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If $f_1$ is the pdf of a $N(\mu,\sigma^2)$ and $f_2$ is the pdf of a $N(\nu,\tau^2)$, can I compute the Chi Squared divergence between $f_1$ and $f_2$ in closed form (as a function of the parameters $\mu, \nu, \sigma^2,\tau^2$)? I've done it with the Kullback Leibler divergence without spending much time, but I can't do the same with the chi-squared.

@John: I've tried to integrate directly $\int \frac{f_1^2}{f_2}dx-1$ or equivalently $\int \frac{{(f_1-f_2)}^2}{f_2}dx$ or $\int f_2 (\frac{f_1}{f_2}-1) dx$. All the problem is reconducible to integrate $\int e^{-(ax^2+bx+c)}dx$ for some $a,b,c$ but this integral does not converge for negative values of $a$. In my calculations, the coefficent $a$ is something like the difference of two variances, so it could be negative.

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What have you tried? –  John Nov 6 '12 at 17:23
    
Hi have you got an answer for your problem? I am also trying to do the same. If you have got an answer please post it. –  user570593 Mar 18 '13 at 15:19

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