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Re: st: nl -choice between alternative parametrisations of sigmoid models
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From |
Steven Joel Hirsch Samuels <[email protected]> |
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To |
[email protected] |
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Subject |
Re: st: nl -choice between alternative parametrisations of sigmoid models |
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Date |
Thu, 25 Oct 2007 17:08:43 -0400 |
Thanks for clarifying, Rosy. I didn't know this about the sigmoid
model.
-Steven
On Oct 25, 2007, at 2:53 PM, Rosy Reynolds wrote:
Steven,
No, I'm not saying that the same data produce two curves.
What I'm saying is that you can have two different sets of
coefficients to describe the same curve.
e.g.
b0= 2, b1 =8, b2 = 1, b3 = 5 will give the same curve as b0= 10, b1
= -8, b2 = -1, b3 = 5
I can show that algebraically, too, as a general case. (At least, I
think I can.)
With the same data, -nl- produces one or other of those sets of
coefficients, depending on the starting values. They are
equivalent; I just have a preference for the set where b2 is positive.
Examples with synthetic data follow, but it applies to real data too.
* Example of identical curves with different coefficients as above.
. set obs 101
obs was 0, now 101
. gen x=(_n-1)/10
. gen y1= 2 + 8/(1 + exp(-(1)*(x-5)))
. gen y2= 10 + -8/(1 + exp(-(-1)*(x-5)))
. assert y1==y2
*Example of -nl- producing models with each of the coefficient
sets, depending on starting values (using x and y1 generated above)
. nl ( y1= {b0} + {b1}/(1 + exp(-{b2}*(x-{b3}))) ), init(b0 1 b1 6
b2 0.5 b3 3)
(obs = 101)
[iterations snipped]
Source | SS df MS
-------------+------------------------------ Number of obs
= 101
Model | 984.148347 3 328.049449 R-squared =
1.0000
Residual | 2.4290e-12 97 2.5042e-14 Adj R-squared =
1.0000
-------------+------------------------------ Root MSE
= 1.58e-07
Total | 984.148347 100 9.84148347 Res. dev. =
-2880.598
----------------------------------------------------------------------
--------
y1 | Coef. Std. Err. t P>|t| [95% Conf.
Interval]
-------------
+----------------------------------------------------------------
/b0 | 2 4.18e-08 4.8e+07 0.000 2 2
/b1 | 8 6.81e-08 1.2e+08 0.000 8 8
/b2 | 1 2.28e-08 4.4e+07 0.000 1 1
/b3 | 5 2.36e-08 2.1e+08 0.000 5 5
----------------------------------------------------------------------
--------
Parameter b0 taken as constant term in model & ANOVA table
. nl ( y1= {b0} + {b1}/(1 + exp(-{b2}*(x-{b3}))) ), init(b0 1 b1 -6
b2 -0.5 b3 3)
(obs = 101)
[iterations snipped]
Source | SS df MS
-------------+------------------------------ Number of obs
= 101
Model | 984.148347 3 328.049449 R-squared =
1.0000
Residual | 2.4290e-12 97 2.5042e-14 Adj R-squared =
1.0000
-------------+------------------------------ Root MSE
= 1.58e-07
Total | 984.148347 100 9.84148347 Res. dev. =
-2880.598
----------------------------------------------------------------------
--------
y1 | Coef. Std. Err. t P>|t| [95% Conf.
Interval]
-------------
+----------------------------------------------------------------
/b0 | 10 4.18e-08 2.4e+08 0.000 10 10
/b1 | -8 6.81e-08 -1.2e+08 -8 -8
/b2 | -1 2.28e-08 -4.4e+07 -1 -1
/b3 | 5 2.36e-08 2.1e+08 0.000 5 5
----------------------------------------------------------------------
--------
Parameter b0 taken as constant term in model & ANOVA table
best wishes
Rosy Reynolds
BSAC Resistance Surveillance Co-ordinator
www.bsacsurv.org
----- Original Message ----- From: "Steven Joel Hirsch Samuels"
<[email protected]>
To: <[email protected]>
Sent: Thursday, October 25, 2007 5:05 PM
Subject: Re: st: nl -choice between alternative parametrisations of
sigmoid models
Rosy, are you saying that the same set of data will produce two
different curves? What conditions vary to do this? Exactly what
are the data? Have you plotted the data? Are you sure that the
curve is monotone?
-Steven
On Oct 25, 2007, at 11:41 AM, Rosy Reynolds wrote:
Hello,
I am fitting 4-parameter logistic (sigmoid Emax) dose-response
models using the built-in -log4- feature of -nl-.
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,
The same curve can actually be produced with two different sets
of these parameters.
In one set, the Hill slope b2 is positive and the other
parameters intuitively have the interpretations above.
In the other set, b2 is negative, the sign of b1 is reversed,
and b0 becomes the outcome at infinitely high dose instead of at
the lowest doses. The lowest-dose outcome is now given by b0+b1.
With our data, -nl- naturally produces the set of coefficients
with negative b2.
For ease of interpretation, I would prefer the set with positive b2.
I can push -nl- into doing that by supplying carefully chosen
starting values close to the desired coefficients. I could even
run -nl- , manipulate the coefficients it obtains, and use those
as starting values. That would always work, I suppose, but it
seems long-winded.
Please could you tell me an easier way to make -nl- parametrise
the model in the preferred way, if you know of one? I haven't
found anything about it in the manual.
Thanks for thinking about this.
best wishes
Rosy Reynolds
BSAC Resistance Surveillance Co-ordinator
www.bsacsurv.org
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Steven Samuels
[email protected]
18 Cantine's Island
Saugerties, NY 12477
Phone: 845-246-0774
EFax: 208-498-7441
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Steven Samuels
[email protected]
18 Cantine's Island
Saugerties, NY 12477
Phone: 845-246-0774
EFax: 208-498-7441
*
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/