# Exercise 4

Exercise 4: Expected log N versus conditional expected log N

Type 'Lab1' at matlab prompt.

When asked for data code, type 1 (this is a Grey-headed albatross time series).

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

Param set 2: set mu = 0.1; s2p = 0.00001; s2np = 0.2

Param set 3: set mu = -0.1; s2p = 0.00001; s2np = 0.2

Questions for Exercise 4

1. (1st panel) With mu = 0, that equals zero growth, and you set process error close to zero also. So why isn't the ML fit of the CDA a straight line across (i.e. 0 growth)?

2. (2nd and 3rd panels) With mu = +/- 0.1 and process error close to zero, the expected change in log N in the CDA is 35 * mu = (3.5 or -3.5). So why are we seeing only a 2.5 (or -2.5) change? The answer is basically the same as for question 1 except now were seeing a larger difference between the expected CDA versus the ML fit.

Type 'Lab1' at matlab prompt.

When asked for data code, type 1 (this is a Grey-headed albatross time series).

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

Param set 2: set mu = 0.1; s2p = 0.00001; s2np = 0.2

Param set 3: set mu = -0.1; s2p = 0.00001; s2np = 0.2

Questions for Exercise 4

1. (1st panel) With mu = 0, that equals zero growth, and you set process error close to zero also. So why isn't the ML fit of the CDA a straight line across (i.e. 0 growth)?

2. (2nd and 3rd panels) With mu = +/- 0.1 and process error close to zero, the expected change in log N in the CDA is 35 * mu = (3.5 or -3.5). So why are we seeing only a 2.5 (or -2.5) change? The answer is basically the same as for question 1 except now were seeing a larger difference between the expected CDA versus the ML fit.

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