> Dear Allstat,
>
> Yesterday I asked the following question and received some useful replies,
> summarised below.
> My thanks to Tim Cole, Stephen Senn, Michael Franklin, Ken Butler and
> Chris Jones.
>
> Regards,
>
> Sean McGuigan
>
> I've come across a paper where the correlation coefficient of final
> measurement (Y) with percentage change from baseline (X), ie: ((Y-X)/X),
> was calculated.
> It's well known that change from baseline (Y-X) is correlated with both
> final and baseline, even when Y and X are uncorrelated (+0.7071, and
> -0.7071, respectively).
>
> What correlation is to be expected between Y and (Y-X)/X, when Y and X are
> uncorrelated?
>
> Boehringer Ingelheim
> Ellesfield Avenue
> Bracknell
> Berkshire RG12 8YS
>
> phone: 01344 74 6766
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>
> ____________________________________________________________
> I make it 1 / sqrt (1 + (CVx/CVy)**2) where CV = sd/mean. So using your
> assumptions of equal sd that appears to give the same result of 1/sqrt(2).
>
> Michael F. Franklin
> ____________________________________________________________
>
> Dear Sean,
>
> You will find this dealt with in my paper:
> baseline distribution and conditional size, Journal of
> Biopharmaceutical Statistics 392) 265-270 (1993). (there are some
> misprints in that paper and you will have to search downstream for
> the correction).
>
> Basically the argument is like this
> 1) The measure (Y-X)/X can be written Y/X -1.
> 2) Therefore we want the correlation between Y/X and Y or X as the
> case may be.
> 3) However if Y and X positive then the rank correlation between
> log X and log Y will be the same as the rank correlation between X
> and Y.
> 4) If the variances are small, the product moment correlation wil be
> similar to the rank correlation.
> 5) The correlation between log(Y)-log(X) and log(Y) or log(X) as the
> case will be the same as the correlation between log(Y/X) and
> log(Y) or log(X).
> 6) Hence the correlation coefficients will again be about root 2 or
> minus root 2.
>
> One can produce a Taylor's expansion argument that gives the
> same (approximate) answer.
>
> Regards
>
> Stephen Senn
> ___________________________________________________________
> (Y-X)/X is Y/X-1, so this is equivalent to finding the correlation between
> Y and Y/X. The fact of dividing by X makes this a non-linear problem,
> which suggests that there might not be a nice general result as with the
> change from baseline.
>
> I would think about simulating the distribution of the correlation:
> generate a random sample of uncorrelated X and Y values, calculate the
> correlation between Y and Y/X, and repeat many times to get a simulated
> distribution of correlations. The results would probably depend on what
> distribution you chose for X and Y, but in any case you could test a
> correlation between Y and % change from baseline from a data set by
> comparing it with your simulated distribution.
>
>
> Ken Butler
> ____________________________________________________________
> Dear Sean,
>
> I don't think it's possible to give an explicit answer, because of the
> form
> of the percent change from baseline. However it ought to be similar to
> your
> simpler case, for the following reason:
>
> Corr(Y, (Y-X)/X ) = Corr(Y, Y/X ) ? Corr(ln Y, ln Y/X ) = Corr(ln Y, ln Y
> -
> ln X ).
>
> So long as Y and ln Y, and X and ln X, are highly correlated (which
> depends
> mainly on the CV of X and Y) this is equivalent to your own example, so
> the
> correlation ought to be near +0.7. You could test it by simulation.
>
> Best wishes,
>
> Tim Cole
> ________________________________________________________________
> If you assume independence, the following is what you get. If you just
> assume
> uncorrelatedness, I fear it gets more difficult.
>
> V((Y/X)-1) = V(Y/X) = E((Y/X)^2) - (E(Y/X))^2
> = E(Y^2)E(1/X^2) - (E(Y))^2 (E(1/X))^2
>
> Cov(Y,(Y/X)-1) = Cov(Y,(Y/X)) = E(Y^2/X) - E(Y) E(Y/X)
> = E(Y^2)E(1/X) - (E(Y))^2 E(1/X) = V(Y) E(1/X)
>
> Thus, corr(Y,(Y/X)-1) = E(1/X) V(Y)
> -----------
> [V(Y) {E(Y^2)E(1/X^2) - (E(Y))^2
> (E(1/X))^2}]^(1/2)
>
> = E(1/X) SD(Y)
> -----------
> [E(Y^2)E(1/X^2) - (E(Y))^2 (E(1/X))^2]^(1/2)
>
>
> Not sure that's a great help but there you are!
>
> Best,
> Chris Jones.
> ________________________________________________________________
>

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