some correlation when time passes prices diverge most of the times as they are not perfectly correlated and even in that case variances are different so after a while because the base prices get far and far although the returns are still correlated as past but the prices lose their correlations –  Amir Yousefi Mar 15 '13 at 22:21
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well, it is absolutely in agreement with theory. the correlation as measured by Pearson's coefficient $\rho$ is linear measure in the sense that the bounds [-1,1] are obtained only when transformations of our variables are linear, so if we have variables $X$ and $Y$ then something like $aX+bY+c$ where $a,b\in\mathbb{R^*}$, $c\in\mathbb{R}$ will have boundaries [-1,1] on correlation coefficient

but

as soon as we drift from linear transformation the boundaries differ and are closer to 0, how close it depends on the type of transformation used. and since brownian motion is not linear transformation of variables of interest the boundaries vanish. as example of this I have attached below a result of my playing with two variables being lognormal distributed:

$X~(0,1)$, and $Y~(0,\sigma^2)$

it can be shown (or here) that low and upper bounds on Pearson $\rho$ in this example are

$\rho_{low}={\frac{e^{-\sigma_X\sigma_Y} -1}{\sqrt{(e^{\sigma_X^2}-1})(e^{\sigma_Y^2}-1)}}$ , $\rho_{high}={\frac{e^{\sigma_X\sigma_Y} -1}{\sqrt{(e^{\sigma_X^2}-1})(e^{\sigma_Y^2}-1)}}$

what is easy to see in my picture and almost identical to your results. how can we deal with this fact? we can use different measures of concordance*, there are many of them, and possibilities are Kendal's tau or Spearmans rho for instance.

  • so what measure of concordance is then? just some function satysfying few axioms, I will refer you again to the links above. correlation is NOT one of them since it doesn't satisfy vi) axiom given by Scarsini(1984) (about pointwise convergence: it doesn't converges when the copula (pointwise) does)

enter image description here

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Yes that is completely right. So my question is that GBM is a self decomposable process by itself but integrating such correlation structure makes the joint process sensitive to desicretization. so if you are simulating a spread option the price you get is dependent on how you discretize your model. Is not it right? and –  Amir Yousefi Mar 15 '13 at 22:25
    
ofcourse, but don't interpret the fact of decay in correlation as decay of relationship between variables. it is just that correlation coefficient is different for nonlinear transformation, so you don't expect then "perfectly correlated/discorrelated" variables to introduce 1/-1 coefficient but some smaller values –  bits_international Mar 15 '13 at 22:44
    
this is why we use different notions: comonotonic (+1) & countermonotonic (-1). spearmans's $\rho$ and kendall's $\tau$ preserve these values while correlation - as we have seen - not. –  bits_international Mar 15 '13 at 22:51
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It depends on the frequency and the horizon. For instance, I got a similar looking chart when I used annual log returns as the input to the log normal distribution and went out 250 years. With daily log returns over a few years, there isn't nearly as much of a decay. However, when you go out 250 years with daily returns you still see the pattern.

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You are right. I was thinking in a daily basis or Monte Carlo simulation context. So the process is not self decomposable? does not it make problem? because all other parameters and also brownian motion itself is not sensitive on the way you discretize that –  Amir Yousefi Mar 14 '13 at 2:07
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