RE: [NMusers] IIV on %CV-Scale - Standard Error (SE)

Thank you all for the clarifications on this,

James previously sent a good explanation outside of nm-users which I am
adding at the end, in case anyone still wants to understand where the
number 2 comes from.

Leonid, you are right there was a typo in my formula and your correction
is correct. What I meant to say was that if one calculates the relative
SE on the variance scale, one has to divide this number by 2 in order to
get the RSE on the appropriate scale. To make this clear to everyone
(stop reading here if you already understand the issue):

If your model is parameterised as:
CL=TVCL*EXP(ETA(1))
And OMEGA11 is estimated to 0.09, this means that the standard deviation
of ETA1 is sqrt(0.09)=0.3. The IIV in CL is then approximately 30%
around the typical CL (TVCL). If the SE of OMEGA11 is estimated to 0.009
the relative SE of OMEGA11 is 0.009/0.09=10% (on the variance scale). =
We
then report IIV in CL as 30% with RSE=5%.

Best

jakob

-----Original Message-----
From: James G Wright [mailto:james_at_wright-dose.com]
Sent: 15 February 2008 13:57
To: Ribbing, Jakob
Subject: RE: [NMusers] IIV on %CV-Scale - Standard Error (SE)

Hi Jakob,

The trick is just knowing that you need to apply Wald's formula because
the SE is a length (that depends on scale) and not a point. Thus, to
get the right length, you need to apply a correction factor that depends
on the rescale. For transformations like x^n, the rescale depends on x,
hence the need to include the derivative. When I last taught calculus,
I explained this as follows (bear with me, and apologises if I have
missed the point of misunderstanding and inadvertently patronize).

Imagine you have a square, with sides of length 10 (x), it has area of
100 (x^2). Now imagine, you increase the length of a side by 1. The
are is now 121=(x+1)^2=x^2+2x+1. Visually, you have added an extra
length on to each side of the square and extra +1 in the corner. The 2x
is the first derivative of x^2, that is how much x^2 changes if you add
1 to x.

What does this have to do with a confidence interval? To calculate the
95% confidence interval you add/subtract 1.96 times the SE to x. You
can estimate the impact this has on x^2, by using the first derivative.
Of course, you have to do all this backwards for the square root case.
For more complex functions, the formula is only approximate because
second-order terms can become important.

That's how I visualise the equations in my head anyway. Best regards,
James

James G Wright PhD
Scientist
Wright Dose Ltd
Tel: 44 (0) 772 5636914
Received on Fri Feb 15 2008 - 11:48:45 EST