# How to transform Ornstein-Uhlenbeck parameters from hourly to daily?

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.

• 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) ? May 13, 2016 at 11:46
• 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? May 13, 2016 at 12:45
• 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. May 13, 2016 at 12:46

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.

• 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? May 13, 2016 at 13:31
• 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. May 13, 2016 at 14:15
• @simmy I have posted a question related to your answer. Kindly have a look: quant.stackexchange.com/questions/26142/… May 21, 2016 at 3:03

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:

$$\begin{split} X^d_{t+\delta}-X^d_t &= (b-X^d_t)(1-e^{-2a\delta})+\int_{t-\delta}^{t}\frac{1}{\delta}\int_{u}^{u+\delta}e^{a(s-u)}dB_s du \\ & = (b-X^d_t)(1-e^{-2a\delta})+\int_{t}^{t+\delta}\frac{1}{\delta}\int_{u-\delta}^{u}e^{a(s-u+\delta)}dB_s du \\ \end{split}$$

• 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? May 14, 2016 at 7:56