Effacer les filtres
Effacer les filtres

95 % confidence interval of parameters estsimation

2 vues (au cours des 30 derniers jours)
boureghda mohammed
boureghda mohammed le 18 Déc 2016
Hi,
I have a discrete nonlinear state space model with 8 state space variables and 9 parameters.
I am using a direct likelihood to estimate the parameters,I used an Uscented Kalman Filter UKF to calculate The log-likelihood expression.
I used the fmincon from MATLAB on the objective function which is the ( negative log-likelihood) function. Because fmincon search for the minimum of the objective function, but what I want is the maximum, so minimize −L(θ) = maximize L(θ) to find the global minimum at the optimal point (the estimated parameters).
I run the program 100 times, Let say that:
The vector of estimated parameters at the first run is parameters = [para1,...,param9].
The vector of estimated parameters at the second run is parameters = [para1,...,param9].
.
.
The vector of estimated parameters at the lats run (100) is parameters = [para1,...,param9].
How can I calculate the 95% confidence interval of the estimated parameters.

Réponses (1)

Star Strider
Star Strider le 18 Déc 2016
With 100 observations, unless there is something about the distribution of your parameters that is unique to Kalman filters, they are probably normally distributed. Use the Statistics and Machine Learning Toolbox norminv funciton. If you don’t have that function, see the documentation for erfcinv, specifically the ‘More About’ section at the end of that page.
You can use the tinv function if you want, but the two results are likely indistinguishable for your data. They should be a little larger than ±1.96, but not by much.

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