Dear All,
Thankyou to all that replied to my email concerning the combined mixed
residual model, and thanks to Leonid for the changes in the $ERROR code. I
have one final question regarding the reporting of the thetas and the
sigmas. Why, when you use thetas in the residual error structure (thus
replacing the epsilons), does it return the standard deviation, whereas
when using epsilons it returns the variance? What algorithm is it using to
obtain each?
Once again, thanks for any help.
Best,
Paul Westwood.
On Feb 15 2008, Ribbing, Jakob wrote:
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:[EMAIL PROTECTED]
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
