Solving stochastic differentials for time series forecasting

In summary, the conversation discusses the attempt to reproduce results from a paper using a model with state variables and Wiener processes. The goal is to estimate the next day's state variables, but the equations involve differential equations and stochastic terms. The conversation offers suggestions for solving the equations numerically or analytically, but notes the difficulty in dealing with the stochastic terms.
  • #1
jjhyun90
8
0
I am trying to reproduce results of a paper. The model is:

[itex]
dS = (v-y-\lambda_1)Sdt + \sigma_1Sdz_1 \\
dy = (-\kappa y - \lambda_2)dt + \sigma_2 dz_2 \\
dv = a((\bar{v}-v)-\lambda_3)dt + \sigma_3 dz_3 \\
dz_1dz_2 = \rho_{12}dt \\
dz_1dz_3 = \rho_{13}dt \\
dz_2dz_3 = \rho_{23}dt \\
[/itex]
where S, y, and v are state variables and [itex] z_1, z_2, z_3 [/itex] are Wiener processes.

The parameters [itex] \kappa, a, \bar{v}, \lambda_1, \lambda_2, \lambda_3, \sigma_1, \sigma_2, \sigma_3, \rho_{12}, \rho_{13}, \rho_{23} [/itex] are given. There is time series data such for each day I can estimate S, y, and v.

The goal is to estimate the next day's S, y, and v. It feels like an easy task, but am not familiar with differential equations or Wiener processes (let alone stochastic calculus). Without understanding the model, I was naively trying to "substitute" dt with Δt, etc, but the given parameters are such that [itex]\rho_{12} > 0, \, \rho_{23} > 0, \, \rho_{13} < 0 [/itex], so I realized I should not do that.

Any help would be greatly appreciated.
 
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  • #2
Are you trying to solve the equations numerically? I think people typically interpret the dt as a small timestep Δt. I guess the tricky part is the stochastic terms - it looks like they should be Gaussian distributed random variables with covariances [itex]\rho_{12}[/itex], etc? Generating those might be tricky? Otherwise everything else looks straightforward.

The equations are simple enough to solve analytically, but if you're not familiar with differential equations that might not be the best avenue for you. But, if you want to try, you can solve the y and v equations using integrating factors, and then plug the solutions into the S equation, which you can solve by dividing out the S and writing [itex]dS/S = d(\ln S)[/itex], then it's a simple matter of integration. (I would write [itex]dz_1/dt = \eta_1(t)[/itex], etc.)
 

Related to Solving stochastic differentials for time series forecasting

1. What is a stochastic differential?

A stochastic differential is a type of differential equation that incorporates random noise or uncertainty into the equation. It is commonly used to model systems that exhibit random behavior, such as financial markets or weather patterns.

2. How can stochastic differentials be used for time series forecasting?

Stochastic differentials can be used to model and analyze time series data by incorporating stochastic processes into the forecasting model. This allows for the prediction of future outcomes with consideration for the random variability in the data.

3. What are the advantages of using stochastic differentials for time series forecasting?

Stochastic differentials allow for a more accurate and realistic representation of time series data by accounting for the inherent randomness and uncertainty in the data. This can lead to more reliable forecasts and better decision making.

4. What are some common methods for solving stochastic differentials?

Some common methods for solving stochastic differentials include the Euler-Maruyama method, the Milstein method, and the Runge-Kutta method. These methods use numerical techniques to approximate the solution to the stochastic differential equation.

5. Are there any limitations to using stochastic differentials for time series forecasting?

While stochastic differentials can provide more accurate forecasts, they can also be more complex and computationally intensive. Additionally, the use of stochastic processes may require more data and assumptions about the underlying system, which may not always be readily available.

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