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I need a probability distribution which is the ratio of two normal distributions or $P=(N_1/N_2)$. The mean of both normals can be assumed to be zero, and the variance is known.

Apparently the distribution I'm looking for is a Cauchy distribution, but I'm having trouble finding exactly which one. I assume that the $x_0$ value I'm looking for is $0$, but what $\gamma$ value am I looking for? in other words given $\sigma_1$ and $\sigma_2$ of two normal distributions what is the $\gamma$ of the distribution of their ratio?

spaceisdarkgreen
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user1492428
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  • The names $\gamma$ and $x_0$ are not standard for the scale and location parameter of the Cauchy distribution. You should either define them or give a reference to wikipedia. Also, please use mathjax for readability. Click on my edit to your question to see how. – spaceisdarkgreen Oct 12 '18 at 23:17

1 Answers1

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The formula for the PDF of the ratio $Z=X/Y$ is $$ f_Z(z) = \int_{-\infty}^\infty |y| f_{X,Y}(zy,y)dy.$$

I assume you mean for the normals to be independent of one another. In this case, the formula gives $$ \int_{-\infty}^\infty |y| f_{X,Y}(zy,y)dy.=\int_{-\infty}^\infty |y| \frac{1}{2\pi\sigma_1\sigma_2}e^{-\frac{z^2y^2}{2\sigma_1^2}}e^{-\frac{y^2}{2\sigma_2^2}}dy\\=\frac{1}{\pi\sigma_1\sigma_2}\int_0^\infty y e^{-\frac{y^2}{2}\left(\frac{z^2}{\sigma_1^2}+\frac{1}{\sigma_2^2}\right)}dy\\=\frac{1}{\pi\sigma_1\sigma_2}\frac{1}{\frac{z^2}{\sigma_1^2}+\frac{1}{\sigma_2^2}}\int_0^\infty ue^{-u^2/2}du\\=\frac{1}{\pi(\sigma_1/\sigma_2)}\frac{1}{\left(\frac{z}{(\sigma_1/\sigma_2)}\right)^2+1}$$

So we have (in the parametrization of Wikipedia), $\gamma = \sigma_1/\sigma_2.$

spaceisdarkgreen
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