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STATEMENT OF PROBLEM:

Suppose $ \left( \begin{array}{ccc} \ Z_1 \\ Z_2 \end{array} \right)$ follows a Bivariate standard normal with covariance $ \rho $

$ \left( \begin{array}{ccc} \ Z_1 \\ Z_2 \end{array} \right) \sim MVN \left( \left( \begin{array}{ccc} \ 0 \\ 0 \end{array} \right) ,\left( \begin{array}{ccc} 1 & \rho \\ \rho & 1 \\ \end{array} \right) \right) $

I am trying to solve for the rejection region c as a function of $\alpha$ such that:

$ P(|Z_1|\ge c) + P(|Z_1|<c , |Z_2| \ge c ) =\alpha $

MY PAST ATTEMPTS:

The solutions is easy when $\rho=0$, as $$ P(|Z_1|\ge c) + P(|Z_1|<c , |Z_2| \ge c ) =\alpha $$ by symmetry of LEFT probability, and by independence of that on the right... $$ =2*[1-\Phi(c)] + P(|Z_1|<c)*P(|Z_2| \ge c ) =\alpha $$ $$ =2*[1-\Phi(c)] + 2*[1-\Phi(c)][\Phi(c)-\Phi(-c)] =\alpha $$ $$ 2*[1-\Phi(c)] + 2*[1-\Phi(c)][2*\Phi(c)-1] =\alpha $$ $$ 4*[1-\Phi(c)]*\Phi(c)=\alpha $$ $$ 4*[\Phi(c)-\Phi(c)^2]=\alpha $$ $$ -4\Phi(c)^2+4\Phi(c)-\alpha=0 $$ which is a quadratic function of $\Phi(c)$, yielding $$ \Phi(c)= \frac{1+\sqrt{1-\alpha}}{2} $$ Giving the solution for c as a function of $\alpha$: $$ \boxed{c = \Phi^{-1} \left( \frac{1+\sqrt{1-\alpha}}{2} \right)} $$

However, when $\rho \ne0$, I cannot find a closed form solution! All i can find is $$ 2*[1-\Phi(c)]+2*\int_{-c}^{c}\Phi\left( \frac{-c+\rho z}{\sqrt{1-\rho^2}}\right)\phi(z) dz =\alpha $$ and solve it numerically. This WORKS, but does not yield the desired closed form solution

The one hint that I was given was to use $ U_1 =Z_1 + Z_2$ and $U_2 = Z_1 - Z_2 $. I tried this, to no avail. Can anyone help??

NOTE

This problem goes on to ask for a similar solution when

$ \left( \begin{array}{ccc} \ Z_1 \\ Z_2 \\Z_3 \end{array} \right) \sim MVN \left( \left( \begin{array}{ccc} \ 0 \\ 0 \\ 0 \end{array} \right) ,\left( \begin{array}{ccc} 1 & \rho & \rho \\ \rho & 1 & \rho \\ \rho & \rho & 1 \\ \end{array} \right) \right) $

Solve for the rejection region c as a function of $\alpha$ such that:

$ P(|Z_1|\ge c) + P(|Z_1|<c , |Z_2| \ge c )+ P(|Z_1|<c , |Z_2|<c , |Z_3| \ge c ) =\alpha $

Lewkrr
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    I believe this cannot generally be put into a closed form. One integration can be done, leaving the integral of a Gaussian times an Error Function: that's as far as one can get. One approach is to study the properties of this function of $(c,\rho)$ sufficiently well that it's as good as any other transcendental function in formulas. – whuber Jun 03 '15 at 19:13
  • If you assume you know $\rho$, you can multiply $(Z_1, Z_2)$ by the inverse of the "square root" ($PP^t = \Sigma$) of the assumed covariance matrix $\Sigma$, thereby obtaining $(U_1, U_2)$, which under the null hypothesis are independent. For the Normal distribution, "independent" is equivalent to "covariance = 0", so your test for the case $\rho = 0$ can then be applied. – jbowman Jun 09 '15 at 18:37
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    In other words, given a vector $\mathbf{Z}$ of $d$ standard normal r.vs with given correlation matrix, find $c>0$ such that the probability $R(c)$ of the square $\{ \max_i |Z_i| \leq c \}$ has fixed value $1- \alpha$. Even for $d=2$, I agree with @whuber saying that no general tractable form will be found. For small $d$ a numerical solution can rely on the R package `mvtnorm` by A. Genz et al. allowing a fast computation of $R(c)$ with a very good precision. This can work with any correlation matrix: compound symmetry or any other specific form being unlikely to lead to a simpler problem. – Yves Jun 10 '15 at 05:49

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