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I am struggling to calculate the following linmit

$$ \lim_{\gamma_- ~\to ~-1} \Phi\left(\frac{-\gamma~ \Phi ^{-1}(\alpha) - \Phi ^{-1}(\beta)} {\sqrt{1- \gamma^2}} \right)$$

where $\alpha,\beta \in [0,1]$ and $\phi$ is the normal CDF. I am particularlly interested in the case where $1/2< \alpha, \beta < 1$ even more when $\alpha, \beta \approx 1$.

In the first case we know that $\Phi ^{-1}(\alpha) , \Phi ^{-1}(\beta) >0$ so the limit would be $-\infty$ by my reasoning.

Although I am not sure about it even less in the second case.

Could someone give me a hand with that please?

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  • $\begingroup$ Note that $0\leq \Phi(x)\leq 1$. $\endgroup$ – Math-fun Jul 2 '15 at 12:54
  • $\begingroup$ an example: $\alpha=\beta=\frac34$ then the limit will be $\frac12$. $\endgroup$ – Math-fun Jul 2 '15 at 12:59
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Here is some heuristical reasoning $$\frac{-\gamma~ \Phi ^{-1}(\alpha) - \Phi ^{-1}(\beta)} {\sqrt{1- \gamma^2}} = \frac{-\gamma~ \Phi ^{-1}(\alpha) - \Phi ^{-1}(\alpha)+ \Phi ^{-1}(\alpha)- \Phi ^{-1}(\beta)} {\sqrt{1- \gamma^2}} = \frac{-(\gamma+1) \Phi ^{-1}(\alpha) + \Phi ^{-1}(\alpha)- \Phi ^{-1}(\beta)} {\sqrt{1- \gamma^2}} = -\sqrt{\frac{1+\gamma}{1-\gamma}} \Phi ^{-1}(\alpha) + \frac{\Phi ^{-1}(\alpha)- \Phi ^{-1}(\beta)} {\sqrt{1- \gamma^2}} =: f_1(\gamma)+f_2(\gamma) $$ We always have $\lim_{\gamma ~\to ~-1} f_1(\gamma)=0$ and if $\alpha=\beta$ then $f_2(\gamma)=0$ and the result is $\Phi(0)=\frac{1}{2}$. If $a\ne b$ we get $f_2(\gamma) \rightarrow \pm \infty$ and the result will be $\Phi(-\infty)=0$ for $b>a$ and $\Phi(+\infty)=1$ for $b<a$.

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