Date:         Mon, 22 Jun 2009 10:19:29 -0700
Reply-To:     Bminer <b_miner@LIVE.COM>
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:         Bminer <b_miner@LIVE.COM>
Organization: http://groups.google.com
Subject:      Re: Deviance versus log likelihood
Comments: To: sas-l@uga.edu
Content-Type: text/plain; charset=ISO-8859-1

On Jun 21, 5:29 pm, art...@NETSCAPE.NET (Arthur Tabachneck) wrote:
> The variable names are different in the two assignments and I'm not sure
> what the -2(etc, etc.) mean.
>
> If you're just multiplying the same number by two, I would think that both
> the means and variance would simply be doubled as compared to the first
> calculation.
>
> Art
> --------
>
> On Sun, 21 Jun 2009 12:17:54 -0700, Bminer <b_mi...@LIVE.COM> wrote:
> >I am used to testing constraints on parameters (two nested models) in
> >logistic regression as either:
>
> >Deviance_reduced - Deviance_full
> >or
> >-2(LL_reduced - LL_full). Both give the same value to compare to
> >critical values of chi-square.
>
> >Using genmod and a normal error, identity link model the two values
> >above are not the same. Why?
>
> >Which should be used in a likelihood ratio test?
> >Thanks!

Let me re-express my question. I was taught that for any model
estimated with ML the difference in deviance  was the same as  -2* the
difference in log likihoods. This was the manner to test the inclusion
of a predictor (compare two nested models) as a chi-square with df
equal to the difference in df between the larger and smaller model.

I notice that when using proc genmod, this is not the case. It appears
to be with link=logit and dist=bin but not others.

I have since read a description that one should use (Deviance_smaller
- Deviance larger)  /   scale from larger. This does not equal  -2*
the difference in log likihoods either.

Can anyone tell me how to take the data from genmod to compute the LR
test?