# An integral involving the normal CDF [closed]

Let $\mu = -a \mathbb{1}_n$ and $\Sigma = \mathbb{1}_n \mathbb{1}_n^T + \operatorname{diag}(\mathbb{1}_n)$. Is there a way to simplify $\Phi_n( - \Sigma^{-1/2} \mu)$ given the very simple structure of $\mu$ and $\Sigma$? Maybe recursively in terms of lower dimensional cdf somehow...? $n$ is potentially very large. Thanks.

## closed as unclear what you're asking by whuber♦May 5 '17 at 17:40

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• $\Phi_n( - \Sigma^{-1/2} \mu)$ is not defined given that $( \Sigma^{-1/2} \mu)$ is a vector. – Xi'an Mar 5 '17 at 14:36
• Since $\Sigma$ is a covariance matrix with the covariances all equal, its inverse square root most likely has a simple structure that might even be found as a theorem in a matrix computation book. @Xi'an $\Phi_n$ is the$n$-dimensional joint CDF (a real-valued function of $n$ real variables) and thus its argument can be regarded as a vector. – Dilip Sarwate Mar 5 '17 at 14:41
• Thanks @Dilip Sarwate. Would you advise some good reference book where I could find such information? More generally I would interested in a comprehensive reference book with algebraic tricks and facts on known distributions? – user79097 Mar 5 '17 at 16:21
• The expression "$\Sigma^{-1/2}$" is not defined because there are many possibilities for the square root (even when $n=1$!). Could you indicate which one is intended? – whuber May 5 '17 at 17:40