# Find the density function of $(U,V)$

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?

## 1 Answer

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 $$U, V$$:

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