CDF of measurement vector with correlated noises Given the following definitions:
$X \sim \mathcal{N}(\bar{x}, \sigma_{0}^{2})$, and $W_i \sim \mathcal{N}(0, \sigma^{2})$, $i \in \{1,2\}$ and $E[W_1W_2]=\rho \sigma^2$. $X$ and $W_i$ are independent.
Two measurements $Z_1$ and $Z_2$ are performed
$$Z_i = X + W_i$$
What is the distribution of vector RV $Z = [Z_1 \ Z_2]^{T}$?
I tried to get the distribution by calculating CDF of $Z$
$$F_Z(z_1,z_2)=P\{Z_1 \leq z_1,Z_2 \leq z_2\} = P\{Z_1 \leq z_1 | Z_2 \leq z_2\} P\{Z_2 \leq z_2\}$$
I know how to calculate $P\{Z_2 \leq z_2\}$
$$Z_2 \sim \mathcal{N}(\bar{x}, \sigma_{0}^{2} + \sigma^{2})$$
I don't know how to calculate conditional probability $P\{Z_1 \leq z_1 | Z_2 \leq z_2\}$.
Is there different approach to find out how $Z$ is distributed?
 A: It seems you assume $X$ is independent of $W_i$, and $W_i$ are jointly normal. Following this fact, $Z_i$ become jointly normal, and you can use the conditional distribution formula to find the joint PDF of $Z_i$. A similar approach for finding the joint PDF would be directly calculating the mean and covariance vector for $[Z_1, Z_2]$ ($Z$ is jointly normal because it's a linear transform of the random vector $[X,W_1,W_2]$). However, multivariate normal CDF has no closed form.
A: Addition to the previous answer. You can approximate a bivariate normal distribution with an arbitrary mean vector ${\bf \mu}$ and covariance matrix $\Sigma$ by the eigen-vector transformation of the covariance matrix (see Appendix A.2 in (n-dimensional normal distribution). This yields an n-dimensional normal distribution with ${\bf \mu}\prime = (0,\ldots,0)^T$ and the covariance matrix $\Sigma \prime =I$ $\; \; -$ the identity matrix.
Using a numerical approximation formula like the bivariate normal integral, you can compute probabilities in the bivariate normal distribution.
