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Let $X_1$ and $X_2$ be bivariate normal with mean $\mu=(0,\mu_2)$, for any $\mu_2$, and correlation $\rho$.

Consider the following inequality: \begin{align*} Pr\left\{|X_1| \ge \Phi^{-1}(1-\alpha/2), |X_2| \ge \Phi^{-1}(1-\alpha/2)\right\} \le 2Pr\left\{|X_1| \ge \Phi^{-1}(1-\alpha/4), |X_2| \ge \Phi^{-1}(1-\alpha/2)\right\} \end{align*}

Is the inequality true for correlation $|\rho| \le 1$?

Are there relevant results on general bivariate normal distributions?

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    $\begingroup$ Are $X_1$ and$X_2-\mu_1$ standard normal random variables? Can we persuade you to edit the problem to use more reasonable notation so that the mean of $X_2$ is denoted by $\mu_2$ rather than $\mu_1$? and perhaps eliminate those left parentheses that follow the commas in your displayed equations? $\endgroup$
    – Dilip Sarwate
    Oct 8 '12 at 20:55
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    $\begingroup$ Although I did not look into the inequality itself, I believe that $|\rho| \leq 1$ is always true since it is a correlation coefficient, which I guess is defined by $\rho = cov(X_1, X_2)\big/ \sqrt{var(X1)}\sqrt{ var(X2)}$...is your question: "is the inequality true?" or do you question the inequlaity under some assumption regarding $\rho$? $\endgroup$
    – julien stirnemann
    Oct 8 '12 at 22:15
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The constants in this problem do not make much sense unless $X_1$ and $X_2$ have variance $1$ so that $X_1$ and $X_2-\mu_2$ are standard normal random variables, an assumption that the OP apparently is unwilling to make since this was asked about in the comments, and the OP did not include the assumption in the revised version of the question.

Assumption: $X_1$ and $X_2$ have variance $1$.

If $X_1$ is a standard normal random variable, then $P\{|X_1| \geq \Phi^{-1}(1-y/2)\} = y$. This result holds for $X_2$ as well if $\mu_2 = 0$. Thus, if $X_1$ and $X_2$ both are independent standard normal random variables, then $$\begin{align} &\quad P\left\{|X_1| \geq \Phi^{-1}(1-\alpha/2), |X_2| \geq \Phi^{-1}(1-\alpha/2)\right\}\\ &= P\left\{|X_1| \geq \Phi^{-1}(1-\alpha/2)\right\} P\left\{|X_2| \geq \Phi^{-1}(1-\alpha/2)\right\}\\ &= \alpha^2 \end{align}$$ while $$\begin{align} &\quad P\left\{|X_1| \geq \Phi^{-1}(1-\alpha/4), |X_2| \geq \Phi^{-1}(1-\alpha/2)\right\}\\ &= P\left\{|X_1| \geq \Phi^{-1}(1-\alpha/4)\right\} P\left\{|X_2| \geq \Phi^{-1}(1-\alpha/2)\right\}\\ &= (\alpha/2)\alpha = \alpha^2/2 \end{align}$$ In short, for the case $\mu_2 = 0$, the conjectured result holds (with equality) for the case $\rho = 0$. Continuing to look at the case $\mu_2 = 0$, if $X_2 = \pm X_1$ (the case when $\rho = \pm 1$), we have $$\begin{align} &\quad P\left\{|X_1| \geq \Phi^{-1}(1-\alpha/2), |X_2| \geq \Phi^{-1}(1-\alpha/2)\right\}\\ &= P\left\{|X_1| \geq \Phi^{-1}(1-\alpha/2)\right\}\\ &= \alpha \end{align}$$ while $$\begin{align} &\quad P\left\{|X_1| \geq \Phi^{-1}(1-\alpha/4), |X_2| \geq \Phi^{-1}(1-\alpha/2)\right\}\\ &= P\left\{|X_1| \geq \Phi^{-1}(1-\alpha/4)\right\}\\ &= \alpha/2 \end{align}$$ and so once again the conjectured result holds (with equality). What happens for other values of $\rho \in [-1,1]$ is not immediately obvious since the bivariate cumulative normal distribution function must be used and the special meanings of $\Phi^{-1}$ are lost. The case $\mu_2 \neq 0$ only exacerbates the messiness of the calculations. Simulation might be the best option to check whether the conjectured bound is reasonable.

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