N(μ, σ²) [Software]

posted by Helmut Homepage – Vienna, Austria, 2017-09-04 15:32 (2397 d 22:10 ago) – Posting: # 17772
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Hi Yong,

❝ ❝ ❝ I prefer the following formula :

❝ ❝ ❝ intersubject CV = sqrt(Var(Sequence*Subject)) / RefLSM

❝ ❝ Why?

❝ Because this is also used for ln_transformed data.


No it isn’t! Back to the basics. The model of a 2×2×2 crossover assumes [sic] IID. For log-transformed data:
\(\log{(\mu_T / \mu_R)} = \mu_T - \mu_R\), which is estimated by the difference of the LSMs of log-transformed data \(\bar{x}_T - \bar{x}_R\). Now it gets interesting (i.e., the assumption!): In the balanced case for simplicity (where n1 = n2 and n = n1 + n2), \(\bar{x}_T - \bar{x}_R\) follows a normal distribution \(N\left ( \log{(\mu_T / \mu_R)}, 2\sigma^2 / n \right )\).
Since σ2 is unknown, it is estimated by the MSE from ANOVA. Then we can estimate \(CV = \sqrt{e^{MSE} - 1}\). Do you see \(\bar{x}_R\) in this derivation? I don’t. Remember that the normal distribution is described by two parameters, μ and σ2, which are independent. If you are interested in the variance component, please leave the mean(s) completely out of it (as it is correctly done in PHX/WNL for log-transformed data).[image]

❝ Like one formula for intrasubject CV for both transformed and none transformed data, I think this formula is generally accepted both for transformed and none transformed data.


I think that for untransformed data dividing MSE by LSMR goes back to Kem Phillips, who wrote*

σ being expressed as a percentage of a reference mean, that is, as a coefficient of variation; values of the difference in means are expressed as percentages of the same reference mean.

That’s unfortunate and IMHO, not correct at all (Wolfgang Pauli would say: That is not only not right; it is not even wrong!). Again: The mean and variance are independent. So, the more I think about it: Even my idea of using the weighted global mean does not make sense. Paraphrasing Ste­phen Senn: Proving that apples are oranges by comparing the weight.



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