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$\mathbf{x}=[x_1,x_2,...,x_m]^{\top}$ is a vector of length $m$ and $\mathbf{y_1}, \mathbf{y_2}, ..., \mathbf{y_n}$ are similarly $n$ vectors of length $m$.
If the elements of $\mathbf{x}$ and $\mathbf{y_1}, ..., \mathbf{y_n}$ are independent and have normal distribution with mean zero and variance one, what is the distribution of $max( r(\mathbf{x},\mathbf{y_1}),r(\mathbf{x},\mathbf{y_2}),...,r(\mathbf{x},\mathbf{y_n}))$ where $r(u,v)$ is the Pearson correlation coefficient between vectors $u$ and $v$.

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What is the "Pearson correlation coefficient between vectors"? – leonbloy Aug 24 at 19:13

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I doubt there is a unique and/or exact answer to your question. When the two variables follow a bivariate normal distribution, the distribution of the Pearson sample correlation coefficient is known, see for example the wiki and wolfram pages.

At the same time, while the exact distribution of the maximum $Y^*$ of $n$ i.i.d random variables $Y_1,...,Y_n$ with distribution function $F()$is in general $F_{Y^*}(y^*) = \left [F_Y(y) \right]^n$, results for the maximum of dependent random variables exist only under various restrictions on the dependence structure, and only asymptotically (J. Galambos has produced results on the subject). And the r.v.'s $ r(\mathbf{x},\mathbf{y_1}),...,r(\mathbf{x},\mathbf{y_n}))$ are dependent due to the common $\mathbf x$ variable. So it appears that your question is rather too general to permit a definite answer.

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Thanks Alecos. I cannot get why a 'unique' answer may not exist. If you numerically generate random numbers and calculate the maximum Pearson correlation, it has a well shaped distribution. Is it at least possible to derive an equation for the expected value of the max Pearson correlation? – mghandi Aug 24 at 17:00
The fact that simulations provide stable results, does not mean that we are able to represent these results in analytical form. The distribution may "exist" in the mathematical sense of the word - but I meant it in the sense of whether we can arrive at a specific exact analytical functional form. The same comments apply for the expected value. You could look up the methods to derive functional forms that approximate the distribution that emerges from simulations. – Alecos Papadopoulos Aug 24 at 17:34

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