Let $(Z_1,Z_2)^T$ be Gaussian random vector with mean 0 and covariance matrix $\Sigma = (1,a;a,1)$ (variance =1 and covariance = a). I would like to prove that function $$\mu_2\mapsto\frac{\mathbb{P}(Z_1>c+\mu_1, Z_2>-\mu_2)}{\mathbb{P}(Z_1>\mu_1, Z_2>-\mu_2)}$$ is increasing in $\mu_2$ over $(0,\infty)$. The constants $\mu_1$ and $c$ are positive ($c\approx 2$, but I do not think it the value is relevant).
Note that this probability is originally a conditional one, but this is a simpler one, I guess.
I had some thoughts to apply a logarithm on the fraction and then derive in order to get something similar to the inverse of Mill's ratio, but never reached an end with it.
Some other thoughts were to look at the quotient of the derivatives and prove that it is increasing in $\mu_2$ (the rule is from https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5424000/), but I could not get anywhere.
I would be very grateful for any proposals.
For anyone who is interested, I was able to make the following proof (sorry for the difference in notations). Note that the references \cite{Pinelis} = L'hospital type rules for monotonicity with applications. And \cite{Sampford} = Some inequalities on mill's ratio and related functions. We show that function $$\nu_1\mapsto h_1(\nu_1) = \frac{\mathbb{P}_{\nu_1<0,\nu_2>0}\left(Z_1>-\nu_1+\sqrt{q_1},Z_2>-\nu_2\right)}{\mathbb{P}_{\nu_1<0,\nu_2>0}\left(Z_1>-\nu_1, Z_2>-\nu_2\right)}$$ , and we show that function $$\nu_2\mapsto h_2(\nu_2) = \frac{\mathbb{P}_{\nu_1<0,\nu_2>0}\left(Z_1>-\nu_1+\sqrt{q_1},Z_2>-\nu_2\right)}{\mathbb{P}_{\nu_1<0,\nu_2>0}\left(Z_1>-\nu_1, Z_2>-\nu_2\right)}$$ is also increasing.\ Let us start with $h_1$. \begin{align*} G(\nu_1) =& \mathbb{P}_{\nu_1<0,\nu_2>0}\left(Z_1>-\nu_1, Z_2>-\nu_2\right) \\ F(\nu_1) =& \mathbb{P}_{\nu_1<0,\nu_2>0}\left(Z_1>-\nu_1+\sqrt{q_1},Z_2>-\nu_2\right) \end{align*} where $Z_1$ and $Z_2$ are two Gaussian random variables $\mathcal{N}(0,1)$ such that Cov$(Z_1,Z_2)=-0.5$. We study the quotient $$\frac{F}{G}(\nu_1) = \frac{\mathbb{P}\left(Z_1>-\nu_1+\sqrt{q}, Z_2>-\nu_2\right)}{\mathbb{P}\left(Z_1>-\nu_1, Z_2>-\nu_2\right)} = \frac{\int_{-\nu_1}^{\infty}\int_{-\nu_2}^{\infty}{\varphi(z_1+\sqrt{q},z_2)dz_2dz_1}}{\int_{-\nu_1}^{\infty}\int_{-\nu_2}^{\infty}{\varphi(z_1,z_2)dz_2dz_1}}$$ We use Propsition 1.1 from \cite{Pinelis}. We study the quotient of the derivatives with respect to $\nu_1$ $$\frac{F'}{G'}(\nu_1) = \frac{\int_{-\nu_2}\varphi(-\nu_1+\sqrt{q},z_2)dz_2}{\int_{-\nu_2}\varphi(-\nu_1,z_2)dz_2}$$ where $$\varphi(x,y) = c\exp\left[-\frac{2}{3}\left(x^2+y^2+xy\right)\right].$$ We need to show that the quotient $F'/G'$ is increasing in $\nu_1$. We calculate the derivatives with respect to $\nu_1$. \begin{align*} F''(\nu_1) = & -\frac{2}{3}\int_{-\nu_2}\left(2\nu_1-2\sqrt{q}-z_2\right)\varphi(-\nu_1+\sqrt{q},z_2)dz_2 \\ G''(\nu_1) = & -\frac{2}{3}\int_{-\nu_2}\left(2\nu_1-z_2\right)\varphi(-\nu_1,z_2)dz_2 \end{align*} We need to calculate $F''G' - G''F'$. We have \begin{align*} F''G' = & -\frac{2}{3}\int_{-\nu_2}\left(2\nu_1-2\sqrt{q}-z_2\right)\varphi(-\nu_1+\sqrt{q},z_2)dz_2\int_{-\nu_2}\varphi(-\nu_1,x_2)dx_2 \\ G''F' & = -\frac{2}{3}\int_{-\nu_2}\left(2\nu_1-x_2\right)\varphi(-\nu_1,x_2)dx_2 \int_{-\nu_2}\varphi(-\nu_1+\sqrt{q},z_2)dz_2 \end{align*} Thus, $$F''G' - G''F' = -\frac{2}{3}\int_{-\nu_2}\int_{-\nu_2}(x_2-z_2-2\sqrt{q})\varphi(-\nu_1+\sqrt{q},z_2)\varphi(-\nu_1,x_2)dx_2dz_2$$ Note that \begin{align*} \varphi(-\nu_1+\sqrt{q},z_2) = & c\exp\left[-\frac{2}{3}\left((-\nu_1+\sqrt{q})^2+z_2^2+(-\nu_1+\sqrt{q})z_2\right)\right] \\ = & c_1\exp\left[-\frac{2}{3}\left(z_2+\frac{\sqrt{q}-\nu_1}{2}\right)^2\right] \\ \varphi(-\nu_1,x_2) = & c_2\exp\left[-\frac{2}{3}\left(x_2-\nu_1/2\right)^2\right] \end{align*} Thus \begin{equation} F''G' - G''F' = -c_1c_2\frac{2}{3}\int_{-\nu_2}\int_{-\nu_2}(x_2-z_2-2\sqrt{q})\exp\left[-\frac{2}{3}\left(z_2+\frac{\sqrt{q}-\nu_1}{2}\right)^2\right]\exp\left[-\frac{2}{3}\left(x_2-\nu_1/2\right)^2\right]dx_2dz_2 \label{eqn:FGGF} \end{equation} We will calculate explicitly these integrals. We integrate first with respect to $x_2$ \begin{multline*} \int_{-\nu_2}(x_2-z_2-2\sqrt{q})\exp\left[-\frac{2}{3}\left(x_2-\nu_1/2\right)^2\right]dx_2 = \\ \int_{-\nu_2}x_2\exp\left[-\frac{2}{3}\left(x_2-\nu_1/2\right)^2\right]dx_2 - (z_2+2\sqrt{q})\int_{-\nu_2}\exp\left[-\frac{2}{3}\left(x_2-\nu_1/2\right)^2\right]dx_2 \\ = A - B(z_2+2\sqrt{q}) \end{multline*} where \begin{align*} A = & \int_{-\nu_2}x_2\exp\left[-\frac{2}{3}\left(x_2-\nu_1/2\right)^2\right]dx_2 \\ = & \int_{-\nu_2-\nu_1/2}(x_2+\nu_1/2)e^{-2x_2^2/3}dx_2 \\ = & \frac{\nu_1}{2}\bar{F}(-\nu_2-\nu_1/2)+\frac{3}{4}e^{-2(\nu_2+\nu_1/2)^2/3} \\ B = & \int_{-\nu_2}\exp\left[-\frac{2}{3}\left(x_2-\nu_1/2\right)^2\right]dx_2 \\ = & \bar{F}(-\nu_2-\nu_1/2) \end{align*} where $\bar{F} = 1-F$ with $F$ the cdf of the Gaussian $\mathcal{N}(0,3/4)$. We integrate now with respect to $z_2$ \begin{align*} \int_{-\nu_2}\left[A - B(z_2+2\sqrt{q})\right]e^{\left[-\frac{2}{3}\left(z_2+\frac{\sqrt{q}-\nu_1}{2}\right)^2\right]}dz_2 = & (A-2\sqrt{q}B)\bar{F}\left(-\nu_2-\nu_1/2+\sqrt{q}/2\right) - \\ & B\int_{-\nu_2}z_2e^{\left[-\frac{2}{3}\left(z_2+\frac{\sqrt{q}-\nu_1}{2}\right)^2\right]}dz_2 \\ = & (A-2\sqrt{q}B)\bar{F}\left(-\nu_2-\nu_1/2+\sqrt{q}/2\right) - \\ & \frac{\sqrt{q}-\nu_1}{2}B\bar{F}\left(-\nu_2-\nu_1/2+\sqrt{q}/2\right) - \\ & \frac{3B}{4} e^{-\frac{2}{3}\left(-\nu_2-\nu_1/2+\sqrt{q}/2\right)^2} \\ = & \left[A-\left(\frac{3}{2}\sqrt{q}+\frac{\nu_1}{2}\right)B\right]\bar{F}\left(-\nu_2-\nu_1/2+\sqrt{q}/2\right) - \\ & \frac{3}{4}e^{-\frac{3}{2}\left(-\nu_2-\nu_1/2+\sqrt{q}/2\right)^2} \bar{F}\left(-\nu_2-\nu_1/2\right) \\ = & \left[\frac{3}{4}e^{-\frac{3}{2}\left(-\nu_2-\frac{\nu_1}{2}\right)^2}-\frac{3}{2}\sqrt{q}\bar{F}(-\nu_2-\frac{\nu_1}{2})\right]\bar{F}\left(-\nu_2-\frac{\nu_1}{2}+\frac{\sqrt{q}}{2}\right) \\ & -\frac{3}{4}e^{-\frac{3}{2}\left(-\nu_2-\frac{\nu_1}{2}+\frac{\sqrt{q}}{2}\right)^2} \bar{F}\left(-\nu_2-\frac{\nu_1}{2}\right) \end{align*} Find a common useful factor \begin{multline*} \int_{-\nu_2}\left[A - B(z_2+2\sqrt{q})\right]e^{\left[-\frac{3}{2}\left(z_2+\frac{\sqrt{q}-\nu_1}{2}\right)^2\right]}dz_2 = \bar{F}\left(-\nu_2-\frac{\nu_1}{2}\right)\bar{F}\left(-\nu_2-\frac{\nu_1}{2}+\frac{\sqrt{q}}{2}\right)\times \\ \left[\frac{3}{4}\frac{e^{-\frac{3}{2}\left(-\nu_2-\frac{\nu_1}{2}\right)^2}}{\bar{F}\left(-\nu_2-\frac{\nu_1}{2}\right)} - \frac{3}{4}\frac{e^{-\frac{3}{2}\left(-\nu_2-\frac{\nu_1}{2}+\frac{\sqrt{q}}{2}\right)^2}}{\bar{F}\left(-\nu_2-\frac{\nu_1}{2}+\frac{\sqrt{q}}{2}\right)} -\frac{3}{2}\sqrt{q} \right] \end{multline*} Note that the Inverse of Mill's ratio defined by $$R(x) = \frac{e^{-x^2/2}}{\int_{x}^{\infty}e^{-z^2/2}dz}$$ is increasing (\cite{Sampford}), thus $$\frac{e^{-\frac{3}{2}\left(-\nu_2-\frac{\nu_1}{2}\right)^2}}{\bar{F}\left(-\nu_2-\frac{\nu_1}{2}\right)} - \frac{e^{-\frac{3}{2}\left(-\nu_2-\frac{\nu_1}{2}+\frac{\sqrt{q}}{2}\right)^2}}{\bar{F}\left(-\nu_2-\frac{\nu_1}{2}+\frac{\sqrt{q}}{2}\right)} = R\left(\sqrt{3}(-\nu_2-\nu_1/2)\right) - R\left(\sqrt{3}(-\nu_2-\nu_1/2+\sqrt{q}/2)\right) < 0.$$ Thus $$\int_{-\nu_2}\left[A - B(z_2+2\sqrt{q})\right]e^{\left[-\frac{3}{2}\left(z_2+\frac{\sqrt{q}-\nu_1}{2}\right)^2\right]}dz_2 < 0$$ and hence $F''G' - G''F'>0$. This proves that function $$\nu_1\mapsto \frac{F'}{G'}(\nu_1)$$ is increasing. We may now use Proposition 1.1 from \cite{Pinelis} to deduce that $h_1(\nu_1) = F/G(\nu_1)$ is increasing in $\nu_1$. We move now to $h_2(\nu_2) = f/g(\nu_2)$. We have \begin{align*} f'(\nu_2) = &\int_{\nu_1}^{\infty}\varphi(z_1+\sqrt{q},-\nu_2)dz_1 \\ g'(\nu_2) = &\int_{\nu_1}^{\infty}\varphi(z_1,-\nu_2)dz_1 \end{align*} We calculate the second derivatives \begin{align*} f''(\nu_2) = &-\frac{2}{3}\int_{\nu_1}^{\infty}(2\nu_2-z_1-\sqrt{q})\varphi(z_1+\sqrt{q},-\nu_2)dz_1 \\ g''(\nu_2) = &-\frac{2}{3}\int_{\nu_1}^{\infty}(2\nu_2-z_1)\varphi(z_1,-\nu_2)dz_1 \end{align*} Thus $$f''g' - f'g''(\nu_2) = -\frac{2}{3}\int_{-\nu_1}\int_{-\nu_1}(x_1-z_1-\sqrt{q})\varphi(x_1,-\nu_2)\varphi(z_1+\sqrt{q},-\nu_2)dx_1dz_1$$ Notice that \begin{align*} \varphi(z_1+\sqrt{q},-\nu_2) & = c_3 \exp\left[-\frac{2}{3}\left(z_1-\nu_2/2\right)^2\right] \varphi(x_1,-\nu_2) & = c_4 \exp\left[-\frac{2}{3}\left(x_1+\sqrt{q}-\nu_2/2\right)^2\right] \end{align*} Therefore, $$f''g' - f'g''(\nu_2) = -\frac{2}{3}c_3c_4\int_{-\nu_1}\int_{-\nu_1}(x_1-z_1-\sqrt{q})\exp\left[-\frac{2}{3}\left(z_1-\nu_2/2\right)^2\right]\exp\left[-\frac{2}{3}\left(x_1+\sqrt{q}-\nu_2/2\right)^2\right]dx_1dz_1$$ This is the same formula as $F''G'-G''F'$, and the only difference is that $\sqrt{q}$ is replace here by $2\sqrt{q}$ and $\nu_1$ is replaced here by $\nu_2$. Thus, similar arguments to what we have shown here above proves that $f''g' - f'g''(\nu_2)>0$ and $f'/g'(\nu_2)$ is increasing. Finally $h_1(\nu_2) = f/g(\nu_2)$ becomes increasing in $\nu_2$ due to Proposition 1.1 from \cite{Pinelis}.