lambda and variance components

questions concerning analysis/theory using program MARK

lambda and variance components

Postby Bryan Hamilton » Sat Nov 25, 2017 8:31 pm

I'm looking for some clarification with pradel models (Pradrec) where lambda is a derived parameter. I should add that I'm actually doing variance components and modeling in RMark.

After rereading chapter 13 of the MARK book, I'd come to the conclusion that variance components does not calculate the correct value for mean lambda. Variance components uses the arithmetic mean and I want the geometric mean so as to not overestimate lambda. This requires doing the calculations by hand.

I tried to follow the chapter 13 examples with my dataset. Now when I compare different ways of calculating overall mean lambda from the lambda values and log(lambda) values (vcv versus geometric mean) all the mean values are different but all fall within a range of 0.0087. This seems trivial and nothing to worry about.

So does the variance components actually calculate mean lambda correctly using geometric means? Or should I still calculate geometric means and variances by hand for lambda in Pradel models?
Bryan Hamilton
 
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Re: lambda and variance components

Postby cooch » Sun Nov 26, 2017 9:43 am

Bryan Hamilton wrote:I'm looking for some clarification with pradel models (Pradrec) where lambda is a derived parameter. I should add that I'm actually doing variance components and modeling in RMark.

After rereading chapter 13 of the MARK book, I'd come to the conclusion that variance components does not calculate the correct value for mean lambda. Variance components uses the arithmetic mean and I want the geometric mean so as to not overestimate lambda. This requires doing the calculations by hand.


No it doesn't. Simply do the variance components on the derived parameter \ln(\lambda). The arithmetic mean of \ln(\lambda) (which is equivalent to the arithmetic mean of r_s, since \ln[\lambda]=r), is the geometric mean.

This is covered in pp 24-25 of Chapter 13. Remember, VC calculates the mean over the shrunk estimates, so will (generally) differ from the simple arithmetic mean of the individual \ln(\lambda) values.
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Re: lambda and variance components

Postby cooch » Sun Nov 26, 2017 1:42 pm

In case it isn't obvious why the mean of individual estimates of r_s is the geometric mean...rewrite the geometric mean

\lambda_S=\bigl(\lambda_1\cdot\lambda_2\cdot\lambda_3\dots\lambda_T\bigr)^{1/T}

by substituting the equivalence that \lambda=e^{r}, yielding:

\lambda_S=\left(e^{r_1}\cdot e^{r_2}\cdot e^{r_3}\dots e^{r_T}\right)^{1/T}.

Taking logs of both sides, this expression can be re-written as

\ln(\lambda_S)=\displaystyle\frac{r_1+r_2+r_3\cdots +r_T}{T}.

So, we can also calculate \ln(\lambda_S) simply as the arithmetic mean \bar{r}. MARK outputs \ln(\lambda) as a derived parameter. Since \ln(\lambda)=r, then the arithmetic mean of \ln(\lambda) is equal to the arithmetic mean of r.
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Re: lambda and variance components

Postby Bryan Hamilton » Sun Nov 26, 2017 4:50 pm

Thanks cooch. That's great stuff.

So exp(arithmetic mean (r)) = the geometric mean (lambda)?

Why aren't other ratios, like survival and recruitment, also best averaged geometrically? Is it because they are bounded from 0-1? I'm trying to understand why lambda is treated differently.
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Re: lambda and variance components

Postby cooch » Sun Nov 26, 2017 7:58 pm

Bryan Hamilton wrote:Thanks cooch. That's great stuff.

So exp(arithmetic mean (r)) = the geometric mean (lambda)?


Correct.
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