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Exercise 3

Exercise 3: Effect of putting the variance in process versus non-process error.

Type 'Lab1' at matlab prompt.
When asked for data code, type 5 (this is a Sharp-tailed grouse time series).

Param set 1: set mu = 0.2; s2p = 0.00001; s2np = 0.3
Param set 2: set mu = 0.2; s2p = 0.3; s2np = 0.00001
Param set 3: set mu = 0; s2p = 0.3; s2np = 0.00001

Questions for Exercise 3:

1. (1st panel) The model doesn't fit the data at all. How can this be the ML fit of the CDA to the data?

2. (2nd panel) Why does the model fit the data perfectly now? What does setting npe really small mean? Why did the negative log likelihood go down?

3. (3rd panel) The CDA fit looks like the same as in the 2nd panel. Why did the negative log likelihood go down?

4. (4th panel) This is the fit using the parameters with the lowest -log L. How can this be the parameter set with the lowest -log L when the fit doesn't look very good?

Created on May 7, 2007 at 03:26:32 PM by eli

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