st: reverse prediction - confidence interval for x at given y in nonlinear model

["Rosy Reynolds" <rr@dandr.demon.co.uk>]

Dear Statalist,
Sigmoid models are customary in pharmacodynamics (dose-response studies).
According to custom, I am using a 4-parameter logistic (sigmoid Emax) model. 
This is very easily done with -nl- as Stata has this model already built in.

The model is  y= b0 + b1/(1 + exp(-b2*(x-b3))) + error

and the coefficients can be interpreted as
b0 = baseline outcome
b1 = Emax i.e. largest change from baseline
b2 = Hill or slope coefficient
b3 = ED50 i.e. value of x (dose) required to produce half-maximal effect, 
that is x required for y=b0 + b1 / 2

As the ED50 is a parameter of the model, -nl- reports it with a standard 
error and confidence interval.
What I would like to do is obtain estimates with standard errors and 
confidence intervals for other similar measures e.g. the ED90, the dose 
required for 90% of maximal effect.
In general, how could I calculate an estimate and confidence interval for 
the x required to achieve any given value of y?
I will be grateful for any guidance anyone can give me.
I hope I'm not missing something very obvious.

best wishes
Rosy Reynolds
BSAC Resistance Surveillance Co-ordinator
www.bsacsurv.org

Below are simulated data and output from -nl- .
In this case, by design, half maximal effect is at y=6, and the ED50 value 
of x to produce that is 5.
-nl- estimates the ED50 (b3) nicely as 4.9 with CI [4.7, 5.1]
I would like to obtain an estimate of x for y=9, for example, with a CI.

* Simulate data with fairly small errors.
. clear
. set obs 101
obs was 0, now 101
. gen x=(_n-1)/10
. local b0 2 // Yo
. local b1 8 // Emax
. local b2 1 // Hill coefficient
. local b3 5 // ED50
. gen y= `b0' + `b1'/(1 + exp(-`b2'*(x-`b3'))) + 0.5*invnormal(uniform())

*Fit model
. nl log4: y x // Stata built-in 4-parameter logistic
(obs = 101)

Iteration 0:  residual SS =   37.0629
Iteration 1:  residual SS =  25.18804
Iteration 2:  residual SS =     22.87
Iteration 3:  residual SS =  22.86999
Iteration 4:  residual SS =  22.86999
Iteration 5:  residual SS =  22.86999

     Source |       SS       df       MS
-------------+------------------------------         Number of obs = 
101
      Model |  1019.30494     3  339.768313         R-squared     = 
0.9781
   Residual |  22.8699928    97  .235773122         Adj R-squared = 
0.9774
-------------+------------------------------         Root MSE      = 
.4855647
      Total |  1042.17493   100  10.4217493         Res. dev.     = 
136.6108

4-parameter logistic function, y = b0 + b1/(1 + exp(-b2*(x - b3)))
------------------------------------------------------------------------------
          y |      Coef.   Std. Err.      t    P>|t|     [95% Conf. 
Interval]
-------------+----------------------------------------------------------------
        /b0 |   1.853862   .1387566    13.36   0.000     1.578469 
2.129256
        /b1 |   8.259274   .2210395    37.37   0.000     7.820572 
8.697977
        /b2 |   .9559409   .0661914    14.44   0.000     .8245693 
1.087312
        /b3 |   4.879619   .0739367    66.00   0.000     4.732876 
5.026363
------------------------------------------------------------------------------
 Parameter b0 taken as constant term in model & ANOVA table 

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