This is related to my previous question (link), but I have a simpler way of expressing it.

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

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

$ P(|Z_1|\ge c) + P(|Z_1|<c , |Z_2| \ge c ) = 1 - P(|Z_1|<c , |Z_2| < c ) = \alpha $

My solution thus far utilizes $ U_1 = Z_1+Z_2, U_2 = Z_1 - Z_2$

$Var(U_1) = 2+2\rho$ , $Var(U_2) = 2-2\rho$ , $Cov(U_1,U_2)=0$

This simplifies the equality to: $=1 - 4*P(0<U_1<c\sqrt2 , 0<U_2 < -U_1 + c\sqrt2 ) = \alpha$

$$P(0<U_1<c\sqrt2 , 0<U_2 < -U_1 + c\sqrt2 ) = \frac{1-\alpha}{4} $$

NOW, all I need to do is solve $P(0<U_1<c\sqrt2 , 0<U_2 < -U_1 + c\sqrt2 )$ as an invertible function of $c$ so that I can solve for $c$ in terms of $\alpha$.

My attempts thus far have involved conditioning on $U_1$, but I haven't been able to push the answer through. Any thoughts??

NOTE: I have seen the discussion about the expected value of a normal CDF when the inside is a linear function, and the discussion about a finite integral of a standard normal CDF when the inside is just the random variable using integration by parts. I havent yet found a solution to the finite integral of a normal CDF when there is a linear function inside.

$############################ MY SOLUTION THUS FAR ############################$

$P(0<U_1<c\sqrt2 , 0<U_2 < -U_1 + c\sqrt2 )$ $$=\int_{0}^{c\sqrt{2}}P(0<U_2 < -u + c\sqrt{2} )f(u) du$$

$$=\int_{0}^{c\sqrt{2}}\left(\Phi\left( \frac{-u + c\sqrt{2}}{\sqrt{2-2\rho}}\right)-\Phi\left( 0\right)\right)f(u) du$$

Normalize U_1: $X = \frac{U_1}{\sqrt{2+2\rho}}$

$$=\int_{0}^{ \frac{c\sqrt{2}}{\sqrt{2+2\rho}} }\left(\Phi\left( \frac{-x\sqrt{2+2\rho} + c\sqrt{2}}{\sqrt{2-2\rho}}\right)-\frac{1}{2}\right)\phi(x) dx$$

$$=\int_{0}^{ \frac{c}{\sqrt{1+\rho}} }\Phi\left( \frac{-x\sqrt{1+\rho} + c}{\sqrt{1-\rho}}\right)\phi(x) dx-\frac{1}{2}\int_{0}^{ \frac{c}{\sqrt{1+\rho}} }\phi(x) dx$$

$$=\int_{0}^{ \frac{c}{\sqrt{1+\rho}} }\Phi\left( \frac{-x\sqrt{1+\rho} + c}{\sqrt{1-\rho}}\right)\phi(x) dx-\frac{1}{2}\Phi\left(\frac{c}{\sqrt{1+\rho}}\right) +\frac{1}{4}$$

So the only part of the problem stopping me here is the portion $\int_{0}^{ \frac{c}{\sqrt{1+\rho}} }\Phi\left( \frac{-x\sqrt{1+\rho} + c}{\sqrt{1-\rho}}\right)\phi(x) dx$, which is of the form $\int_{0}^{ w }\Phi\left( \frac{ax + b}{d}\right)\phi(x) dx$. It's stopping me dead in my tracks. I'm hoping that if I can solve it, the expression will boil down to a solution for c that is a function of both $\alpha$ and $\rho$ by way of some quadratic function of $\Phi\left(\frac{c}{\sqrt{1+\rho}}\right)$, similar to my solution when $\rho=0$.

Any thoughts? Even a different approach to conditioning on $U_1$ would be welcome.

Thanks for your help!

  • $\begingroup$ What exactly would it mean to "solve for $c$"? You ask to compute the bivariate Normal inverse CDF. There are no general closed-form solutions (except when $\rho\in \{-1,0,1\}$). Usually the calculations are done numerically. What, then, would constitute an acceptable answer to your question? At the end, what do you mean by "there is a linear function inside"? That does not seem connected to this problem at all. $\endgroup$ – whuber Jan 29 '15 at 22:55
  • $\begingroup$ @whuber - See my solution for when $\rho=0$ on my related question (stats.stackexchange.com/questions/127258/…). In that instance, it's easy to see the $c$ is a function of $\alpha$, $ c = \Phi^{-1} \left( \frac{1+\sqrt{1-\alpha}}{2} \right) $. I am hoping to find a closed form solution for when rho is NOT equal to 0. $\endgroup$ – Lewkrr Jan 29 '15 at 23:18
  • $\begingroup$ Also, my solution (when conditioning on $U_1$) dead ends at a place that I cannot solve for lack of calculus skills. If there is a closed form solution to $\int_{w}^{\infty}\Phi\left( \frac{ax+b}{d}\right)\phi(x) dx $ then I might have the answer. Would it help if I posted my work thus far?? $\endgroup$ – Lewkrr Jan 29 '15 at 23:24

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