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Suppose $(Z_1, Z_2)$ is the zero-mean bivariate normal distribution with covariance $\left( \begin{matrix} 1 & \rho; \\ \rho & 1\end{matrix} \right)$ with positive $\rho > 0$. What I want to know a valuable tighy lower bound of the probability $$\mathrm{P} \left( Z_1 > z_1, Z_2 < z_2\right)$$ where $z_1, z_2$ are some positive constants.

PS: There are various valuable lower bound for the tail probability $\mathrm{P} \left( Z_1 > z_1, Z_2 > z_2\right)$ with positive $\rho > 0$, see p495-p499 in Continuous Bivariate Distributions. Can we transform $\mathrm{P} \left( Z_1 > z_1, - Z_2 > -z_2\right)$ to $\mathrm{P} \left( Z_1 > z_1, Z_2 > z_2\right)$ keeping the correlation coefficient positive?

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  • $\begingroup$ Instead of $\rho>1$ (which can never happen), did you mean $\rho>0$? $\endgroup$
    – Iosif Pinelis
    Mar 24 at 17:38
  • $\begingroup$ @IosifPinelis . Yes $\rho > 0$! Thanks! $\endgroup$
    – 香结丁
    Mar 24 at 17:40
  • $\begingroup$ Also, "valuable tighy" should apparently be "valuable tight". Even so, what do you mean by that, specifically? $\endgroup$
    – Iosif Pinelis
    Mar 24 at 17:42
  • $\begingroup$ Thanks! It is a typo. “Tight” means I want the lower bound is relatively sharp. $\endgroup$
    – 香结丁
    Mar 24 at 23:25

1 Answer 1

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Let $h:=z_1$, $k:=z_2$, and $r:=\rho\ge0$. We want to lower-bound $P(Z_1>h,Z_2<k)$. Formula (2.11) in the paper by Willink (cited the book you linked) gives the following upper bound on $P(Z_1>h,Z_2>k)$:
$$P(Z_1>h,Z_2>k) \\ \le\Phi(-h)\Big[\Phi\Big(\frac{rh-k}{\sqrt{1-r^2}}\Big) +re^{(h^2-k^2)/2}\,\Phi\Big(\frac{rk-h}{\sqrt{1-r^2}}\Big)\Big] \tag{1}\label{1}$$ for $h>0$ and states certain optimality properties of this bound; here, as usual, $\Phi$ denotes the standard normal cdf.

Since $P(Z_1>h,Z_2<k)=1-\Phi(h)-P(Z_1>h,Z_2>k)=\Phi(-h)-P(Z_1>h,Z_2>k)$, immediately from \eqref{1} we get the following lower bound on $P(Z_1>h,Z_2<k)$:
$$P(Z_1>h,Z_2<k) \\ \ge\Phi(-h)\Big[\Phi\Big(\frac{k-rh}{\sqrt{1-r^2}}\Big) -re^{(h^2-k^2)/2}\,\Phi\Big(\frac{rk-h}{\sqrt{1-r^2}}\Big)\Big]. \tag{2}\label{2}$$ Inequality \eqref{2} will turn into the equality when $r=0$.

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  • $\begingroup$ Thank you for your response. However, the right-hand side of the inequality in equation (2) may be negative if $\Phi\left(\frac{k-r h}{\sqrt{1-r^2}}\right) < r e^{\left(h^2-k^2\right) / 2} \Phi\left(\frac{r k-h}{\sqrt{1-r^2}}\right)$. If this is the case, is there an alternative formula that we can use instead? $\endgroup$
    – 香结丁
    Mar 24 at 23:31
  • $\begingroup$ @香结丁 : If the lower bound in (2) on $P(Z_1>h,Z_2<k)$ is $\le0$, then of course this lower bound is trivial. But then the upper bound on $P(Z_1>h,Z_2>k)$ in (1) will be greater than the trivial upper bound $\Phi(-h)=P(Z_1>h)$ on $P(Z_1>h,Z_2>k)$. You seemed to say that the bounds like that in (1) are valuable. So, the bound in (2) is equally valuable. Clearly, one cannot get an upper bound on $P(Z_1>h,Z_2<k)$ that will be tight for all $h,k$ -- except for the tautological bound, the probability $P(Z_1>h,Z_2<k)$ itself. $\endgroup$
    – Iosif Pinelis
    Mar 26 at 0:53
  • $\begingroup$ I am good now! Thanks so much! I just checked this website messages. $\endgroup$
    – 香结丁
    Apr 3 at 20:51

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