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I have a ratio of two Gaussians, $$l(z) = \frac{p[z|+]}{p[z|-]}$$, where $$p[z|+]\sim N(\mu^+, \sigma)$$ and $$p[z|-]\sim N(\mu^-, \sigma)$$. I want to find the integral $$\int_{-\infty}^{\infty}l(z)dz$$. As far as I know, the ratio of normal distributions follows a Cauchy distribution. However, the textbook I'm reading claims that the integral can be expressed as $$\frac{1}{2}erfc(\frac{\mu^- - \mu^+}{2\sigma})$$, where $$erfc(x)=\frac{2}{\sqrt{\pi}}\int_{x}^{\infty}exp(-y^2)dy$$. I'm not sure how the z can be integrated out from the original integral, what kind of trick should I be using here?

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The ratio of two Gaussian random variable is Cauchy, https://en.wikipedia.org/wiki/Ratio_distribution but you are dealing with likelihood ratio (the ratio of two probability distribution function). If you substitute Gaussian pdf expressions in the ratio, you'll figure out that $l(z)$ will be another Gaussian as well and that's why you have $erf(.)$ as the integral.

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  • $\begingroup$ Right, that is my initial ideal as well. However, after substituting the Gaussian PDF, I get $$I(z)=\int_{-\infty}^{\infty}exp(\frac{(z-\mu_-)^2-(z-\mu_+)^2}{2\sigma^2})dz$$. The numerator in the exponent can be reduced to $$(\mu_-^2-\mu_+^2)-2z(\mu_- - \mu_+)$$, I suppose some kind of completing the square is needed? But I couldn't think of a proper way to get that numerator into a square.. $\endgroup$
    – Asy
    Oct 11, 2017 at 23:52
  • $\begingroup$ This integral is not bounded! Which means that those pdfs were not Gaussian! You should probably need to make sure what kind of pdf you are dealing with. More over, we write $z|+ \sim \mathcal{N}(\cdot)$ in contrast with $p(z|+) = \mathcal{N}(\cdot)$. $\endgroup$ Oct 12, 2017 at 20:19

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