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Let $Z_1$ and $Z_2$ follow $ N(\delta, \alpha)$ and covariance between them is $\beta$. How to calculate $P(Z_1>0, Z_2>0)?$

Satya Prakash
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    Your subject line says "bivariate normal" but your posted question says "Let $Z_1$ and $Z_2$ follow $ N(\delta, \alpha)$ and covariance between them is $\beta$." To say "Let $Z_1$ and $Z_2$ follow $ N(\delta, \alpha)$ and covariance between them is $\beta$." does not imply that the pair $(Z_1,Z_2)$ has a bivariate normal distribution. It wouldn't be a bad idea to say that explicitly in the question. – Michael Hardy Aug 29 '17 at 21:26
  • Sure you are not in fact interested in the case $\delta=0$? – Did Aug 29 '17 at 22:07
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    And the lack of context is chilling, especially in view of the fact that you seem to be asking again and again variants of the same question about joint normal distributions, to which the answer is always "No explicit formula in general". Sorry but I do not know where all this is going... – Did Aug 29 '17 at 22:13

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The correlation between $Z_1$ and $Z_2$ will be $\beta /(\sigma_1\sigma_2) = \beta/\alpha^2$. Then the probability density function is $$ f(z_1,z_2) = \frac1{2\pi\alpha^2\sqrt{1-\beta/\alpha^2}} e^{-\frac1{2(1-\beta/\alpha^2)}}\left( \frac{(z_1-\delta)^2+(z_2-\delta)^2-2\beta/\alpha^2(z_1-\delta)(z2-\delta)}{\alpha^2}\right) $$ The probability wanted is then $$ \int_{z_1=0}^\infty \int_{z_2=0}^\infty f(z_1,z_2) dz_2\,dz_1 $$ Now for fixed $z_1$ the $z_2$ integral will give an erf result (in fact, it would even if the covariance were zero, because of the non-zero $\delta$) so the answer yo seek is an integral of an erf. Although this is quite tractable numerically, I know of no familiar function that represents it in closed form.

Mark Fischler
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  • Thanks. So it is clear that we will not have an explicit form. If $\alpha = \frac{1+\gamma}{\gamma(1-\gamma)}$ and $\beta = 1/\gamma$ for $0<\gamma<1$. My main interest was to know, for what value of $\gamma$, $P(Z_1>0, Z_2>0)$ maximized?. I did simulations and it looks that it is true for $\gamma = 1/2$. I am trying to prove it using Slepian's inequality which is based on stochastic ordering. Please, any further thoughts.. – Satya Prakash Aug 29 '17 at 22:30