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# Subject: st: RE: RE: risk ratio

 From jhilbe@aol.com To statalist@hsphsun2.harvard.edu Subject Subject: st: RE: RE: risk ratio Date Mon, 22 Mar 2010 18:09:29 -0400

It is definitely possible to use shortcut code like Nick showed. I tend to prefer, however, for pedagogical purposes, to specify separately the linear predictor, inverse link function, and pseudo-random number generator for a given distribution. I believe that it makes the logic of synthetic model creation more clear. When the models become more complex it helps not to make mistakes. And for binomial models where there is no special inverse link function; eg loglog and cloglog, you must specify it separately. The code I give in the article, and in response to the query on Statalist, is constructed to be as clear as possible. Please keep in mind though that it is the case that for several of the synthetic models discussed, the code can be compacted in the manner Nick suggests. This is particularly the case for embedding the linear predictor within the appropriate pseudo-random number generator.
```
Joseph Hilbe

=====================================
As a footnote to this, note a few equivalences:

invlogit(x) <=> 1 / (1 + exp(-x))

rnormal() <=> invnorm(runiform())

An alternative to

gen x1 = invnorm(runiform())
gen xb = 2 + 0.75*x1
gen exb = 1/(1+exp(-xb))
gen by = rbinomial(1, exb)

is thus

gen x1 = rnormal()
gen by = rbinomial(1, invlogit(2 + 0.75*x1))

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

Joseph Hilbe

I have an article coming out in the next Stata Journal that details
how to create synthetic models for a wide  variety
of discrete response regression models. For your problem though, I
think that the best approach is to create a synthetic
binary logistic model with a single predictor - as you specified. Then
model the otherwise logistic data as
Poisson with a robust variance estimator. And the coefficient must be
exponentiated. It can be interpreted as
a relative risk ratio.

Below is code to create a simple binary logistic model. Then model as
mentioned above. You asked for a
continuous pseudo-random variate, so I generated it from a normal
distribution. I normally like to use pseudo-random
uniform variates rather normal variates when creating these types of
models, but it usually makes little difference.
Recall that without a seed the model results will differ each time
run. If you want the same results, pick a
seed. I used my birthday.

I hope that this is what you were looking for.

Joseph Hilbe

clear
set obs 50000
set seed 1230
gen x1 = invnorm(runiform())
gen xb = 2 + 0.75*x1
gen exb = 1/(1+exp(-xb))
gen by = rbinomial(1, exb)
glm by x1, nolog fam(bin 1)
glm by x1, nolog fam(poi) eform robust

. glm by x1, nolog fam(bin 1)
Generalized linear models                          No. of obs      =
50000
Optimization     : ML                              Residual df     =
49998
Scale parameter =
1
Deviance         =  37672.75548                    (1/df) Deviance =
.7534852
Pearson          =  49970.46961                    (1/df) Pearson  =
.9994494
Variance function: V(u) = u*(1-u)                  [Bernoulli]
Link function    : g(u) = ln(u/(1-u))              [Logit]
AIC             =
.7535351
Log likelihood   = -18836.37774                    BIC             =
- -503294.5
```
- ------------------------------------------------------------------------
```- ------
|                 OIM
by |      Coef.   Std. Err.      z    P>|z|     [95% Conf.
Interval]
```
- -------------+----------------------------------------------------------
```- ------
x1 |   .7534291   .0143134    52.64   0.000     .7253754
.7814828
_cons |   1.993125   .0149177   133.61   0.000     1.963887
2.022363
```
- ------------------------------------------------------------------------
```- ------

. glm by x1, nolog fam(poi) eform robust
Generalized linear models                          No. of obs      =
50000
Optimization     : ML                              Residual df     =
49998
Scale parameter =
1
Deviance         =  12673.60491                    (1/df) Deviance =
.2534822
Pearson          =   7059.65518                    (1/df) Pearson  =
.1411988
Variance function: V(u) = u                        [Poisson]
Link function    : g(u) = ln(u)                    [Log]
AIC             =
1.970592
Log pseudolikelihood = -49262.80246                BIC             =
- -528293.7
```
- ------------------------------------------------------------------------
```- ------
|               Robust
by |        IRR   Std. Err.      z    P>|z|     [95% Conf.
Interval]
```
- -------------+----------------------------------------------------------
```- ------
x1 |   1.104476   .0021613    50.78   0.000     1.100248
1.10872
```
- ------------------------------------------------------------------------
```- ------
.

Tomas Lind wrote:
Does anyone know how to generate fake data for a dichotomous outcome (0,
1)
that is dependent on a continuous exposure variable in an
epidemiological
relative risk context. I know how to use the logit transformation but in
that case exposure is proportional to log(ods) and not to risk.

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```