Define $f(x,y)=\frac{1}{2}(\phi(x,y,\rho)+\phi(x,y,-\rho))$ as a joint distribution

Question

Define $f(x,y)=\frac{1}{2}(\phi(x,y,\rho)+\phi(x,y,-\rho))$ as a joint distribution of $(X,Y)$, where $\phi(x,y,\rho)$ be the pdf of Bivariate Normal$(0,0,1,1, \rho)$ Find the marginal distributions of $X,Y$ and find $\rho_{X,Y}$ It is clear that the marginals of $f(x,y)$ would be $\phi(x)$ and $\phi(y)$ respectively where $\phi(x)$ is the pdf of standard normal distribution. But will $\rho_{X,Y}=\rho$? How will I show that?

Answer 1

$\newcommand{\Cov} {\operatorname{Cov}} \newcommand{\E} {\mathbb{E}} $By definition $$\rho_{X, Y} = \frac{\Cov(X, Y)} {\sigma_X\sigma_Y} $$ In this case $\sigma_X=\sigma_Y=1$. So we only need to find $\Cov(X, Y) $: \begin{align} \Cov(X, Y) & = \E[(X-\E[X]) (Y-\E[Y]) = \E[XY] \\ &=\iint_\mathbb{R^2} \frac{xy} {2}(\phi(x,y,\rho)+\phi(x,y,-\rho)) \, dx dy\\ &=\iint_\mathbb{R^2} \frac{xy} {2}\phi(x,y,\rho) \, dx dy+ \iint_\mathbb{R^2} \frac{xy} {2}\phi(x,y,-\rho) \, dx dy \end{align} You can now evaluate the integrals and get the answer. But there is a faster way than that. See the first integral as $\E[UV] /2$ and the second one as $\E[U'V']/2$ with $(U, V)\sim N(0,0,1,1, \rho)$ and $ (U', V') \sim N(0,0,1,1,-\rho)$. By using the formula $\Cov(U, V) =\E[UV] - \E[U] \E[V] $ and noting that $\E[U] =\E[V]=0 $ we get $\E[UV] = \Cov(U, V)$. We also know that \begin{align} \rho=\rho_{U, V} = \frac{\Cov(U, V)} {\sigma_U\sigma_V} =\Cov(U, V) \end{align} So $\E[UV] = \rho$ and using the same argument we get $\E[U'V'] =-\rho$. Combine these two things and get $$\rho_{X, Y} = \frac 1 2 \rho + \frac 1 2(-\rho) =0$$