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# RE: st: Can we use the standard binary choice model?

 From Nick Cox To "'statalist@hsphsun2.harvard.edu'" Subject RE: st: Can we use the standard binary choice model? Date Mon, 24 Jan 2011 13:20:49 +0000

```Consider graphs like

. twoway function invlogit(5+ (x - 2) - 5*(x-2)^2), ra(-3 5)

These to me suggest that if this model is realistic it can be approximated using a quadratic in the predictor and a logit link. Models like this are common in ecology whenever there is some optimum for organisms (e.g. in moisture or temperature) and abundance is expected to be highest at that optimum.

In other words, the model is equivalent to one symmetric around (x1 + x2)/2.

There is no hint here why the theory suggests sharp thresholds at x1 and x2. In practice I'd expect those to be fuzzy, so I wouldn't feel guilty about not taking the model very literally.

Nick
n.j.cox@durham.ac.uk

On Behalf Of Maarten buis

--- On Mon, 24/1/11, Quang Nguyen wrote:
> We have a theoretical model predicing the relationship between
> a binary variable "y" and a continous variable in the following
> pattern: y=1 if  y is in the range of [x1,x2] and y=0 if y is
> smaller than x1 or greater than x2. Where x1 and x2 are some
> threshold determined by the theoretical model's parameters.
>
> Can you suggest an empirical model to verify the above
> model's prediction?

Let me fix some terminology and notation:

y* is a latent continuous dependent/explained variable.
y  is the observed binary dependent/explained variable.
xb is the linear predictor: b0 + b1 x1 + b2 x2 + b3 x3, the xs
are independent/explanatory variables not your theshholds.

y* = xb

y  = 1 if a1 <= y* <= a2
y  = 0 if y* < a1 | y* > a2

This would be relatively easy to solve if a1 = - a2:

y*^2 = xb^2

y  = 1 if y*^2 <= a2^2
y  = 0 if y*^2 >  a2^2

y  = 1 if y*^2 - a2^2 <= 0
y  = 0 if y*^2 - a2^2 >  0

So now it is just a regular binary choice model whith a
lot of extra variables (squares and products of variables),
and a hard to interpret constant.

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