Exchangeability of a Joint PDF

Question

I'm wondering why the exchangeability of the bivariate normal pdf, allows me to immediately write down the distribution of Y2, having found that of Y1.

Answer 1

Rewrite the joint density as $$ f_{Y_1,Y_2}(y_1,y_2)=g(z_1;z_2;\rho),\qquad z_k=(y_k,\mu_k,\sigma_k^2), $$ for some function $g$ symmetric with respect to $(z_1,z_2)$, that is, such that, for every $(z_1,z_2)$, $$ g(z_1;z_2;\rho)=g(z_2;z_1;\rho). $$ Then, $f_{Y_1}(y_1)=h(z_1)$ where $z_1=(y_1,\mu_1,\sigma_1^2)$, the function $h$ does not depend on $(y_2,\mu_2,\sigma_2^2,\rho)$ and $$ h(z_1)=\int g(z_1;y_2,\mu_2,\sigma_2^2;\rho)\,\mathrm dy_2. $$ (The fact that $h$ does not depend on $\rho$ can be seen on the expression of $f_{Y_1}$ but it is not crucial to the argument below.) Likewise, by definition, $$ f_{Y_2}(y_2)=\int f_{Y_1,Y_2}(y_1,y_2)\,\mathrm dy_1, $$ hence $$ f_{Y_2}(y_2)=\int g(z_1;z_2;\rho)\,\mathrm dy_1=\int g(z_2;z_1;\rho)\,\mathrm dy_1=\int g(z_2;y_1,\mu_1,\sigma_1^2;\rho)\,\mathrm dy_1=h(z_2), $$ where the first identity comes from the definition of $f_{Y_2}$ as the second marginal of $f_{Y_1,Y_2}$ and the definition of $g$, the second identity comes from the symmetry of $g$, the third identity comes from the definition of $z_1$, and the fourth identity comes from the definition of $h$. To sum up, if $$ f_{Y_1}(y_1)=h(y_1,\mu_1,\sigma_1^2), $$ then $$ f_{Y_2}(y_2)=h(y_2,\mu_2,\sigma_2^2), $$ which is the symmetry you noticed.