3
$\begingroup$

In treatment studies it is common to report multiple outcome measures from the same subjects. The treatment effects on these outcomes are typically correlated so this should be taken into account when modeling the data. However, papers only publish means and standard deviations before and after the intervention. So given for example a report of three tests:

data.frame(mean.pre=c(1,5,100), sd.pre=c(0.2, 1, 10), mean.post=c(3, 2, 80), sd.post=c(0.3, 0.9, 11), rownames=c('test1','test2','test3'))
      mean.pre sd.pre mean.post sd.post
test1        1    0.2         3     0.3
test2        5    1.0         2     0.9
test3      100   10.0        80    11.0

... would it be possible to calculate the 3 x 3 covariance matrix of the standardized mean differences d = (mean.post - mean.pre)/sd.pre between test1, test2, and test3?

Bonus info: sometimes the standard deviation of the change sd(mean.post-mean.pre) can be obtained through t-statistics and p-values and sometimes the correlation cor(mean.pre, mean.post) is also known. Does that add anything?

If the above is not sufficient to calculate covariances, can we at least put a bound on the possible covariances given this data?

$\endgroup$
3
$\begingroup$

The best bound you can get is due to a variant of the Cauchy-Schwarz inequality: $$|Cov(X,Y)| \leq \sqrt{Var(X) \cdot Var(Y)}$$

This is of course very broad, as the covariance can be negative but the right hand side is always positive. It ensures that the $2 \times 2$-covariance matrix between $X$ and $Y$ is positive semidefinite. I'm afraid that the bonus info doesn't improve much, because now you have a $4 \times 4$-covariance matrix where you have the $2 \times 2$ diagonal blocks known but on the other side now 4 unknown covariances that can be chosen arbitrarily as long as the whole covariance matrix remains positive semidefinite. (I'll think of a proof.)

I would suggest to ask the authors. If they can't answer it, they published poor science anyway and you can discard even means and standard deviations.

$\endgroup$
  • 3
    $\begingroup$ You should probably write the two-sided bound: $-\sqrt{Var(X) \cdot Var(Y)} \leq Cov(X,Y)\leq \sqrt{Var(X) \cdot Var(Y)}$ $\endgroup$ – Silverfish Jan 20 '15 at 12:08
  • 1
    $\begingroup$ Thanks! Well, in this case 28 out of 28 studies included in the meta-analysis did not publish covariances and the 9 authors who responded did not provide it or the data necessary to calculate it. Poor science indeed. I will leave the question for a day and accept this answer later if nothing magical pops up. Silverfish's explicitation is good so you may want to update the answer with that or write $|Cov(X,Y)|$ on the left hand side. $\endgroup$ – Jonas Lindeløv Jan 20 '15 at 12:15
  • $\begingroup$ (+1) You can do a little better when there are more than two variables: the covariance matrix must be positive semidefinite. $\endgroup$ – whuber Jan 20 '15 at 12:53
  • $\begingroup$ @whuber Sounds interesting. Could you elaborate since this is the usual case? Would this constraint break the symmetry around 0? $\endgroup$ – Jonas Lindeløv Jan 20 '15 at 18:02
  • 1
    $\begingroup$ It's actually not terribly helpful, Jonas, although it does narrow the possibilities. It also does break the symmetry: after all, it's pretty hard to make three variables all be perfectly inversely related. If $A$ and $B$ go in opposite directions and $B$ and $C$ go in opposite directions, that practically forces $A$ and $C$ to go in the same direction. You can get details by searching our site. The answers for three variables at stats.stackexchange.com/questions/72790 seem particularly instructive. $\endgroup$ – whuber Jan 20 '15 at 18:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.