$$d(P,Q)=\frac{\left(\mu_1-\mu_2\right)^2+2{\left(\sigma_1-\sigma_2\right)}^2}{\left(\mu_1-\mu_2\right)^2+2{\left(\sigma_1+\sigma_2\right)}^2}$$ How do you go about proving that the formula has the triangle inequality property? That is, for any normal distributions $$P\sim N(\mu_1,\sigma_1^2)$$ $$Q\sim N(\mu_2,\sigma_2^2)$$ $$R\sim N(\mu_3,\sigma_3^2)$$ have$$d(P,R)\leq d(P,Q)+d(Q,R).$$
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$\begingroup$ Please provide a detailed proof of the process, thank you! $\endgroup$– YuklamCommented Oct 6 at 15:47
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$\begingroup$ Wondering where this proposed metric came from, I see that $$(\mu_1-\mu_2)^2 + 2(\sigma_1^2 + \sigma_2^2)$$ is $4$ times the variance of the average of two Gaussians. But you have $$(\mu_1-\mu_2)^2 + 2(\sigma_1^2 + 2\sigma_1\sigma_2 + \sigma_2^2).$$ $\endgroup$– Michael HardyCommented Oct 7 at 16:46
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$\begingroup$ $\qquad\uparrow\qquad$To be clear, I meant averaging the two probability distributions, not averaging the two random variables. $\endgroup$– Michael HardyCommented Oct 7 at 16:57
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$\begingroup$ @Yuklam : Do you have a response to the answer below? $\endgroup$– Iosif PinelisCommented Oct 9 at 1:45
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$\newcommand\si\sigma$This is not true. For instance, suppose that $\mu_1=\mu_2=\mu_3=0$ and
$(\si_1,\si_2,\si_3)=(1,2,6).$
Then $d(P, Q) + d(Q, R) - d(P, R)=-263/1764<0$.
answered Oct 6 at 16:07