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Normal Ratio Distribution


GaussianRatioDistribution

The ratio X/Y of independent normally distributed variates with zero mean is distributed with a Cauchy distribution. This can be seen as follows. Let X and Y both have mean 0 and standard deviations of sigma_x and sigma_y, respectively, then the joint probability density function is the bivariate normal distribution with rho=0,

 f(x,y)=1/(2pisigma_xsigma_y)e^(-[x^2/(2sigma_x^2)+y^2/(2sigma_y^2)]).
(1)

From ratio distribution, the distribution of U=X/Y is

P(u)=int_(-infty)^infty|y|f(uy,y)dy
(2)
=1/(2pisigma_xsigma_y)int_(-infty)^infty|y|e^(-[y^2/(2sigma_y^2)+u^2y^2/(2sigma_x^2)])dy
(3)
=1/(pisigma_xsigma_y)int_0^inftyyexp[-y^2(1/(2sigma_y^2)+(u^2)/(2sigma_x^2))]dy.
(4)

But

 int_0^inftyye^(-ay^2)dy=1/(2a),
(5)

so

P(u)=1/(pisigma_xsigma_y)1/(2(1/(2sigma_y^2)+(u^2)/(2sigma_x^2)))
(6)
=1/pi(sigma_xsigma_y)/(u^2sigma_y^2+sigma_x^2)
(7)
=1/pi((sigma_x)/(sigma_y))/(u^2+((sigma_x)/(sigma_y))^2),
(8)

which is a Cauchy distribution.

A more direct derivative proceeds from integration of

P_(X/Y)(u)=int_(-infty)^inftyint_(-infty)^infty(e^(-x^2/(2sigma_x^2)))/(sigma_xsqrt(2pi))(e^(-y^2/(2sigma_y^2)))/(sigma_ysqrt(2pi))delta(x/y-u)dxdy
(9)
=1/pi((sigma_x)/(sigma_y))/(u^2+((sigma_x)/(sigma_y))^2),
(10)

where delta(x) is a delta function.


See also

Cauchy Distribution, Normal Difference Distribution, Normal Distribution, Normal Product Distribution, Normal Sum Distribution

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Cite this as:

Weisstein, Eric W. "Normal Ratio Distribution." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/NormalRatioDistribution.html

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