From: j.h.proost

Date: Thu, 10 Dec 2015 19:28:13 +0100

Dear Dennis,

Apart from bootstrap, as Leonid suggested, a solution may be found in

the law of the propagation of errors.

If your function is Y = a + b . X

the variance in Y due to variance in a and b may be obtained from the

law of the propagation of errors (assuming that X is error-free):

var(Y) = [ dY/da ]^2 . var(a) + [ dY/db ]^2 . var(b) + dY/da . dY/db

. covar(a,b)

where var(a) is the square of SE(a) and covar(a,b) is the covariance

between a and b.

Since dY/da = 1 and dY/db = X, it follows:

var(Y) = var(a) + X^2 . var(b) + X . covar(a,b)

SE(Y) is the square root of var(Y).

This approach provides a quick and easy estimate of the SE of Y for any

value of X. It seems reasonable to assume that this estimate of SE(Y) is

approximately 'as good as the estimates of SE(a) and SE(b)'. If you

trust SE(a) and SE(b), you may trust SE(Y).

best regards,

Johannes H. Proost

Dept. Pharmacokinetics, Toxicology and Targeting

University of Groningen

the Netherlands

Op 10-12-2015 om 17:23 schreef Fisher Dennis:

*> Colleagues
*

*>
*

*> I have fit an exposure response model using NONMEM — the optimal model
*

*> is a segmented two-part regression with Cp on the x-axis and response
*

*> on the y-axis. The two regression lines intercept at the cutpoint.
*

*> The parameters are:
*

*> slope of the left regression
*

*> cutpoint between regressions
*

*> “intercept” — y value at the cutpoint
*

*> slope of the right regression (fixed at zero; models in which the
*

*> value was estimated yielded similar values for the objective function)
*

*>
*

*> I have been asked to calculate the confidence interval for the
*

*> response at various Cp values.
*

*>
*

*> Above the cutpoint, this seems straightforward:
*

*> a. if NONMEM yielded standard errors, the only relevant parameter is
*

*> the y value at the cutpoint and its standard error
*

*> b. if NONMEM did not yield standard errors, the confidence interval
*

*> could come from either likelihood profiles or bootstrap
*

*>
*

*> My concern is calculating at Cp values below the cutpoint, for which
*

*> both slope and intercept come into play. Any thoughts as to how to do
*

*> this in the presence or absence of NONMEM standard errors?
*

*> The reason that I mention with / without presence of SE’s is that this
*

*> model was fit to two different datasets, one of which yielded SE’s,
*

*> the other not.
*

*>
*

*> Any thoughts on this would be appreciated.
*

*>
*

*> Dennis
*

*>
*

*> Dennis Fisher MD
*

*> P < (The "P Less Than" Company)
*

*> Phone: 1-866-PLessThan (1-866-753-7784)
*

*> Fax: 1-866-PLessThan (1-866-753-7784)
*

*> www.PLessThan.com <http://www.plessthan.com/>
*

*>
*

*>
*

*>
*

Received on Thu Dec 10 2015 - 13:28:13 EST

Date: Thu, 10 Dec 2015 19:28:13 +0100

Dear Dennis,

Apart from bootstrap, as Leonid suggested, a solution may be found in

the law of the propagation of errors.

If your function is Y = a + b . X

the variance in Y due to variance in a and b may be obtained from the

law of the propagation of errors (assuming that X is error-free):

var(Y) = [ dY/da ]^2 . var(a) + [ dY/db ]^2 . var(b) + dY/da . dY/db

. covar(a,b)

where var(a) is the square of SE(a) and covar(a,b) is the covariance

between a and b.

Since dY/da = 1 and dY/db = X, it follows:

var(Y) = var(a) + X^2 . var(b) + X . covar(a,b)

SE(Y) is the square root of var(Y).

This approach provides a quick and easy estimate of the SE of Y for any

value of X. It seems reasonable to assume that this estimate of SE(Y) is

approximately 'as good as the estimates of SE(a) and SE(b)'. If you

trust SE(a) and SE(b), you may trust SE(Y).

best regards,

Johannes H. Proost

Dept. Pharmacokinetics, Toxicology and Targeting

University of Groningen

the Netherlands

Op 10-12-2015 om 17:23 schreef Fisher Dennis:

Received on Thu Dec 10 2015 - 13:28:13 EST