I get the parameters (long-term mean, volatility, mean-reversion speed, correlation) of two correlated Ornstein-Uhlenbeck processes via a likelihood estimation from hourly data. If I want to transform these to use them to create a daily - instead of hourly - simulation (tree or Monte Carlo), what do I have to do? Thanks in advance.
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$\begingroup$ could you be more precised ? do you want to simulate the mean over one day ? or just a daily simulation of hourly price (for example the 3pm one) ? $\endgroup$ – MJ73550 May 13 '16 at 11:46
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$\begingroup$ So the processes are correlated with dW1 * dW2 = rho* dt. The two O-U processes are dX1 = k1*(mu1-X1)*dt + sigma1 * dW1 and dX2 = k2*(mu2-X2)*dt + sigma2 * dW2. I get the parameters and I want to build a quadrinomial lattice (2-dim binomial tree) that has a resolution of days (= one node describes the probability of the two prices being at a certain point at that day) instead of hours. I.e. I want neither the mean over one day nor the daily simulation of the 3 pm price but the simulation of, for example, the expected mean price for each day over 365 days. Does this help? $\endgroup$ – LenaH May 13 '16 at 12:45
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$\begingroup$ Hourly or daily, the parameters should not change. As @MJ73550 pointed out, it is similar to a simulation with hourly or daily time steps, where the parameters are held the same. $\endgroup$ – Gordon May 13 '16 at 12:46
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You can aggregate your starting hourly data to obtain daily data and re-estimate the parameters, then simulate. Alternatvely, with your parameters already obtained, you can simulate hourly data and make a post-simulation aggregation to have daily data.
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$\begingroup$ There is no way to work via the parameters? A process that is depicted in hourly steps does not have the same mean reversion speed and volatility as the same process that is depicted in daily steps, for example? $\endgroup$ – LenaH May 13 '16 at 13:31
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$\begingroup$ I don't think so. The dynamics of an hourly process can be very different from the daily's. So, in my opinion, you should act as I said before. $\endgroup$ – simmy May 13 '16 at 14:15
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$\begingroup$ @simmy I have posted a question related to your answer. Kindly have a look: quant.stackexchange.com/questions/26142/… $\endgroup$ – Polar Bear May 21 '16 at 3:03
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Let $X^h$ be your hourly process
Let $X^d$ be your daily process
Let $\delta$ be one day
you have
$$X^d_t=\frac{1}{\delta}\int_{t-\delta}^{t}X^h_s ds$$
$$dX^h_t = a(b-X^h_t)dt + \sigma dB_t$$
$$\Delta X^d_t := X^d_{t+\delta}-X^d_t =\frac{1}{\delta}\int_{t-\delta}^t\left(X^h_{u+\delta}-X^h_{u}\right)du$$
so it is a gaussian random variable by knowns results on OU.
You can express it and compute $Cov(\Delta X^{d}_{k\delta},\Delta X^d_{j\delta})$
You will then be able to conclude.
Details
by known results :
$$X^h_{t+\delta}-X^h_t=(b-X_{t})(1-e^{-a\delta})+\int_{t}^{t+\delta}e^{a(u-t)}dB_u$$
so:
$$\begin{split} X^d_{t+\delta}-X^d_t &= (b-X^d_t)(1-e^{-2a\delta})+\int_{t-\delta}^{t}\frac{1}{\delta}\int_{u}^{u+\delta}e^{a(s-u)}dB_s du \\ & = (b-X^d_t)(1-e^{-2a\delta})+\int_{t}^{t+\delta}\frac{1}{\delta}\int_{u-\delta}^{u}e^{a(s-u+\delta)}dB_s du \\ \end{split} $$
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$\begingroup$ Thank you. Hm.. What is X in the last line? (It's not daily or hourly?..) Unfortunately I'm not that strong in stochastic calculus.. would you mind sharing the derivation? $\endgroup$ – LenaH May 14 '16 at 7:56
