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jbrajkovic
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Variance of Schwartz 1 factor model

November 27th, 2009, 6:47 pm

Scwartz has a model given by: (Link to the paper). Using Ito lemma it is transformed to Ornstein Uhlenbeck process for log of price. I want to find numerically, using FD, the value of real option written on the underlying following Schwartze's model. For this I need the value for variance. He shows what the variance of the log of price is. Can I use this variance? (equations 1 to 5, page 926)
 
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jbrajkovic
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Variance of Schwartz 1 factor model

November 27th, 2009, 7:17 pm

Or should I just use the following logic. Var(S)=E[S^2]-[E(S)^2]. I know that S=exp(S) and E=E[exp(x)]. Perhaps that's the way to do it?
 
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jbrajkovic
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Variance of Schwartz 1 factor model

November 30th, 2009, 6:31 am

Perhaps the question is a bit vague. So, I have a starting equation (Schwartz model) given by: . To estimate the parameters of the equation, first I use x=lnp, and using Ito lemma I transform this equation into an OU process given by: . For OU I estimate all the parameters. Now I need to go back to the initial equation. As I know E[x_t], I also know that E[p_t]=exp[E[x_t]]. Then I need to calculate the variance of p_t. I know V(p_t)=E[p_t-E[p_t]]^2=E[p_t^2]-(E[p_t])^2. Now even though at first this looked easy, I believe there are some 'tricks' involved. Just like in the case of geometric Brownian motion where a few transformations are needed to get the variance of p_t following GBM. On the following page (pdf) the author shows how to get the same thing for GBM, page 11 (Link). But I don't think this is going to work for Schwartz model, so I would need a bit different approach.
 
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jbrajkovic
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Variance of Schwartz 1 factor model

December 2nd, 2009, 8:12 pm

Perhaps I have made progress here. I have written out a short note on how to solve the problem. I would appreciate any comments/suggestions.
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Stale
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Variance of Schwartz 1 factor model

December 3rd, 2009, 6:13 am

Hi J,This model by Schwartz is an geometric Ornstein-Uhlenbeck process. A such, it is assumed that the spot could be written as the log of the underlying OU-process + seasonal variations, i.e. S_t=log(X_t+f(t)). You want to find the sigma for the process X_t which is supposed to model the stochastic behaviour. To do this, you would estimate the variance of exp(S_t)-f(t). To find this last expression you must apply Ito, of course, and have a solution for the process X_t. Then the estimate for sigma could be used anywhere, just remember that when simulating you must use the same dt, i.e. time-step.HTHStale
 
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jbrajkovic
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Variance of Schwartz 1 factor model

December 3rd, 2009, 7:56 am

Stale hello!Thanks for the suggestion. Let me break this problem down so it is easier to see what is bugging me: 1. At this stage I assume there are no seasonal variations. 2. I assume level of dark spread (p) follows Schwartz model. 3. Using ito lemma I get it into OU process, x=lnp: dx=k(a-sigma2/(2*k)-x)dt+sigmadz. 4. I estimate the parameters k, a, sigma (eg using the same procedure as on SITMO website). Now I want to determine: 5. Value of the project following Schwartz mode where q is quantity and p is level of dark spread: 6. Use finite difference to value option to invest in such a project and determine optimal exercise boundary. For this I need parameters estimated under 4. My initial confusion was: A) I thought I had to estimate variance of dark spread for finite difference. Actually, I believe I need to use sigma of OU process. B). When determining expected value of dark spread which is lognormally distributed, what to put as t in equation for expected dark spread. I guess, because I am using weekly data, t should be 1. Anyhow, my posts may seem rather chaotic, but I guess that's a part of learning process.
 
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jbrajkovic
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Variance of Schwartz 1 factor model

December 3rd, 2009, 10:23 am

On a related note, there is another issue which is terribly confusing. I want to calculate the value of project given by: . Here p is lognormally distributed. So I know that E[p]=exp[E[x]+0.5*Var[x]], where x=lnp and follows OU process. Expected value of x is given by: E[x]=x_0*exp[-k*t]+m*(1-exp[-k*t]). Using weekly data I estimate k=0.366, standard deviation of x=0.484, m=3.03, q=1, xo=ln(20) and r=0.1/52. To get the value of the project I substitute value of E[x] into E[p]. Because I want to value a project for a period of 30 years, and because I am using weekly values, T in the integral would equal=30*52=1560. So I integrate: . I get that the value of the project is 11 495. Now I want to use annual values, which in the end should give me very similar results. So to go from weekly to annual values I have: k=0.366*52=19.032, r=0.1, q=52, T=30, and sigma should be 0.484*sqrt(52)=3.49. Now the integral becomes: exp[-0.1t] This equals 4.51641*10^6, which is totally unreasonable. Can anyone spot a problem in my calculations?
 
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jbrajkovic
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Variance of Schwartz 1 factor model

December 3rd, 2009, 10:24 am

Quotemaybe this is helpful regarding FD and OU?Outrun thanks!Luckily, FD is not an issue now, but rather inputs I need to use.
 
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jbrajkovic
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Variance of Schwartz 1 factor model

January 19th, 2010, 1:56 pm

Transforming Schwartz model into log form, i.e. I get the values of the parameters. Any idea on how to transform the mean reverting factor k to be able to use it with level prices in (how to get k_L)? thanks
 
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deskquant
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Variance of Schwartz 1 factor model

December 8th, 2010, 1:01 am

curious if anyone had thought on this one here orfe.princeton.edu/~rcarmona/download/fe/convenienceyield.pdf
 
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jbrajkovic
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Variance of Schwartz 1 factor model

December 8th, 2010, 6:11 am

thanks for the link.
 
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deskquant
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Joined: October 24th, 2010, 2:54 am

Variance of Schwartz 1 factor model

December 8th, 2010, 6:54 pm

sure sure, i would not mind discussing this a little bit, if you guys thought it improved the valuations a lot, over the time invariant case.
 
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Edgey
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Variance of Schwartz 1 factor model

December 10th, 2010, 1:28 pm

Be careful not to mix parameters and statistics. Just because Sigma=Standard Deviation for the normal distribution it doesn't mean that sigma is always the standard deviation. You can't use the square root of time rule for mean reverting processes. The stronger the mean reversion, the quicker variance of the process stabilizes to a constant value. The sigma in the Schwartz model is the instantaneous log volatility. See Clewlow and Strickland for how to project the variance forward.
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