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I stumbled upon this identity which should be "really" straightforward:

We are given that: \begin{equation} A(y) = \frac{\mu_1 - \mu_2 }{\nu_{1} \sqrt{1- \rho^2}} + \frac{(\rho \nu_1 -\nu_2 ) }{\nu_{1} \sqrt{1- \rho^2}} y \equiv a + by \end{equation} This means A(y) is a linear function of y.

OK, so I want to prove:

\begin{equation} \int_{\infty}^{\infty} N \left( A(y+ \rho \nu_1 ) + \sqrt{1-\rho^2}\nu_1 \right) n(y) dy = \int_{\infty}^{\infty} N \left( A(y) \right) n(y; -\rho \nu_1 , 1)dy \end{equation}

where $N(\cdot)$ is the normal cdf and $n(\cdot)$ is the normal pdf. $n(y; -\rho \nu_1 , 1)$ is a normal pdf, with mean $ -\rho \nu_1 $ and variance $1$. All the greek letters are constants and $\rho$ is a correlation coefficient.

Thank you!

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