that’s a tough problem! You’ll have to assume some distribution of the timestamp within the interval -e.g. uniform?- and then go through some horrible equations. Maybe it’s best to start out with assuming that the data comes from the mid of each period?

How do you treat messy data? My data doesn’t have periodic intervals or same quantity for every, hour, week or month. Basically I have data where I only know which month it was taken on. Do you have any good reference on how to manage bad data sets?

regards and congrats on the excelent work. I’m anxious to implement your OU model and compare it to other models.

Gustavo

Interesting. That’s good to hear. I’ve indeed got an Excel file for you (see below)!

It’s limited to 4 tenors (I used it to construct the plots here). Having this reference implementation should hopefully get you going and act as a basic of constructing the sheet you need? Feel free to ask questions if you run into things.

I can also help you structurally and build the tool you need, but that would then be in a consultant role.

Here’s the sheet:
SitmoVasicekInterpolation.xls

Enjoy!

Yes, you’re right! I Fixed it right away, ..thanks for mentioning this! It’s much appreciated.

In this article, the very first formula for mean does not seem to be correct — integrand should just be x*f(x) instead of (x – mu)*f(x).

As written, this integral of single power of x about the mean is always zero

I have been doing a similar test with the Cox-Ingersoll-Ross model and I was getting wierd results.

Thanks again!

What I wanted to say is that if you calibrate the model to some data set, and have time labels in either units of years, or number of months.. and if you *then* use the same time unit in simulation, then the results will be exactly the same.

This means that you can just set dt to 1 in both calibration and simulation (if the time-step in both cases is the same).

If that is the case then with you’re example above, shouldn’t you have changed the mean reversion rate and the volatility to an annual value before using it in the exact solution of the SDE with dt as 0.25?

=S_*EXP(-lambda_*dt_)+mean_*(1-EXP(-lambda_*dt_))+sigma_*SQRT((1-EXP(-2*lambda_*dt_))/(2*lambda_))*N_

The dimension of the mean reversion rate and volatility would then be monthly too.

Just have another simple question:

If I have a series of historical data that is recorded monthly for 5 years with units % per annum. So I have 60 pieces of data each with units % per annum. If I use the above calibration will the resulting calibrated parameters be monthly or annual values % per annum?

Thanks again

I just a query regarding the data in table 1. When I input delta=0.25, lambda=3.0, mu=1.0, sigma=0.50, S0 = 3, N(0,1) = -1.0268 into the exact solution of the SDE, I get 1.39 as S1. Why am I getting different values? Are the parameters annualized?

Thanks!

The approach I have taken is to model the spread as the difference between two lognormal assest (in this case the spread can e.g. be negative)

I learned something from you. Thx. Mao

The idea is that you calculate the expected payoff at expiration with a double integral
Value = int int ProbDensity(S1,S2)*Payoff(S1,S2) dS1 dS2

The payoff part is simple, it’s 1 if S1-S2>=K and 0 otherwise

The double integral can be reduced to a single integral by solving the inner integral. That’s where the cumulative Normal comes from (solving the inner integral).

The remaining single integral is then solved numerically with a technique called Gaussian Quadrature. That is very similar to solving an integral with a Riemann sum, but it converges much quicker for smooth functions. That the “sum” part you’re seeing.

So the pointers are:
- values is (present value) of the expected payoff at expiration
- expected payoff at expiration is the double integral of the 2d probability density function of the two underlyings times the payoff
- a change of variables will allow you to solve the inner integral
- the remaining integral can be solved numerically with either a simple Riemann sum (converges slowly) or a Gaussian Quadrature (much better convergence)

Would you kindly give some answers or hints? Thank you. Mao

If you have a dividend yield of 5% and an interest rate of 3%, you”l have do subtract the dividend yield from the interest rate, and fill in -0.02 for the drift.

I’ve fixed the errors, the equations above are now updates.

In the factor multiplied to exp(…), rho – > rho^2 for both original and alternative
expressions.
For the exponent of the exponential function in the original bivariate, a negative
sign is missing.