Covariance in terms of independent random variables If I understood everything in statistics correctly, the covariance of two random variables should be given by
$$\mathbb{E}[\phi(X_1)\phi(X_2)] = \iint \phi(x_1)\phi(x_2) p(x_1, x_2) \mathrm{d}x_1 \mathrm{d}x_2,$$
where $\phi$ is some function, $X_1$ and $X_2$ are some random variables and $p$ is the probability density function of the joint distribution.
I recently found this paper, where a claim is made (in equation 5) that comes down to the fact that one could also write
$$\mathbb{E}[\phi(X_1)\phi(X_2)] = \iint \phi\big(x_1\big)\phi\left(c x_1 + \sqrt{1 - c^2} x_2\right) p(x_1) p(x_2) \mathrm{d}x_1 \mathrm{d}x_2,$$
with $c$ the correlation coefficient, in the specific setting of the paper (I believe $X_1$ and $X_2$ come from the same Gaussian distribution) if I am not mistaken. I have no idea how to get to this result, however.
Did I miss something or could anybody enlighten me on how to get to this second formulation and what assumptions are necessary to get there?
 A: Unfortunately your summary of the problem does not accurately reflect the relevant part of the paper.  In particular, you use the same random variables in your two equations, but in the paper they are different.  Effectively what is occurring in the integral in $(5)$ in the paper is that the authors are finding the expected value $\mathbb{E}[\phi(U_1) \phi(U_2)]$ where the normal random vector $\mathbf{U}$ is defined as a linear transform of an underlying bivariate standard normal random vector:
$$\mathbf{Z} = \begin{bmatrix} Z_1 \\ Z_2 \end{bmatrix} \sim \text{N}(\boldsymbol{0}, \boldsymbol{I}).$$
The authors use the following transformation:
$$\mathbf{U} 
= \begin{bmatrix} U_1 \\ U_2 \end{bmatrix} = \mathbf{t} \mathbf{Z}
\quad \quad \quad \quad \mathbf{t} = \begin{bmatrix} \sqrt{q_{11}} & 0 \\ \sqrt{q_{22}} c_{12} & \sqrt{q_{22}} \sqrt{1-c_{12}^2} \end{bmatrix} . $$
Applying the rules for linear transformations of normal random vectors we see that the transformed random vector is distributed as:
$$\mathbf{U} 
 \sim \text{N}(\boldsymbol{0}, \mathbf{t} \mathbf{t}^\text{T}) \quad 
\quad \quad \quad \mathbf{t} \mathbf{t}^\text{T} = \begin{bmatrix} q_{11} & \sqrt{q_{11} q_{22}} c_{12} \\ \sqrt{q_{11} q_{22}} c_{12} & q_{22} \end{bmatrix}.$$
As can be seen from the variance matrix for this new random vector, the values $q_{11}$ and $q_{22}$ are the variances of $U_1$ and $U_2$ respectively, and $c_{12}$ is the correlation between them.  So effectively what the authors are doing is to write out the expectation of a product of functions of this general normal random vector in terms of an underlying standard normal random vector.  That is, they are effectively saying that:
$$\begin{equation} \begin{aligned}
\mathbb{E}[\phi(U_1) \phi(U_2)] 
&= \int \phi(u_1) \phi(u_2) p(u_1, u_2) du_1 du_2 \\[6pt]
&= \int \phi(\sqrt{q_{11}} z_1) \phi(\sqrt{q_{22}} c_{12} z_1 + \sqrt{q_{22}} \sqrt{1-c_{12}^2} z_2) p(z_1, z_2) dz_1 dz_2, \\[6pt]
\end{aligned} \end{equation}$$
where in the latter integral we use the joint density $p(z_1, z_2) = \text{N}(z_1|0,1) \text{N}(z_2|0,1)$ for standard normal values.  All that is going on in this part of the paper is that the authors are re-expressing an integral with respect to a general normal random vector in terms of an underlying standard normal random vector.
