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I am trying to solve part (b) of this question, but I am having trouble with something. With some help, I have been able to solve the following:

$$ \begin{align*} \max_{a \neq 0, b \neq 0}\operatorname{Cor}(a^\top X_1, b^\top X_2) &= \max_{a \neq 0, b \neq 0} \frac{a^\top \Sigma_{12} b}{\sqrt{a^\top \Sigma_{11} a} \sqrt{b^\top \Sigma_{22} b}} \\ &= \max_{a \neq 0, b \neq 0} \frac{a^\top \Sigma_{11}^{1/2}\Sigma_{11}^{-1/2} \Sigma_{12} \Sigma_{22}^{1/2}\Sigma_{22}^{-1/2} b}{\sqrt{a^\top \Sigma_{11}^{1/2}\Sigma_{11}^{1/2} a} \sqrt{b^\top \Sigma_{22}^{1/2} \Sigma_{22}^{1/2} b}} \\ &= \max_{u \neq 0, v \neq 0} \frac{u^\top M v}{\sqrt{u^\top u}\sqrt{v^\top v}} \end{align*} $$ using $u = \Sigma_{11}^{1/2} a$ and $v = \Sigma_{22}^{1/2}b$.

However, I don't know how I have to get to the square of this. Surely just squaring the outcome will not help, because then you also have the square of the correlation. Can someone help me with this?

Thanks in advance.

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  • $\begingroup$ my guess is that it's a typo and they meant Cov rather than Cor. $\endgroup$
    – mark leeds
    Commented Jan 16 at 14:08
  • $\begingroup$ @markleeds Do you think they meant Cov in both (a) and (b) or only in (b)? $\endgroup$
    – Lucius Aelius Seianus
    Commented Jan 16 at 14:45
  • $\begingroup$ actually, it's gotta be cor since they are dividing by the variances. My mistake but something is weird there because I don't see your error. Cor^2 would be R^2 so maybe they mean R^2 ? don't want to lead you down wrong path. Maybe someone else can check it out because I'm not around today. Whatever they mean, it's probably both (a) and (b) since (b) is derived from (a). $\endgroup$
    – mark leeds
    Commented Jan 17 at 16:00
  • $\begingroup$ Hi: ook up the derivation of principal components in a multivariate text ? ( johnson and wichern or mardia and bibby ) . They do a similar thing in that algorithm so maybe that would give you some hint. It's possible you are doing something wrong but I don't see it. $\endgroup$
    – mark leeds
    Commented Jan 17 at 16:03
  • $\begingroup$ Note that cor is between zero and one so, maximizing corr is the same as maximizing corr^2. Maybe that's what they are using in order to justify the squaring ? I'm at a loss and, now that I think about it, you can look up the principal components but I don't think it's gonna help. Very strange what's going on there. $\endgroup$
    – mark leeds
    Commented Jan 17 at 16:21

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