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I posted recently a related question about the convergence rate of a Pearson correlation coefficient, here.

Now, I am interested in the matrix version.

Let $X_1,\ldots,X_N \sim \mathcal{N}(0,1)$ be correlated random variables. Let $\hat{\rho}(X_i,X_j)$ be the estimated correlation between $X_i$ and $X_j$ from a sample of size $T$.

I derived (cf. the previous post) the following tail bound.

For $t > 0$,

$$\mathbb{P}\left(\left|\hat{\rho}(X_i,X_j) - \rho(X_i,Y_j)\right| \geq t \right) \leq \exp\left\{ -\frac{1}{2}T \left(\sqrt{1 + \frac{2t}{1-\rho}} + 1 + \frac{t}{1-\rho} \right) + \ln 2 \right\}$$

Now, I wish a rather sharp tail bound for

$$\mathbb{P}\left(\sup_{ij} \left|\hat{\rho}(X_i,X_j) - \rho(X_i,Y_j)\right| \geq t \right)$$

An obvious one is

$$\mathbb{P}\left(\sup_{ij} \left|\hat{\rho}(X_i,X_j) - \rho(X_i,Y_j)\right| \geq t \right) \leq N^2 \exp\left\{ -\frac{1}{2}T \left(\sqrt{1 + \frac{2t}{1-\rho}} + 1 + \frac{t}{1-\rho} \right) + \ln 2 \right\}$$

Can we do better?

My guess is that we can leverage the correlation between the (correlation) coefficients to improve this bound.

There are some results for "simultaneous confidence interval estimation", but I do not know if it can be applied here.

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  • $\begingroup$ Would Wishart distribution help? $\endgroup$ – Aksakal Aug 27 '15 at 13:32
  • $\begingroup$ Thanks for the hint. I am aware that there are concentration bounds for the covariance matrix in spectral norm (arxiv.org/pdf/1212.5860.pdf) but I need bounds for the sup norm. $\endgroup$ – mic Aug 27 '15 at 13:45
  • $\begingroup$ " we can leverage the correlation between the (correlation) coefficients to improve this bound." - what is the intuition behind this thought? $\endgroup$ – Aksakal Aug 27 '15 at 14:07
  • $\begingroup$ Just meant that we may obtain a better bound, for instance in spectral norm it is the case (a factor $N$ instead of $N^2$). In my case, I am considering block-diagonal correlation matrix with uniform correlation $\rho$ inside each block (clusters). $\endgroup$ – mic Aug 27 '15 at 14:28
  • $\begingroup$ I would start with $N=3$, instead of the general case $N$. In some diff geometry problems $N=2$ is not extendable to a general case, but $N=3$ is. $\endgroup$ – Aksakal Aug 27 '15 at 14:36

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