Pooled model-averaged abundance estimate with CI

[jlaufenb]

Can someone point me to a resource that explains the correct way to calculate a pooled model-averaged N-hat and CI, using unconditional variances, for 2 groups (i.e., males and females)? Maybe I'm not looking in the right places.

Thanks
Jared
[/i]

[jlaake]

One way is to construct using a log-normal interval on f0 (estimated number not caught) and then addin M_t+1 to interval end points. For an estimate x, a log-normal interval is constructed as x/C,x*C where C=exp[1.96*sqrt(ln(1+cv^2))] where sqrt is square root, ln is the natural log and cv^2 =var(x)/x^2. I think you can find a description of a log-normal CI in the Distance Sampling book and in AFS Monograph 5.

--jeff

[cooch]

Can someone point me to a resource that explains the correct way to calculate a pooled model-averaged N-hat and CI, using unconditional variances, for 2 groups (i.e., males and females)? Maybe I'm not looking in the right places.

Thanks
Jared

Step 1 - calculate model averaged N (trivial) and model averaged variance (relatively easy) for each group (see closed abundance chapter)

Step 2 - delta method for variance of a sum (easy).

The only tricky bit (potentially) is the CI calculation - since its not simply a linear function of the variance (see chapter 14). I suspect you simply apply the approach discussed in 14.9.1, substituting in pooled M(t+1) and calculated summed f(0) to get C. Never tried this, but it should be relatively straightforward.

[cooch]

One way is to construct using a log-normal interval on f0 (estimated number not caught) and then addin M_t+1 to interval end points. For an estimate x, a log-normal interval is constructed as x/C,x*C where C=exp[1.96*sqrt(ln(1+cv^2))] where sqrt is square root, ln is the natural log and cv^2 =var(x)/x^2. I think you can find a description of a log-normal CI in the Distance Sampling book and in AFS Monograph 5.

--jeff

This is more or less what is described in the relevant sections of Chapter 14 (section 14.9, to be precise). Good to know about alternate references for this. Also in the Williams, Nichols & Conroy book, although their equation for C is incorrect (this is noted in a footnote in Chapter 14).

[jlaufenb]

Thanks for the quick response gentlemen. I was under the same impression of how to do this until I was checking the numbers from the Kendall et al. 2009 Demography and genetics of grizzly bears paper (JWM 73(1), table 2). They report gender specific Mt+1 and model-averaged N-hat and SE. However, the pooled SE appears to not have been calculated using the delta method for variance of the summed group-specific model-averaged N-hats (i.e., [SE-female^2 + SE-male^2]^.5 ).

The only other way I can think of doing this, which you can tell me if it is valid, is to sum the conditional gender-specific N-hats and variances (using delta method) for model, model-average these summed values using [/i]w's, and calculate the estimated unconditional summed variance and log-based CI using equations from Chapter 14.

Let me know what you think.

Thanks
Jared

[jlaake]

The variance of a sum is only the sum of the variances IF the components are independent which means in this case that they have NO parameters in common in all of the models used in the model averaging. I think that is unlikely in that paper but don't know for sure. You need to compute the model averaged v-c matrix. If you are using RMark, it will compute that for you. Then the variance of your sum is the sum of the 2 variances + 2 * covariance. This is the same as summing all the elements of the v-c matrix which is easier when there are 3 or more in the sum.

--jeff

[cooch]

The variance of a sum is only the sum of the variances IF the components are independent which means in this case that they have NO parameters in common in all of the models used in the model averaging. I think that is unlikely in that paper but don't know for sure. You need to compute the model averaged v-c matrix. If you are using RMark, it will compute that for you. Then the variance of your sum is the sum of the 2 variances + 2 * covariance. This is the same as summing all the elements of the v-c matrix which is easier when there are 3 or more in the sum.

--jeff

Jeff is correct - I'll push him for one clarification, though. IN the instance when the two samples are treated as separate groups (males, females) then even if the candidate model sets used for the two groups are identical (which implies complete overlap in terms of parameters), then in fact you could argue that there is no sampling covariance between estimates of male and female abundance. At which point, it drops out of the equation for the variance of a sum, and away you go. The basic point as Jeff notes is that car of a sum as a simply sum of the variances works only if the cov matrix is an identity matrix (or, if cov(X,Y)=0, which is the same thing). If you sample males and females from the same population, using the same technique, and the same estimation models, there is a reasonable question as to whether assuming cov=0 (in fact, it probably isn't). However, in order to get cov, you'd need to have males and females as groups in the same analysis. If you analyze males as a single analysis, then females in another analysis, you won't have an estimate of the sampling covariance between them, and woudln't be able to do much better than cov=0, even though in point of fact its probably wrong to assume such.

[jlaake]

Jeff is correct - I'll push him for one clarification, though. IN the instance when the two samples are treated as separate groups (males, females) then even if the candidate model sets used for the two groups are identical (which implies complete overlap in terms of parameters), then in fact you could argue that there is no sampling covariance between estimates of male and female abundance. At which point, it drops out of the equation for the variance of a sum, and away you go. The basic point as Jeff notes is that car of a sum as a simply sum of the variances works only if the cov matrix is an identity matrix (or, if cov(X,Y)=0, which is the same thing). If you sample males and females from the same population, using the same technique, and the same estimation models, there is a reasonable question as to whether assuming cov=0 (in fact, it probably isn't). However, in order to get cov, you'd need to have males and females as groups in the same analysis. If you analyze males as a single analysis, then females in another analysis, you won't have an estimate of the sampling covariance between them, and woudln't be able to do much better than cov=0, even though in point of fact its probably wrong to assume such.

Evan makes 2 points here that need clarification. With regard to a model with all common parameters, you can only get a 0 covariance if they are treated as a single group. In particular, if N is in the likelihood then each group would have a separate f0 but common p's and they would have a covariance. But even if N isn't in the likelihood you are making 2 predictions from the same model and they will have a covariance. However, if you ignored the group structure then you are okay because you would only have a total estimate rather than separate ones that were being summed.

In the second point he is referrring to covariance that would result from dependence of fates and that is not included in the estimated covariances. I think you'll find that if you fit p(g*t) the covariance of the N's will be 0.

But what is being described here is summing model averaged estimates . If you model averaged a p(g*t) and a p(t) model then the model averaged estimates will have a covariance term as a result of the second model.

It would be useful if either Bill Kendall chimed in or you contact him individually so you can get some clarification on what he did in that paper.

[cooch]

It would be useful if either Bill Kendall chimed in or you contact him individually so you can get some clarification on what he did in that paper.

Different Kendall (Kathy Kendall if I recall correctly), but beside this egregious error , everything else Jeff has in his note is correct.

[jlaake]

Oops! Not familiar with the paper so maybe that Kendall will reply or Bill can chime in if he wants!