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Kullback in his "Information theory and statistics" gives the symmetrized divergence as follows

$J(1,2)=\iint(f(x,y)-g(x)h(y))log{\frac{f(x,y)}{g(x)h(y)}}$

Later (p.8), he states that symmetrized divergence between a bivariate normal distribution $f(x,y)=N(\boldsymbol{\mu,\boldsymbol{\Sigma}})$ and two univariate normal distributions$g(x),h(y)=N(0,\sigma)$. (i.e. means for all distributions are zero)

is $J(1,2)=\frac{\rho^2}{1-\rho^2}$.

I proved that the divergence $I(f(x,y):g(x),h(y))=-\frac{1}{2} log{(1-\rho^2)}$ and I can see that $I(g(x),h(y):f(x,y))$ has to be $\frac{1}{2} log{(1-\rho^2)}+\frac{\rho^2}{1-\rho^2}$ but I can't prove the second part.

$I(g(x),h(y):f(x,y))=\frac{1}{2} log{(1-\rho^2)}+\iint{\{g(x)h(y)-\frac{1}{2}[\frac{x^2}{\sigma_x}+\frac{y^2}{\sigma_y}]+\frac{1}{2(1-\rho^2)}[\frac{x^2}{\sigma_x}+\frac{y^2}{\sigma_y}-2\rho\frac{xy}{\sigma_x\sigma_y}]\}dxdy}$

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