Date:         Wed, 7 Jul 2010 00:02:21 -0400
Reply-To:     Ryan Black <ryan.andrew.black@GMAIL.COM>
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:         Ryan Black <ryan.andrew.black@GMAIL.COM>
Subject:      Re: NLMIXED (Logistic Regression) - Estimate Statements & 95% CIs
In-Reply-To:  <435327.54145.qm@web32408.mail.mud.yahoo.com>
Content-Type: text/plain; charset=ISO-8859-1

Dale,

You are a life saver! Your feedback, as always, is exactly what I
needed. I really hope you have time to respond to two follow-up
questions:

(1) Let's say the distribution of x is roughly normal for both groups,
but there is little overlap between the distributions (e.g. range of x
is 0 to 10 for group=0 and 8 to 18 for group=1). In general, when
estimating the difference in probabilities would you select a value
that is representative of both groups? (e.g. average of shared range =
(10+8)/2=9)?

(2) I've opted to treat x as a covariate. Let's say the covariate were
a sort of exposure variable (e.g. length of time exposed), how might
you treat this covariate within the context of a logistic regression?
Would you agree with my decision to treat it as a covariate? If you
were to consider this as an offset variable, how would you model it?

I realize these questions are a bit off topic for SAS-L, but I'm very
interested in your expert opinion.

As always, I understand if you're too busy to respond.

Hope you're having a good evening.

Ryan

On Tue, Jul 6, 2010 at 11:20 PM, Dale McLerran <stringplayer_2@yahoo.com> wrote:
> --- On Tue, 7/6/10, Ryan Black <ryan.andrew.black@GMAIL.COM> wrote:
>
>> From: Ryan Black <ryan.andrew.black@GMAIL.COM>
>> Subject: NLMIXED (Logistic Regression) - Estimate Statements & 95% CIs
>> To: SAS-L@LISTSERV.UGA.EDU
>> Date: Tuesday, July 6, 2010, 7:40 PM
>> Will the estimate statements below yield the correct estimates?
>>
>> proc nlmixed data=mydata;
>>    eta = b0 + b1*group + b2*x + u;
>>    exp_eta = exp(eta);
>>    p = exp_eta / (1 + exp_eta);
>>    model y ~ binary(p);
>>    estimate 'x-adjusted probability for group=0'
>>                            exp(b0+b2) / (1+exp(b0+b2));
>>    estimate 'x-adjusted probability for group=1'
>>                            exp(b0+b1+b2) / (1+exp(b0+b1+b2));
>>    estimate 'x-adjusted probability difference between groups'
>>                            (exp(b0+b2) / (1 + exp(b0+b2)))
>>                          - (exp(b0+b1+b2) / (1 + exp(b0+b1+b2)));
>>    random u ~ normal(0,s2u) subject=subject;
>>    run;
>>
>> Also, how might I obtain the correct 95% CIs for these
>> three estimates?
>>
>> Thanks,
>>
>> Ryan
>>
>
> Ryan,
>
> Whether the above estimate statements produce correct point
> estimates depends on a precise statement of what you are
> trying to estimate.  It could also depend on the distribution
> of the covariate X.
>
> My guess is that the above statements are not what you want.
> You probably want estimates of the probability of the response
> for groups 0 and 1 at the mean value for the predictor X.
> Your estimate statements produce point estimates for the
> probability of the response for each of the two groups at the
> value X=1.  If the data have mean(X)=1 or if there is some
> interpretability which is inherent to the value X=1, then
> your estimate statements would be correct.
>
> However, most of the time, one would want to know the mean
> value of X in the data that will be employed to fit the model
> prior to executing your NLMIXED code.  Suppose that you know
> that the mean value is 3.28.  Then you would construct your
> estimate statements as:
>
>   estimate 'x-adjusted probability for group=0'
>                       exp(b0+b2*3.28) / (1+exp(b0+b2*3.23));
>   estimate 'x-adjusted probability for group=1'
>                       exp(b0+b1+b2*3.28) / (1+exp(b0+b1+b2*3.28));
>   estimate 'x-adjusted probability difference between groups'
>                       (exp(b0+b2*3.28) / (1 + exp(b0+b2*3.28)))
>                     - (exp(b0+b1+b2*3.28) / (1 + exp(b0+b1+b2*3.28)));
>
>
> Equivalent statements which are a little more computationally
> efficient would be:
>
>   estimate 'x-adjusted probability for group=0'
>                       1 / (1+exp(-(b0 + b2*3.23)));
>   estimate 'x-adjusted probability for group=1'
>                       1 / (1+exp(-(b0 + b1 + b2*3.28)));
>   estimate 'x-adjusted probability difference between groups'
>                       1 / (1 + exp(-(b0 + b2*3.28))))
>                     - 1 / (1 + exp(-(b0 + b1 + b2*3.28))));
>
>
> As for "correct confidence intervals", I presume that for
> the probabilities of group=0 and group=1, you want to construct
>
>  1 / (1 + exp(-(b0 + b2*3.28 - 1.96*SE(b0 + b2*3.28)))
>  1 / (1 + exp(-(b0 + b2*3.28 + 1.96*SE(b0 + b2*3.28)))
>
>  1 / (1 + exp(-(b0 + b1 + b2*3.28 - 1.96*SE(b0 + b1 + b2*3.28)))
>  1 / (1 + exp(-(b0 + b1 + b2*3.28 + 1.96*SE(b0 + b1 + b2*3.28)))
>
>
> In order to compute these statistics, you would need to write
> estimate statements for
>
>  estimate "Logit(group=0)" b0 + b2*3.28;
>  estimate "Logit(group=1)" b0 + b1 + b2*3.28;
>
> You can then apply the link function to the confidence limits
> of the logit terms.  Confidence intervals for the difference
> between the two proportions can be gotten employing various
> formulae.  I really don't know what formula you wish to use,
> so cannot really comment on what is the appropriate method
> to use.  However, unless your probability estimates are near
> 0 or 1, then there will probably not be much difference in
> the confidence limits obtained for different formulae.  The
> confidence intervals computed from the estimate statement
> for the difference of the two proportions should be pretty
> close to CI's obtained employing just about any formula
> subject to the caveat about probabilities which are near the
> extremes of the distribution.
>
> Dale
>
> ---------------------------------------
> Dale McLerran
> Fred Hutchinson Cancer Research Center
> mailto: dmclerra@NO_SPAMfhcrc.org
> Ph:  (206) 667-2926
> Fax: (206) 667-5977
> ---------------------------------------
>