[R-sig-ME] glmmTMB: how to calculate posterior prob. of structural zero?

[Aaron Mackey]

Nevermind, I think I see my mistake -- the conditional expectation gives
the new mean, and with the dispersion you get a P value for having seen
zero, given the mean and dispersion. Sorry to be daft. If I wanted the
probability of some value other than zero, then the formula would contain
that actual value. thanks again! -Aaron

On Fri, Oct 5, 2018 at 8:29 PM Aaron Mackey <ajmackey using gmail.com> wrote:

> Right -- so that's not what we need, is it? We need the probability of
> having actually seen a zero, given the conditional model (regardless of
> what the expected count might be). The probability of seeing a zero in the
> other part of the mixture is 1.0, and the prior for both of the two models
> is the mixture coefficient (and 1-coef). But we still need the likelihood
> of the zero, given the conditional model, not the likelihood of seeing the
> expected value.
>
> -Aaron
>
> On Fri, Oct 5, 2018 at 4:30 PM Ben Bolker <bbolker using gmail.com> wrote:
>
>> On Fri, Oct 5, 2018 at 4:07 PM Aaron Mackey <ajmackey using gmail.com> wrote:
>> >
>> > Thanks for this, it seems to provide sensible numbers; but just to
>> confirm, does predict(model, type="conditional") actually use the Y
>> variable observed count? Or does it generate the expected log count from
>> the provided X variables (as in the case when newdata is provided)?
>>
>>   It generates the expected  *count* (not the log count) *of the
>> conditional part of the model only* (i.e. not including structural
>> zeros), based on the estimated parameters and a model matrix (either
>> the one from the model or a newly generated one, if newdata is
>> specified)
>>
>> >
>> > -Aaron
>> >
>> > On Wed, Oct 3, 2018 at 3:01 PM Ben Bolker <bbolker using gmail.com> wrote:
>> >>
>> >>
>> >>   If you're fitting a zero-inflated Poisson model, it would be
>> >> especially easy because the zero probability is simply exp(-lambda), so
>> >> I believe the "posterior"(ish) probability that the zero is due to the
>> >> structural rather than conditional part of the model would be
>> >>
>> >>  P(zero|structural)/P(zero) =
>> >> P(zero|structural)/(P(zero_structural)+P(zero_conditional))
>> >>
>> >> or
>> >>
>> >> prob_struc <- function(model) {
>> >>    strucprob <- predict(model, type="zprob")
>> >>    condprob <- exp(-predict(model, type="conditional"))
>> >>    return(strucprob/(strucprob+condprob))
>> >> }
>> >>
>> >> For a negative binomial ("nbinom2", or quadratic parameterization) the
>> >> zero probability is (k/(k+mu))^k (I think ... e.g.
>> >> dnbinom(0,mu=0.5,size=0.5) is sqrt(0.5)).  sigma(model) returns the
>> >> overdispersion parameter k, so a function as above but with
>> >>
>> >>   k <- sigma(model)
>> >>   condprob <- (k/(k+predict(model,type="conditional")))^k
>> >>
>> >>  inserted should do the trick.  (Please test these yourself and make
>> >> sure they are sensible before proceeding!)
>> >>
>> >> On 2018-10-03 02:18 PM, Aaron Mackey wrote:
>> >> > I'm happily using glmmTMB to fit zero-inflated count models with my
>> data,
>> >> > but I'd like to also know which zeroes in my data are more likely
>> (or not)
>> >> > to be structural vs. expected from the conditional distribution. I
>> know how
>> >> > to use Bayes formula to calculate the posterior, and predict(zinb,
>> >> > type="zprob") gives me the prior probabilites for each data point
>> being
>> >> > structural or not (respecting the zero inflation part of the model),
>> and
>> >> > the likelihoods for the structural components are 1 (if the data
>> point is a
>> >> > zero) or 0 (if the data point is not a zero) -- but is there a way to
>> >> > extract the likelihood for each zero data point with respect to the
>> >> > conditional part of the model?
>> >> >
>> >> > thanks,
>> >> > -Aaron
>> >> >
>> >> >       [[alternative HTML version deleted]]
>> >> >
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>> >> >
>> >>
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>

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