Ratio distribution

Template:Short description A ratio distribution (also known as a quotient distribution) is a probability distribution constructed as the distribution of the ratio of random variables having two other known distributions. Given two (usually independent) random variables Template:Mvar and Template:Mvar, the distribution of the random variable Template:Mvar that is formed as the ratio Template:Math is a ratio distribution.

An example is the Cauchy distribution (also called the normal ratio distribution), which comes about as the ratio of two normally distributed variables with zero mean. Two other distributions often used in test-statistics are also ratio distributions: the Template:Student's t-distribution arises from a Gaussian random variable divided by an independent chi-distributed random variable, while the F-distribution originates from the ratio of two independent chi-squared distributed random variables. More general ratio distributions have been considered in the literature.Template:SfnpTemplate:SfnpTemplate:SfnpTemplate:SfnpTemplate:SfnpTemplate:Sfnp

Often the ratio distributions are heavy-tailed, and it may be difficult to work with such distributions and develop an associated statistical test. A method based on the median has been suggested as a "work-around".Template:Sfnp

Algebra of random variables

Template:Main The ratio is one type of algebra for random variables: Related to the ratio distribution are the product distribution, sum distribution and difference distribution. More generally, one may talk of combinations of sums, differences, products and ratios. Many of these distributions are described in Melvin D. Springer's book from 1979 The Algebra of Random Variables.Template:Sfnp

The algebraic rules known with ordinary numbers do not apply for the algebra of random variables. For example, if a product is Template:Tmath and a ratio is Template:Tmath it does not necessarily mean that the distributions of Template:Tmath and Template:Tmath are the same. Indeed, a peculiar effect is seen for the Cauchy distribution: The product and the ratio of two independent Cauchy distributions (with the same scale parameter and the location parameter set to zero) will give the same distribution.Template:Sfnp This becomes evident when regarding the Cauchy distribution as itself a ratio distribution of two Gaussian distributions of zero means: Consider two Cauchy random variables, Template:Tmath and Template:Tmath, each constructed from two Gaussian distributions Template:Tmath and Template:Tmath. Then $$\frac{C_1}{C_2}=\frac{G_1/G_2}{G_3/G_4}=\frac{G_1{G}_4}{G_2{G}_3}=\frac{G_1}{G_2}\times \frac{G_4}{G_3}=C_1\times C_3,$$ where Template:Tmath. The first term is the ratio of two Cauchy distributions while the last term is the product of two such distributions.

Derivation

A way of deriving the ratio distribution of Template:Tmath from the joint distribution of the two other random variables Template:Tmath, Template:Tmath, with joint Template:Abbr Template:Tmath, is by integration of the following formTemplate:Sfnp $$p_Z(z)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}|y|p_{X,Y}(zy,y)dy.$$

If the two variables are independent then $p_{XY}(x,y)=p_X(x)p_Y(y)$ and this becomes $$p_Z(z)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}|y|p_X(zy)p_Y(y)dy.$$

This may not be straightforward. By way of example take the classical problem of the ratio of two standard Gaussian samples. The joint Template:Abbr is $$p_{X,Y}(x,y)=\frac{1}{2\pi }\exp (-\frac{x^2}{2})\exp (-\frac{y^2}{2}).$$

Defining $Z=X/Y$ we have $$\begin{array}{cc}{p}_Z(z)& =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|y|\exp (-\frac{{\left(zy\right)}^2}{2})\exp (-\frac{y^2}{2})dy\\ & =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|y|\exp (-\frac{y^2(z^2+1)}{2})dy.\end{array}$$ Using the known definite integral ${\int }_0^{\mathrm{\infty }}x\exp (-c{x}^2)dx=\frac{1}{2c}$ we get $$p_Z(z)=\frac{1}{\pi (z^2+1)},$$ which is the Cauchy distribution, or Template:Student's t-distribution with Template:Tmath.

The Mellin transform has also been suggested for derivation of ratio distributions.Template:Sfnp

In the case of positive independent variables, proceed as follows. The diagram shows a separable bivariate distribution $f_{x,y}(x,y)=f_x(x)f_y(y)$ which has support in the positive quadrant $x,y>0$ and we wish to find the pdf of the ratio Template:Tmath. The hatched volume above the line Template:Tmath represents the cumulative distribution of the function Template:Tmath multiplied with the logical function Template:Tmath. The density is first integrated in horizontal strips; the horizontal strip at height Template:Tmath extends from Template:Tmath to Template:Tmath and has incremental probability $f_y(y)dy{\int }_0^{Ry}{f}_x(x)dx$.

Secondly, integrating the horizontal strips upward over all Template:Tmath yields the volume of probability above the line $$F_R(R)={\int }_0^{\mathrm{\infty }}{f}_y(y)({\int }_0^{Ry}{f}_x(x)dx)dy$$ Finally, differentiate $F_R(R)$ with respect to $R$ to get the Template:Abbr Template:Tmath. $$f_R(R)=\frac{d}{dR}[{\int }_0^{\mathrm{\infty }}{f}_y(y)({\int }_0^{Ry}{f}_x(x)dx)dy]$$ Move the differentiation inside the integral: $$f_R(R)={\int }_0^{\mathrm{\infty }}{f}_y(y)(\frac{d}{dR}{\int }_0^{Ry}{f}_x(x)dx)dy$$ and since $$\frac{d}{dR}{\int }_0^{Ry}{f}_x(x)dx=y{f}_x(Ry)$$ then $$f_R(R)={\int }_0^{\mathrm{\infty }}{f}_y(y)f_x(Ry)ydy.$$ As an example, find the Template:Abbr of the ratio Template:Tmath when $$\begin{array}{cc}{f}_x(x)& =\alpha e^{-\alpha x},\\ f_y(y)& =\beta e^{-\beta y},\\ x,y& \ge 0.\end{array}$$

Evaluating the cumulative distribution of a ratio

We have $${\int }_0^{Ry}{f}_x(x)dx=-{e^{-\alpha x}|}_0^{Ry}=1-e^{-\alpha Ry}$$ thus $$\begin{array}{cc}{F}_R(R)& ={\int }_0^{\mathrm{\infty }}{f}_y(y)(1-e^{-\alpha Ry})dy\\ & ={\int }_0^{\mathrm{\infty }}\beta e^{-\beta y}(1-e^{-\alpha Ry})dy\\ & =1-\frac{\alpha R}{\beta +\alpha R}\\ & =\frac{R}{\frac{\beta }{\alpha }+R}.\end{array}$$ Differentiation with respect to Template:Tmath yields the Template:Abbr of Template:Tmath: $$f_R(R)=\frac{d}{dR}\left(\frac{R}{\frac{\beta }{\alpha }+R}\right)=\frac{\frac{\beta }{\alpha }}{{(\frac{\beta }{\alpha }+R)}^2}.$$

Moments of random ratios

From Mellin transform theory, for distributions existing only on the positive half-line Template:Tmath, we have the product identity $$\mathrm{E}[(UV{)}^p]=\mathrm{E}[U^p]\cdot \mathrm{E}[V^p]$$ provided Template:Tmath, Template:Tmath are independent. For the case of a ratio of samples like Template:Tmath, in order to make use of this identity it is necessary to use moments of the inverse distribution. Set $1/Y=Z$ such that Template:Tmath. Thus, if the moments of $X^p$ and $Y^{-p}$ can be determined separately, then the moments of $X/Y$ can be found. The moments of $Y^{-p}$ are determined from the inverse Template:Abbr of Template:Tmath, often a tractable exercise. At simplest, Template:Tmath.

To illustrate, let $X$ be sampled from a standard Gamma distribution $$x^{\alpha -1}{e}^{-x}/\mathrm{\Gamma }(\alpha )$$ whose Template:Tmath-th moment is Template:Tmath.

$Z=Y^{-1}$ is sampled from an inverse Gamma distribution with parameter $\beta$ and has Template:Abbr Template:Tmath. The moments of this PDF are $$\mathrm{E}[Z^p]=\mathrm{E}[Y^{-p}]=\frac{\mathrm{\Gamma }(\beta -p)}{\mathrm{\Gamma }(\beta )},p<\beta .$$

Multiplying the corresponding moments gives $$\mathrm{E}[(X/Y{)}^p]=\mathrm{E}[X^p]\cdot \mathrm{E}[Y^{-p}]=\frac{\mathrm{\Gamma }(\alpha +p)}{\mathrm{\Gamma }(\alpha )}\frac{\mathrm{\Gamma }(\beta -p)}{\mathrm{\Gamma }(\beta )},p<\beta .$$

Independently, it is known that the ratio of the two Gamma samples $R=X/Y$ follows the beta prime distribution: $$f_{{\beta }^{\prime }}(r,\alpha ,\beta )=\mathrm{B}(\alpha ,\beta {)}^{-1}{r}^{\alpha -1}(1+r{)}^{-(\alpha +\beta )}$$ whose moments are $$\mathrm{E}[R^p]=\frac{\mathrm{B}(\alpha +p,\beta -p)}{\mathrm{B}(\alpha ,\beta )}.$$

Substituting $\mathrm{B}(\alpha ,\beta )=\frac{\mathrm{\Gamma }(\alpha )\mathrm{\Gamma }(\beta )}{\mathrm{\Gamma }(\alpha +\beta )}$ we have $$\begin{array}{cc}\mathrm{E}[R^p]& =\frac{\mathrm{\Gamma }(\alpha +p)\mathrm{\Gamma }(\beta -p)}{\mathrm{\Gamma }(\alpha +\beta )}/\frac{\mathrm{\Gamma }(\alpha )\mathrm{\Gamma }(\beta )}{\mathrm{\Gamma }(\alpha +\beta )}\\ & =\frac{\mathrm{\Gamma }(\alpha +p)\mathrm{\Gamma }(\beta -p)}{\mathrm{\Gamma }(\alpha )\mathrm{\Gamma }(\beta )},\end{array}$$ which is consistent with the product of moments above.

Means and variances of random ratios

In the Product distribution section, and derived from Mellin transform theory (see section above), it is found that the mean of a product of independent variables is equal to the product of their means. In the case of ratios, we have $$\mathrm{E}(X/Y)=\mathrm{E}(X)\cdot \mathrm{E}(1/Y)$$ which, in terms of probability distributions, is equivalent to $$\mathrm{E}(X/Y)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x{f}_x(x)dx\times {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{y}^{-1}{f}_y(y)dy.$$

Note that Template:Tmath; i.e., Template:Tmath.

The variance of a ratio of independent variables is $$\begin{array}{cc}\mathrm{Var}(X/Y)& =\mathrm{E}([X/Y{]}^2)-{\mathrm{E}}^{\mathrm{2}}(X/Y)\\ & =\mathrm{E}(X^2)\cdot \mathrm{E}(1/Y^2)-{\mathrm{E}}^2(X)\cdot {\mathrm{E}}^2(1/Y).\end{array}$$

Normal ratio distributionsTemplate:Anchor

Uncorrelated central normal ratio

When Template:Tmath and Template:Tmath are independent and have a Gaussian distribution with zero mean, the form of their ratio distribution is a Cauchy distribution. This can be derived by setting Template:Tmath, then showing that Template:Tmath has circular symmetry. For a bivariate uncorrelated Gaussian distribution we have $$\begin{array}{cc}p(x,y)& =\frac{1}{\sqrt{2\pi }}\exp (-\frac{x^2}{2})\times \frac{1}{\sqrt{2\pi }}\exp (-\frac{y^2}{2})\\ & =\frac{1}{2\pi }\exp (-\frac{x^2+y^2}{2})\\ & =\frac{1}{2\pi }\exp (-\frac{r^2}{2}),\text{for}{r}^2=x^2+y^2.\end{array}$$

If Template:Tmath is a function only of Template:Tmath, then Template:Tmath is uniformly distributed on Template:Tmath with density Template:Tmath, so the problem reduces to finding the probability distribution of Template:Tmath under the mapping $$Z=X/Y=\tan \theta .$$

We have, by conservation of probability $$p_z(z)|dz|=p_{\theta }(\theta )|d\theta |,$$ and since Template:Tmath, $$p_z(z)=\frac{p_{\theta }(\theta )}{\left|\frac{dz}{d\theta }\right|}=\frac{1}{2\pi }{\cos }^2\theta .$$ Setting $${\cos }^2\theta =\frac{1}{1+(\tan \theta {)}^2}=\frac{1}{1+z^2},$$ we get $$p_z(z)=\frac{1/(2\pi )}{1+z^2}.$$ There is a spurious factor of 2 here. Actually, two values of $\theta$ spaced by $\pi$ map onto the same value of Template:Tmath, the density is doubled, and the final result is $$p_z(z)=\frac{1/\pi }{1+z^2},-\mathrm{\infty }<z<\mathrm{\infty }.$$

When either of the two normal distributions is non-central, then the result for the distribution of the ratio is much more complicated and is given below in the succinct form presented by David Hinkley.Template:Sfnp The trigonometric method for a ratio does, however, extend to radial distributions like bivariate normals or a bivariate Template:Student's t-distribution, in which the density depends only on radius Template:Tmath. It does not extend to the ratio of two independent Template:Student's t-distributions, which give the Cauchy ratio shown in a section below for one degree of freedom.

Uncorrelated noncentral normal ratio

In the absence of correlation (Template:Tmath), the probability density function of the ratio Template:Tmath of two normal variables $X=𝒩({\mu }_X,{\sigma }_X^2)$ and $Y=𝒩({\mu }_Y,{\sigma }_Y^2)$ is given exactly by the following expression, derived in several sources:Template:Sfnp $$p_Z(z)=\frac{e^{-c/2}}{a^2(z)\cdot 2\pi {\sigma }_x{\sigma }_y}(\sqrt{2\pi }\frac{b(z)}{a(z)}\exp \left(\frac{b^2(z)}{2a^2(z)}\right)\mathrm{e}\mathrm{r}\mathrm{f}\left(\frac{b(z)}{\sqrt{2}a(z)}\right)+2),$$ where $$\begin{array}{cc}a(z)& =\sqrt{\frac{1}{{\sigma }_x^2}{z}^2+\frac{1}{{\sigma }_y^2}},\\ b(z)& =\frac{{\mu }_x}{{\sigma }_x^2}z+\frac{{\mu }_y}{{\sigma }_y^2},\\ c& =\frac{{\mu }_x^2}{{\sigma }_x^2}+\frac{{\mu }_y^2}{{\sigma }_y^2}.\end{array}$$

Under several assumptions (usually fulfilled in practical applications), it is possible to derive a highly accurate solid approximation to the Template:Abbr. Main benefits include reduced formulae complexity, closed-form Template:Abbr, simple defined median, and well defined error management. For the sake of simplicity introduce parameters: Template:Tmath, Template:Tmath, and Template:Tmath. Then the so-called solid approximation $p_Z^{†}(z)$ to the uncorrelated noncentral normal ratio PDF is expressed by equationTemplate:Sfnp $$p_Z^{†}(z)=\frac{1}{\sqrt{\pi }}\frac{p}{\mathrm{e}\mathrm{r}\mathrm{f}[q]}\frac{1}{r}\frac{1+\frac{p^2}{q^2}\frac{z}{r}}{{(1+\frac{p^2}{q^2}{\left[\frac{z}{r}\right]}^2)}^{\frac{3}{2}}}\exp \left(-\frac{p^2{(\frac{z}{r}-1)}^2}{1+\frac{p^2}{q^2}{\left[\frac{z}{r}\right]}^2}\right).$$

Under certain conditions, a normal approximation is possible, with variance:Template:Sfnp $${\sigma }_z^2=\frac{{\mu }_x^2}{{\mu }_y^2}(\frac{{\sigma }_x^2}{{\mu }_x^2}+\frac{{\sigma }_y^2}{{\mu }_y^2}).$$

Correlated central normal ratio

The above expression becomes more complicated when the variables Template:Tmath and Template:Tmath are correlated. If Template:Tmath Template:Tmath but Template:Tmath and Template:Tmath, the more general Cauchy distribution is obtained, $$p_Z(z)=\frac{1}{\pi }\frac{\beta }{(z-\alpha {)}^2+{\beta }^2},$$ where Template:Tmath is the correlation coefficient between Template:Tmath and Template:Tmath, and $$\begin{array}{cc}\alpha & =\rho \frac{{\sigma }_x}{{\sigma }_y},\\ \beta & =\frac{{\sigma }_x}{{\sigma }_y}\sqrt{1-{\rho }^2}.\end{array}$$

The complex distribution has also been expressed with Kummer's confluent hypergeometric function or the Hermite function.Template:Sfnp

Correlated noncentral normal ratio

This was shown in Springer.Template:Sfnp.

A transformation to the log domain was suggested by KatzTemplate:Sfnp Template:See below. Let the ratio be $$T\sim \frac{{\mu }_x+𝒩(0,{\sigma }_x^2)}{{\mu }_y+𝒩(0,{\sigma }_y^2)}=\frac{{\mu }_x+X}{{\mu }_y+Y}=\frac{{\mu }_x}{{\mu }_y}\frac{1+\frac{X}{{\mu }_x}}{1+\frac{Y}{{\mu }_y}}.$$

Take the logarithm of both sides to get $$\ln (T)=\ln \left(\frac{{\mu }_x}{{\mu }_y}\right)+\ln (1+\frac{X}{{\mu }_x})-\ln (1+\frac{Y}{{\mu }_y}).$$ Observing that Template:Tmath, then asymptotically $$\begin{array}{cc}\ln (T)& \approx \ln \left(\frac{{\mu }_x}{{\mu }_y}\right)+\frac{X}{{\mu }_x}-\frac{Y}{{\mu }_y}\\ & \sim \ln \left(\frac{{\mu }_x}{{\mu }_y}\right)+𝒩(0,\frac{{\sigma }_x^2}{{\mu }_x^2}+\frac{{\sigma }_y^2}{{\mu }_y^2}-2\frac{\mathrm{Cov}(X,Y)}{{\mu }_x{\mu }_y}).\end{array}$$

Alternatively, Geary suggested in 1930 that $$t\approx \frac{{\mu }_{y}T-{\mu }_x}{\sqrt{{\sigma }_y^2{T}^2-2\rho {\sigma }_x{\sigma }_{y}T+{\sigma }_x^2}}$$ has approximately a standard Gaussian distribution;Template:Sfnp This transformation has been called the Geary–Hinkley transformation.Template:Sfnp The approximation is good if Template:Tmath is unlikely to assume negative values, i.e., Template:Tmath.

Exact correlated noncentral normal ratio

This is developed by SpringerTemplate:Sfnp and Hinkley.Template:Sfnp Geary showed how the correlated ratio Template:Tmath could be transformed into a near-Gaussian form and developed an approximation for Template:Tmath dependent on the probability of negative denominator values Template:Tmath being vanishingly small. Fieller's later correlated ratio analysis is exact but care is needed when combining modern math packages with verbal conditions in the older literature.Template:Cn Pham-Ghia has exhaustively discussed these methods.Template:Cn Hinkley's correlated results are exact but it is shown below that the correlated ratio condition can also be transformed into an uncorrelated one so only the simplified Hinkley equations above are required, not the full correlated ratio version.

Let the ratio be: $$z=\frac{x+{\mu }_x}{y+{\mu }_y}$$ in which Template:Tmath are zero-mean correlated normal variables with variances Template:Tmath and Template:Tmath have means Template:Tmath. Write $x^{\prime }=x-\rho y\frac{{\sigma }_x}{{\sigma }_y}$ such that Template:Tmath become uncorrelated and Template:Tmath has standard deviation Template:Tmath. Then the ratio $$z=\frac{x^{\prime }+\rho y\frac{{\sigma }_x}{{\sigma }_y}+{\mu }_x}{y+{\mu }_y}$$ is invariant under this transformation and retains the same Template:Abbr.

The Template:Tmath term in the numerator appears to be made separable by expanding $$x^{\prime }+\rho y\frac{{\sigma }_x}{{\sigma }_y}+{\mu }_x=x^{\prime }+{\mu }_x-\rho {\mu }_y\frac{{\sigma }_x}{{\sigma }_y}+\rho (y+{\mu }_y)\frac{{\sigma }_x}{{\sigma }_y}$$ to get $$z=\frac{x^{\prime }+{{\mu }_x}^{\prime }}{y+{\mu }_y}+\rho \frac{{\sigma }_x}{{\sigma }_y}$$ in which $\mu {\text{'}}_x={\mu }_x-\rho {\mu }_y\frac{{\sigma }_x}{{\sigma }_y}$ and Template:Tmath has now become a ratio of uncorrelated non-central normal samples with an invariant Template:Tmath-offset.Template:Efn.

Finally, to be explicit, the PDF of the ratio Template:Tmath for correlated variables is found by inputting the modified parameters Template:Tmath, Template:Tmath and Template:Tmath into the Hinkley equation above which returns the PDF for the correlated ratio with a constant offset Template:Tmath on Template:Tmath.

The figures above show an example of a positively correlated ratio with Template:Tmath Template:Tmath, Template:Tmath, Template:Tmath, and Template:Tmath, in which the shaded wedges represent the increment of area selected by given ratio Template:Tmath which accumulates probability where they overlap the distribution. The theoretical distribution, derived from the equations under discussion combined with Hinkley's equations, is highly consistent with a simulation result using 5,000 samples. In the top figure it is clear that for a ratio Template:Tmath the wedge has almost bypassed the main distribution mass altogether and this explains the local minimum in the theoretical PDF Template:Tmath. Conversely as Template:Tmath moves either toward or away from one the wedge spans more of the central mass, accumulating a higher probability.

Complex normal ratio

The ratio of correlated zero-mean circularly symmetric complex normal distributed variables was determined by Baxley et al.Template:Sfnp and has since been extended to the nonzero-mean and nonsymmetric case.Template:Sfnp In the correlated zero-mean case, the joint distribution of Template:Tmath is $$f_{x,y}(x,y)=\frac{1}{{\pi }^2|\mathrm{\Sigma }|}\exp (-{\left[\begin{array}{c}x\\ y\end{array}\right]}^H{\mathrm{\Sigma }}^{-1}\left[\begin{array}{c}x\\ y\end{array}\right])$$ where $$\begin{array}{cc}\mathrm{\Sigma }& =\left[\begin{array}{cc}{\sigma }_x^2& \rho {\sigma }_x{\sigma }_y\\ {\rho }^{\ast }{\sigma }_x{\sigma }_y& {\sigma }_y^2\end{array}\right];\\ x& =x_r+i{x}_i;\\ y& =y_r+i{y}_i;\end{array}$$ $(\cdot {)}^H$ is an Hermitian transpose; and $$\rho ={\rho }_r+i{\rho }_i=\mathrm{E}\left(\frac{x{y}^{\ast }}{{\sigma }_x{\sigma }_y}\right)\in \left|ℂ\right|\le 1.$$

The Template:Abbr of Template:Tmath is found to be

$$\begin{array}{cc}{f}_z(z_r,z_i)& =\frac{1-|\rho {|}^2}{\pi {\sigma }_x^2{\sigma }_y^2}{(\frac{|z{|}^2}{{\sigma }_x^2}+\frac{1}{{\sigma }_y^2}-2\frac{{\rho }_r{z}_r-{\rho }_i{z}_i}{{\sigma }_x{\sigma }_y})}^{-2}\\ & =\frac{1-|\rho {|}^2}{\pi {\sigma }_x^2{\sigma }_y^2}{({|\frac{z}{{\sigma }_x}-\frac{{\rho }^{\ast }}{{\sigma }_y}|}^2+\frac{1-|\rho {|}^2}{{\sigma }_y^2})}^{-2}.\end{array}$$ In the usual event that ${\sigma }_x={\sigma }_y$ we get $$f_z(z_r,z_i)=\frac{1-{\left|\rho \right|}^2}{\pi {({|z-{\rho }^{\ast }|}^2+1-{\left|\rho \right|}^2)}^2}.$$

Further closed-form results for the Template:Abbr are also given.

The ratio distribution of correlated complex variables, Template:Math.

The graph shows the pdf of the ratio of two complex normal variables with a correlation coefficient of Template:Tmath. The PDF peak occurs at roughly the complex conjugate of a scaled down Template:Tmath.

Ratio of log-normal

The ratio of independent or correlated log-normals is log-normal. This follows, because if $X_1$ and $X_2$ are log-normally distributed, then $\ln (X_1)$ and $\ln (X_2)$ are normally distributed. If they are independent or their logarithms follow a bivarate normal distribution, then the logarithm of their ratio is the difference of independent or correlated normally distributed random variables, which is normally distributed.Template:Efn

This is important for many applications requiring the ratio of random variables that must be positive, where joint distribution of $X_1$ and $X_2$ is adequately approximated by a log-normal. This is a common result of the multiplicative central limit theorem, also known as Gibrat's law, when $X_i$ is the result of an accumulation of many small percentage changes and must be positive and approximately log-normally distributed.Template:Efn

Uniform ratio distribution

With two independent random variables following a uniform distribution, e.g., $$p_X(x)=\{\begin{array}{cc}1,& 0<x<1\\ 0,& \text{otherwise}\end{array}$$ the ratio distribution becomes $$p_Z(z)=\{\begin{array}{cc}\frac{1}{2},& 0<z<1\\ \frac{1}{2z^2},& z\ge 1\\ 0,& \text{otherwise}.\end{array}$$

Cauchy ratio distribution

If two independent random variables, Template:Tmath and Template:Tmath each follow a Cauchy distribution with median equal to zero and shape factor Template:Tmath, $$p_X(x|a)=\frac{a}{\pi (a^2+x^2)},$$ then the ratio distribution for the random variable Template:Tmath isTemplate:Sfnp $$p_Z(z|a)=\frac{1}{{\pi }^2(z^2-1)}\ln (z^2).$$ This distribution does not depend on Template:Tmath and the result stated by SpringerTemplate:Sfnp is not correct.

The ratio distribution is similar to but not the same as the product distribution of the random variable Template:Tmath:Template:Sfnp $$p_W(w|a)=\frac{a^2}{{\pi }^2(w^2-a^4)}\ln \left(\frac{w^2}{a^4}\right).$$

More generally, if two independent random variables Template:Tmath and Template:Tmath each follow a Cauchy distribution with median equal to zero and shape factor Template:Tmath and Template:Tmath respectively, then:

1. The ratio distribution for the random variable Template:Tmath isTemplate:Sfnp $$p_Z(z|a,b)=\frac{ab}{{\pi }^2(b^2{z}^2-a^2)}\ln \left(\frac{b^2{z}^2}{a^2}\right).$$
2. The product distribution for the random variable Template:Tmath isTemplate:Sfnp $$p_W(w|a,b)=\frac{ab}{{\pi }^2(w^2-a^2{b}^2)}\ln \left(\frac{w^2}{a^2{b}^2}\right).$$

The result for the ratio distribution can be obtained from the product distribution by replacing Template:Tmath with Template:Tmath.

Ratio of standard normal to standard uniform

Template:Main If Template:Tmath has a standard normal distribution and Template:Tmath has a standard uniform distribution, then Template:Tmath has a distribution known as the slash distribution, with probability density function $$p_Z(z)=\{\begin{array}{cc}\frac{[\phi (0)-\phi (z)]}{z^2},& z\ne 0\\ \frac{\phi (0)}{2},& z=0,\end{array}$$ where Template:Tmath is the probability density function of the standard normal distribution.

Chi-squared, Gamma distributions

Let Template:Tmath be a normal distribution Template:Tmath, Template:Tmath and Template:Tmath be chi-squared distributions with Template:Tmath and Template:Tmath degrees of freedom respectively, all independent, with $$f_{\chi }(x,k)=\frac{x^{(\frac{k}{2}-1)}{e}^{-\frac{x}{2}}}{2^{k/2}\mathrm{\Gamma }\left(\frac{k}{2}\right)}.$$ Then

• $\frac{G}{\sqrt{Y/m}}\sim t_m$ is the Template:Student's t-distribution;
• $\frac{Y/m}{Z/n}=F_{m,n},$ i.e. Fisher's [[F-test|Template:Itco-test]] distribution;
• $\frac{Y}{Y+Z}\sim \beta (\frac{m}{2},\frac{n}{2})$ is the beta distribution; and
• $\frac{Y}{Z}\sim {\beta }^{\prime }(\frac{m}{2},\frac{n}{2})$ is the standard beta prime distribution.

If Template:Tmath, a noncentral chi-squared distribution, Template:Tmath, and Template:Tmath is independent of Template:Tmath, then $$\frac{V_1/k_1}{V_2/k_2}\sim F{\text{'}}_{k_1,k_2}(\lambda ),$$ a [[noncentral F-distribution|noncentral Template:ItcoF-distribution]].

$\frac{m}{n}F{\text{'}}_{m,n}={\beta }^{\prime }(\frac{m}{2},\frac{n}{2})$ or $F{\text{'}}_{m,n}={\beta }^{\prime }(\frac{m}{2},\frac{n}{2},1,\frac{n}{m})$ defines Template:Tmath, Fisher's Template:Itco density distribution, the Template:Abbr of the ratio of two chi-squared distributions with Template:Tmath degrees of freedom.Template:Efn

If we enter an Template:Itco-test table with Template:Tmath, Template:Tmath and 5% probability in the right tail, the critical value is found to be 6.59. This coincides with the integral $$F_{3,4}(6.59)={\int }_{6.59}^{\mathrm{\infty }}{\beta }^{\prime }(x;\frac{m}{2},\frac{n}{2},1,\frac{n}{m})dx=\mathrm{0.05.}$$

For gamma distributions Template:Tmath and Template:Tmath with arbitrary shape parameters Template:Tmath and Template:Tmath, and their scale parameters both set to unity (that is, Template:Tmath, Template:Tmath, where Template:Tmath), then

$$\begin{array}{cccc}\frac{U}{U+V}& \sim \beta ({\alpha }_1,{\alpha }_2),& \text{ expectation }& =\frac{{\alpha }_1}{{\alpha }_1+{\alpha }_2};\\ \frac{U}{V}& \sim {\beta }^{\prime }({\alpha }_1,{\alpha }_2),& \text{ expectation }& =\frac{{\alpha }_1}{{\alpha }_2-1},{\alpha }_2>1;\\ \frac{V}{U}& \sim {\beta }^{\prime }({\alpha }_2,{\alpha }_1),& \text{ expectation }& =\frac{{\alpha }_2}{{\alpha }_1-1},{\alpha }_1>1.\end{array}$$

If Template:Tmath, then $$\theta U\sim \mathrm{\Gamma }(x;\alpha ,\theta )=\frac{x^{\alpha -1}{e}^{-x/\theta }}{{\theta }^k\mathrm{\Gamma }(\alpha )}.$$ Note that here Template:Tmath is a scale parameter, rather than a rate parameter.

If Template:Tmath, Template:Tmath, then by rescaling the Template:Tmath parameter to unity we have $$\begin{array}{cc}\frac{\frac{U}{{\theta }_1}}{\frac{U}{{\theta }_1}+\frac{V}{{\theta }_2}}& =\frac{{\theta }_{2}U}{{\theta }_{2}U+{\theta }_{1}V}\sim \beta ({\alpha }_1,{\alpha }_2)\\ \frac{\frac{U}{{\theta }_1}}{\frac{V}{{\theta }_2}}& =\frac{{\theta }_2}{{\theta }_1}\frac{U}{V}\sim {\beta }^{\prime }({\alpha }_1,{\alpha }_2).\end{array}$$ Thus $$\begin{array}{cc}\frac{U}{V}& \sim {\beta }^{\prime }({\alpha }_1,{\alpha }_2,1,\frac{{\theta }_1}{{\theta }_2})\\ \mathrm{E}\left[\frac{U}{V}\right]& =\frac{{\theta }_1}{{\theta }_2}\frac{{\alpha }_1}{{\alpha }_2-1},\end{array}$$ in which ${\beta }^{\prime }(\alpha ,\beta ,p,q)$ represents the generalised beta prime distribution.

In the foregoing it is apparent that if $X\sim {\beta }^{\prime }({\alpha }_1,{\alpha }_2,1,1)\equiv {\beta }^{\prime }({\alpha }_1,{\alpha }_2)$ then $\theta X\sim {\beta }^{\prime }({\alpha }_1,{\alpha }_2,1,\theta )$. More explicitly, since $${\beta }^{\prime }(x;{\alpha }_1,{\alpha }_2,1,R)=\frac{1}{R}{\beta }^{\prime }(\frac{x}{R};{\alpha }_1,{\alpha }_2)$$ if $U\sim \mathrm{\Gamma }({\alpha }_1,{\theta }_1),V\sim \mathrm{\Gamma }({\alpha }_2,{\theta }_2)$ then $$\frac{U}{V}\sim \frac{1}{R}{\beta }^{\prime }(\frac{x}{R};{\alpha }_1,{\alpha }_2)=\frac{{\left(\frac{x}{R}\right)}^{{\alpha }_1-1}}{{(1+\frac{x}{R})}^{{\alpha }_1+{\alpha }_2}}\cdot \frac{1}{RB({\alpha }_1,{\alpha }_2)},$$ where $$\begin{array}{cc}x& \ge 0;\\ R& =\frac{{\theta }_1}{{\theta }_2};\\ B({\alpha }_1,{\alpha }_2)& =\frac{\mathrm{\Gamma }({\alpha }_1)\mathrm{\Gamma }({\alpha }_2)}{\mathrm{\Gamma }({\alpha }_1+{\alpha }_2)}.\end{array}$$

Beta distributions

First developed by T. Pham-GiaTemplate:Sfnp who found that the PDF of a Beta ratio can be expressed in terms of hypergeometric functions. These are calculated numerically as an infinite discrete sum or as an integral. The latter is straightforward such that, if $Y$ is the ratio of two independent Beta samples $Y=X_1/X_2$, each with PDF $B(\alpha ,\beta {)}^{-1}{x}^{\alpha -1}(1-x{)}^{\beta -1}$ proceeds in two ranges $0\le y<1$ and $1<y<\mathrm{\infty }$ thus

$$\begin{array}{cc}& f_A(y|0<y\le 1)=\frac{1}{B({\alpha }_1,{\beta }_1)({\alpha }_2,{\beta }_2)}{y}^{{\alpha }_1-1}{\int }_0^{1}dp{p}^{\alpha 1+{\alpha }_2-1}(1-p{)}^{{\beta }_2-1}(1-py{)}^{{\beta }_1-1}\\ & f_B(y|1<y<\mathrm{\infty })=\frac{1}{B({\alpha }_1,{\beta }_1)({\alpha }_2,{\beta }_2)}\frac{1}{y^{{\alpha }_2+1}}{\int }_0^{1}dp{p}^{\alpha 1+{\alpha }_2-1}(1-p{)}^{{\beta }_1-1}{(1-\frac{y}{p})}^{{\beta }_2-1}\end{array}$$

Rayleigh Distributions

If X, Y are independent samples from the Rayleigh distribution $f_r(r)=(r/{\sigma }^2)e^{-r^2/2{\sigma }^2},r\ge 0$, the ratio Z = X/Y follows the distributionTemplate:Sfnp

$$f_z(z)=\frac{2z}{(1+z^2{)}^2},z\ge 0$$ and has cdf $$F_z(z)=1-\frac{1}{1+z^2}=\frac{z^2}{1+z^2},z\ge 0$$ The Rayleigh distribution has scaling as its only parameter. The distribution of $Z=\alpha X/Y$ follows $$f_z(z,\alpha )=\frac{2\alpha z}{(\alpha +z^2{)}^2},z>0$$ and has cdf $$F_z(z,\alpha )=\frac{z^2}{\alpha +z^2},z\ge 0$$

Fractional gamma distributions (including chi, chi-squared, exponential, Rayleigh and Weibull)

The generalized gamma distribution is $$f(x;a,d,r)=\frac{r}{\mathrm{\Gamma }(d/r)a^d}{x}^{d-1}{e}^{-(x/a{)}^r}x\ge 0;a,d,r>0$$ which includes the regular gamma, chi, chi-squared, exponential, Rayleigh, Nakagami and Weibull distributions involving fractional powers. Note that here a is a scale parameter, rather than a rate parameter; d is a shape parameter.

Modelling a mixture of different scaling factors

In the ratios above, Gamma samples, U, V may have differing sample sizes ${\alpha }_1,{\alpha }_2$ but must be drawn from the same distribution $\frac{x^{\alpha -1}{e}^{-x/\theta }}{{\theta }^k\mathrm{\Gamma }(\alpha )}$ with equal scaling $\theta$.

In situations where U and V are differently scaled, a variables transformation allows the modified random ratio pdf to be determined. Let $X=\frac{U}{U+V}=\frac{1}{1+B}$ where $U\sim \mathrm{\Gamma }({\alpha }_1,\theta ),V\sim \mathrm{\Gamma }({\alpha }_2,\theta ),\theta$ arbitrary and, from above, $X\sim \mathrm{B}\mathrm{e}\mathrm{t}\mathrm{a}({\alpha }_1,{\alpha }_2),B=V/U\sim {\mathrm{B}\mathrm{e}\mathrm{t}\mathrm{a}}^{\prime }({\alpha }_2,{\alpha }_1)$.

Rescale V arbitrarily, defining $Y\sim \frac{U}{U+\phi V}=\frac{1}{1+\phi B},0\le \phi \le \mathrm{\infty }$

We have $B=\frac{1-X}{X}$ and substitution into Y gives $Y=\frac{X}{\phi +(1-\phi )X}$, $dY/dX=\frac{\phi }{(\phi +(1-\phi )X{)}^2}$

Transforming X to Y gives $f_Y(Y)=\frac{f_X(X)}{|dY/dX|}=\frac{\beta (X,{\alpha }_1,{\alpha }_2)}{\phi /[\phi +(1-\phi )X{]}^2}$

Noting $X=\frac{\phi Y}{1-(1-\phi )Y}$ we finally have $$f_Y(Y,\phi )=\frac{\phi }{[1-(1-\phi )Y{]}^2}\beta (\frac{\phi Y}{1-(1-\phi )Y},{\alpha }_1,{\alpha }_2),0\le Y\le 1$$

Thus, if $U\sim \mathrm{\Gamma }({\alpha }_1,{\theta }_1)$ and $V\sim \mathrm{\Gamma }({\alpha }_2,{\theta }_2)$
then $Y=\frac{U}{U+V}$ is distributed as $f_Y(Y,\phi )$ with $\phi =\frac{{\theta }_2}{{\theta }_1}$

The distribution of Y is limited here to the interval [0,1]. It can be generalized by scaling such that if $Y\sim f_Y(Y,\phi )$ then

$$\mathrm{\Theta }Y\sim f_Y(Y,\phi ,\mathrm{\Theta })$$

where $f_Y(Y,\phi ,\mathrm{\Theta })=\frac{\phi /\mathrm{\Theta }}{[1-(1-\phi )Y/\mathrm{\Theta }{]}^2}\beta (\frac{\phi Y/\mathrm{\Theta }}{1-(1-\phi )Y/\mathrm{\Theta }},{\alpha }_1,{\alpha }_2),0\le Y\le \mathrm{\Theta }$ Template:Block indent

Reciprocals of samples from beta distributions

Though not ratio distributions of two variables, the following identities for one variable are useful:

• If $X\sim \beta (\alpha ,\beta )$ then $𝐱=\frac{X}{1-X}\sim {\beta }^{\prime }(\alpha ,\beta )$
• If $𝐘\sim {\beta }^{\prime }(\alpha ,\beta )$ then $y=\frac{1}{𝐘}\sim {\beta }^{\prime }(\beta ,\alpha )$ combining the latter two equations yields
• If $X\sim \beta (\alpha ,\beta )$ then $𝐱=\frac{1}{X}-1\sim {\beta }^{\prime }(\beta ,\alpha )$.
• If $𝐘\sim {\beta }^{\prime }(\alpha ,\beta )$ then $y=\frac{𝐘}{1+𝐘}\sim \beta (\alpha ,\beta )$

Corollary

$$\frac{1}{1+𝐘}=\frac{{𝐘}^{-1}}{{𝐘}^{-1}+1}\sim \beta (\beta ,\alpha )$$

Template:Block indent If $U\sim \mathrm{\Gamma }(\alpha ,1),V\sim \mathrm{\Gamma }(\beta ,1)$ then $\frac{U}{V}\sim {\beta }^{\prime }(\alpha ,\beta )$ and $$\frac{U/V}{1+U/V}=\frac{U}{V+U}\sim \beta (\alpha ,\beta )$$

Further results can be found in the Inverse distribution article.

• If Template:Mvar, Template:Mvar are independent exponential random variables with mean Template:Mvar, then Template:Math is a double exponential random variable with mean 0 and scale Template:Mvar.

Binomial distribution

This result was derived by Katz et al.Template:Sfnp

Suppose $X\sim \mathrm{Binomial}(n,p_1)$ and $Y\sim \text{Binomial}(m,p_2)$ and Template:Tmath, $Y$ are independent. Let Template:Tmath.

Then $\log (T)$ is approximately normally distributed with mean $\log (p_1/p_2)$ and variance Template:Tmath.

The binomial ratio distribution is of significance in clinical trials: if the distribution of T is known as above, the probability of a given ratio arising purely by chance can be estimated, i.e. a false positive trial. A number of papers compare the robustness of different approximations for the binomial ratio.Template:Citation needed

Poisson and truncated Poisson distributions

In the ratio of Poisson variables Template:Math there is a problem that Template:Mvar is zero with finite probability so Template:Mvar is undefined. To counter this, consider the truncated, or censored, ratio Template:Math where zero sample of Template:Mvar are discounted. Moreover, in many medical-type surveys, there are systematic problems with the reliability of the zero samples of both Template:Mvar and Template:Mvar and it may be good practice to ignore the zero samples anyway.

The probability of a null Poisson sample being Template:Tmath, the generic Template:Abbr of a left truncated Poisson distribution is $${\tilde{p}}_x(x;\lambda )=\frac{1}{1-e^{-\lambda }}\frac{e^{-\lambda }{\lambda }^x}{x!},x\in 1,2,3,\cdots$$ which sums to unity. Following Cohen,Template:Sfnp for Template:Mvar independent trials, the multidimensional truncated Template:Abbr is $$\tilde{p}(x_1,x_2,\dots ,x_n;\lambda )=\frac{1}{{(1-e^{-\lambda })}^n}\prod _{i=1}^n\frac{e^{-\lambda }{\lambda }^{x_i}}{x_i!},x_i\in 1,2,3,\cdots$$ and the log-likelihood becomes $$L=\ln (\tilde{p})=-n\ln (1-e^{-\lambda })-n\lambda +\ln (\lambda )\sum _{i=1}^n{x}_i-\ln \prod _{i=1}^n(x_i!),x_i\in 1,2,3,\cdots$$ On differentiation we get $$\frac{dL}{d\lambda }=\frac{-n}{1-e^{-\lambda }}+\frac{1}{\lambda }\sum _{i=1}^n{x}_i$$ and setting to zero gives the maximum likelihood estimate ${\hat{\lambda }}_{\text{ML}}$ $$\frac{{\hat{\lambda }}_{\text{ML}}}{1-e^{-{\hat{\lambda }}_{\text{ML}}}}=\frac{1}{n}\sum _{i=1}^n{x}_i=\overline{x}$$

Note that as $\hat{\lambda }\to 0$ then $\overline{x}\to 1$ so the truncated maximum likelihood $\lambda$ estimate, though correct for both truncated and untruncated distributions, gives a truncated mean $\overline{x}$ value which is highly biassed relative to the untruncated one. Nevertheless it appears that $\overline{x}$ is a sufficient statistic for $\lambda$ since ${\hat{\lambda }}_{ML}$ depends on the data only through the sample mean $\overline{x}=\frac{1}{n}\sum _{i=1}^n{x}_i$ in the previous equation which is consistent with the methodology of the conventional Poisson distribution.

Absent any closed form solutions, the following approximate reversion for truncated $\lambda$ is valid over the whole range Template:Tmath. $$\hat{\lambda }=\overline{x}-e^{-(\overline{x}-1)}-0.07(\overline{x}-1)e^{-0.666(\overline{x}-1)}+\epsilon ,|\epsilon |<0.006$$ which compares with the non-truncated version which is simply $\hat{\lambda }=\overline{x}$. Taking the ratio $R={\hat{\lambda }}_X/{\hat{\lambda }}_Y$ is a valid operation even though ${\hat{\lambda }}_X$ may use a non-truncated model while ${\hat{\lambda }}_Y$ has a left-truncated one.

The asymptotic large-$n\lambda$ variance of $\hat{\lambda }$ (and Cramér–Rao bound) is $$𝕍𝕒𝕣(\hat{\lambda })\ge -{\left(𝔼{\left[\frac{{\delta }^{2}L}{\delta {\lambda }^2}\right]}_{\lambda =\hat{\lambda }}\right)}^{-1}$$

in which substituting Template:Mvar gives $$\frac{{\delta }^{2}L}{\delta {\lambda }^2}=-n[\frac{\overline{x}}{{\lambda }^2}-\frac{e^{-\lambda }}{(1-e^{-\lambda }{)}^2}]$$ Then substituting $\overline{x}$ from the equation above, we get Cohen's variance estimate $$𝕍𝕒𝕣(\hat{\lambda })\ge \frac{\hat{\lambda }}{n}\frac{(1-e^{-\hat{\lambda }}{)}^2}{1-(\hat{\lambda }+1)e^{-\hat{\lambda }}}$$

The variance of the point estimate of the mean Template:Tmath, on the basis of Template:Mvar trials, decreases asymptotically to zero as Template:Mvar increases to infinity. For small $\lambda$ it diverges from the truncated pdf variance in SpringaelTemplate:Sfnp for example, who quotes a variance of $$𝕍𝕒𝕣(\lambda )=\frac{\lambda /n}{1-e^{-\lambda }}[1-\frac{\lambda e^{-\lambda }}{1-e^{-\lambda }}]$$ for n samples in the left-truncated pdf shown at the top of this section. Cohen showed that the variance of the estimate relative to the variance of the pdf, $𝕍𝕒𝕣(\hat{\lambda })/𝕍𝕒𝕣(\lambda )$, ranges from 1 for large $\lambda$ (100% efficient) up to 2 as $\lambda$ approaches zero (50% efficient).

These mean and variance parameter estimates, together with parallel estimates for X, can be applied to Normal or Binomial approximations for the Poisson ratio. Samples from trials may not be a good fit for the Poisson process; a further discussion of Poisson truncation is by Dietz and BohningTemplate:Sfnp and there is a Zero-truncated Poisson distribution Wikipedia entry.

Double Lomax distribution

This distribution is the ratio of two Laplace distributions.Template:Sfnp Let X and Y be standard Laplace identically distributed random variables and let z = X / Y. Then the probability distribution of z is

$$f(x)=\frac{1}{2{(1+|z|)}^2}$$

Let the mean of the X and Y be a. Then the standard double Lomax distribution is symmetric around a.

This distribution has an infinite mean and variance.

If Z has a standard double Lomax distribution, then 1/Z also has a standard double Lomax distribution.

The standard Lomax distribution is unimodal and has heavier tails than the Laplace distribution.

For 0 < a < 1, the a-th moment exists.

$$\mathrm{\Lambda }=\frac{|𝐗|}{|𝐗+𝐘|}$$$$E(Z^a)=\frac{\mathrm{\Gamma }(1+a)}{\mathrm{\Gamma }(1-a)}$$

where Template:Math is the gamma function.

Ratio distributions in multivariate analysis

Ratio distributions also appear in multivariate analysis.Template:Sfnp If the random matrices Template:Math and Template:Math follow a Wishart distribution then the ratio of the determinants $$\phi =\frac{|𝐗|}{|𝐘|}$$ is proportional to the product of independent F random variables. In the case where Template:Math and Template:Math are from independent standardized Wishart distributions then the ratio $$\mathrm{\Lambda }=\frac{|𝐗|}{|𝐗+𝐘|}$$ has a Wilks' lambda distribution.

Ratios of quadratic forms involving Wishart matrices

In relation to Wishart matrix distributions, if $S\sim W_p(\mathrm{\Sigma },\nu +1)$ is a sample Wishart matrix, and vector $V$ is arbitrary, but statistically independent, then corollary 3.2.9 of MuirheadTemplate:Sfnp states $$\frac{V^{T}SV}{V^T\mathrm{\Sigma }V}\sim {\chi }_{\nu }^2.$$

The discrepancy of one in the sample numbers arises from estimation of the sample mean when forming the sample covariance, a consequence of Cochran's theorem. Similarly, $$\frac{V^T{\mathrm{\Sigma }}^{-1}V}{V^T{S}^{-1}V}\sim {\chi }_{\nu -p+1}^2,$$ which is Theorem 3.2.12 of Muirhead.Template:Sfnp

See also

• Relationships among probability distributions
• Inverse distribution (also known as reciprocal distribution)
• Product distribution
• Ratio estimator
• Slash distribution

Works cited

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External links

• Ratio Distribution at MathWorld
• Normal Ratio Distribution at MathWorld
• Ratio Distributions at MathPages