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Let $X=(X_1,X_2)^T$ be a random vector with 2-dimensional normal distribution, $E(X_1)=E(X_2)=0 , \operatorname{Var}(X_1)=\operatorname{Var}(X_2)=1$ and $\operatorname{Cov}(X_1, X_2)= \nu$ with $|\nu| <1$. And let $Z \sim \mathrm{Bin}(1,\alpha)$ be independent from $(X_1,X_2)$ and $(U,V)^T := Z(X_1,X_2)^T+(1-Z)(-X_1,X_2)^T.$

I know that $(-X_1,X_2)$ has also normal distribution. Now I want to find the density function of $(U,V)$.

As a hint: I need to use the formula of total probability.

Futher I would like to find the marginal distribution of $U$ und $V$.

Can you help me, please?

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Hint: consider, from LOTP $$P(U=u, V=v) = P(U=u, V=v \lvert Z = 1) P(Z=1) + P(U=u, V=v \lvert Z = 0)P(Z=0)$$

Also remember that two-dimensional vector of two normals is: $$(X_1, X_2) \sim N( (\mu_1, \mu_2), \begin{pmatrix} \operatorname{Var} X_1 & \operatorname{Cov}(X_1, X_2) \\ \operatorname{Cov}(X_1, X_2) & \operatorname{Var}(X_2) \end{pmatrix} )$$, which means that: $$ p(x_1, x_2) \propto \exp( - (x_1 - \mu_1, x_2 - \mu_2)^T \Sigma^{-1} (x_2 - \mu_1, x_2 - \mu_2)) $$ where $\Sigma$ is covariance matrix from second equation. Overall, $\mu_i$ mean respective expected values. In your case these are going to be zero.

Then integrate to find marginal distributions of $U, V$:

$$p(u) = \int p(u, v) dv$$

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