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
