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 X and Y, the distribution of the random variable Z that is formed as the ratio Z = X/Y 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 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.^{[1][2][3][4][5][6][7][8][9]}

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".^[10]

Algebra of random variables

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.^[8]

The algebraic rules known with ordinary numbers do not apply for the algebra of random variables. For example, if a product is C = AB and a ratio is D=C/A it does not necessarily mean that the distributions of D and B 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.^[8] 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, ${\displaystyle C_{1}}$ and ${\displaystyle C_{2}}$ each constructed from two Gaussian distributions ${\displaystyle C_{1}=G_{1}/G_{2}}$ and ${\displaystyle C_{2}=G_{3}/G_{4}}$ then

${\displaystyle {\frac {C_{1}}{C_{2}}}={\frac {{G_{1}}/{G_{2}}}{{G_{3}}/{G_{4}}}}={\frac {G_{1}G_{4}}{G_{2}G_{3}}}={\frac {G_{1}}{G_{2}}}\times {\frac {G_{4}}{G_{3}}}=C_{1}\times C_{3},}$

where ${\displaystyle C_{3}=G_{4}/G_{3}}$. 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 ${\displaystyle Z=X/Y}$ from the joint distribution of the two other random variables X , Y , with joint pdf ${\displaystyle p_{X,Y}(x,y)}$, is by integration of the following form^[3]

${\displaystyle p_{Z}(z)=\int _{-\infty }^{+\infty }|y|\,p_{X,Y}(zy,y)\,dy.}$

If the two variables are independent then ${\displaystyle p_{XY}(x,y)=p_{X}(x)p_{Y}(y)}$ and this becomes

${\displaystyle p_{Z}(z)=\int _{-\infty }^{+\infty }|y|\,p_{X}(zy)p_{Y}(y)\,dy.}$

This may not be straightforward. By way of example take the classical problem of the ratio of two standard Gaussian samples. The joint pdf is

${\displaystyle p_{X,Y}(x,y)={\frac {1}{2\pi }}\exp \left(-{\frac {x^{2}}{2}}\right)\exp \left(-{\frac {y^{2}}{2}}\right)}$

Defining ${\displaystyle Z=X/Y}$ we have

${\displaystyle {\begin{aligned}p_{Z}(z)&={\frac {1}{2\pi }}\int _{-\infty }^{\infty }\,|y|\,\exp \left(-{\frac {\left(zy\right)^{2}}{2}}\right)\,\exp \left(-{\frac {y^{2}}{2}}\right)\,dy\\&={\frac {1}{2\pi }}\int _{-\infty }^{\infty }\,|y|\,\exp \left(-{\frac {y^{2}\left(z^{2}+1\right)}{2}}\right)\,dy\end{aligned}}}$

Using the known definite integral ${\textstyle \int _{0}^{\infty }\,x\,\exp \left(-cx^{2}\right)\,dx={\frac {1}{2c}}}$ we get

${\displaystyle p_{Z}(z)={\frac {1}{\pi (z^{2}+1)}}}$

which is the Cauchy distribution, or Student's t distribution with n = 1

The Mellin transform has also been suggested for derivation of ratio distributions.^[8]

In the case of positive independent variables, proceed as follows. The diagram shows a separable bivariate distribution ${\displaystyle f_{x,y}(x,y)=f_{x}(x)f_{y}(y)}$ which has support in the positive quadrant ${\displaystyle x,y>0}$ and we wish to find the pdf of the ratio ${\displaystyle R=X/Y}$. The hatched volume above the line ${\displaystyle y=x/R}$ represents the cumulative distribution of the function ${\displaystyle f_{x,y}(x,y)}$ multiplied with the logical function ${\displaystyle X/Y\leq R}$. The density is first integrated in horizontal strips; the horizontal strip at height y extends from x = 0 to x = Ry and has incremental probability ${\textstyle f_{y}(y)dy\int _{0}^{Ry}f_{x}(x)\,dx}$.
Secondly, integrating the horizontal strips upward over all y yields the volume of probability above the line

${\displaystyle F_{R}(R)=\int _{0}^{\infty }f_{y}(y)\left(\int _{0}^{Ry}f_{x}(x)dx\right)dy}$

Finally, differentiate ${\displaystyle F_{R}(R)}$ with respect to ${\displaystyle R}$ to get the pdf ${\displaystyle f_{R}(R)}$.

${\displaystyle f_{R}(R)={\frac {d}{dR}}\left[\int _{0}^{\infty }f_{y}(y)\left(\int _{0}^{Ry}f_{x}(x)dx\right)dy\right]}$

Move the differentiation inside the integral:

${\displaystyle f_{R}(R)=\int _{0}^{\infty }f_{y}(y)\left({\frac {d}{dR}}\int _{0}^{Ry}f_{x}(x)dx\right)dy}$

and since

${\displaystyle {\frac {d}{dR}}\int _{0}^{Ry}f_{x}(x)dx=yf_{x}(Ry)}$

then

${\displaystyle f_{R}(R)=\int _{0}^{\infty }f_{y}(y)\;f_{x}(Ry)\;y\;dy}$

As an example, find the pdf of the ratio R when

${\displaystyle f_{x}(x)=\alpha e^{-\alpha x},\;\;\;\;f_{y}(y)=\beta e^{-\beta y},\;\;\;x,y\geq 0}$

Evaluating the cumulative distribution of a ratio

We have

${\displaystyle \int _{0}^{Ry}f_{x}(x)dx=-e^{-\alpha x}\vert _{0}^{Ry}=1-e^{-\alpha Ry}}$

thus

${\displaystyle {\begin{aligned}F_{R}(R)&=\int _{0}^{\infty }f_{y}(y)\left(1-e^{-\alpha Ry}\right)dy=\int _{0}^{\infty }\beta e^{-\beta y}\left(1-e^{-\alpha Ry}\right)dy\\&=1-{\frac {\alpha R}{\beta +\alpha R}}\\&={\frac {R}{{\tfrac {\beta }{\alpha }}+R}}\end{aligned}}}$

Differentiation wrt. R yields the pdf of R

${\displaystyle f_{R}(R)={\frac {d}{dR}}\left({\frac {R}{{\tfrac {\beta }{\alpha }}+R}}\right)={\frac {\tfrac {\beta }{\alpha }}{\left({\tfrac {\beta }{\alpha }}+R\right)^{2}}}}$

Moments of random ratios

From Mellin transform theory, for distributions existing only on the positive half-line ${\displaystyle x\geq 0}$, we have the product identity ${\displaystyle \operatorname {E} [(UV)^{p}]=\operatorname {E} [U^{p}]\;\;\operatorname {E} [V^{p}]}$ provided ${\displaystyle U,\;V}$ are independent. For the case of a ratio of samples like ${\displaystyle \operatorname {E} [(X/Y)^{p}]}$, in order to make use of this identity it is necessary to use moments of the inverse distribution. Set ${\displaystyle 1/Y=Z}$ such that ${\displaystyle \operatorname {E} [(XZ)^{p}]=\operatorname {E} [X^{p}]\;\operatorname {E} [Y^{-p}]}$. Thus, if the moments of ${\displaystyle X^{p}}$ and ${\displaystyle Y^{-p}}$ can be determined separately, then the moments of ${\displaystyle X/Y}$ can be found. The moments of ${\displaystyle Y^{-p}}$ are determined from the inverse pdf of ${\displaystyle Y}$ , often a tractable exercise. At simplest, ${\textstyle \operatorname {E} [Y^{-p}]=\int _{0}^{\infty }y^{-p}f_{y}(y)\,dy}$.

To illustrate, let ${\displaystyle X}$ be sampled from a standard Gamma distribution

${\displaystyle x^{\alpha -1}e^{-x}/\Gamma (\alpha )}$ whose ${\displaystyle p}$-th moment is ${\displaystyle \Gamma (\alpha +p)/\Gamma (\alpha )}$.

${\displaystyle Z=Y^{-1}}$is sampled from an inverse Gamma distribution with parameter ${\displaystyle \beta }$ and has pdf ${\displaystyle \;\Gamma ^{-1}(\beta )z^{1+\beta }e^{-1/z}}$. The moments of this pdf are

${\displaystyle \operatorname {E} [Z^{p}]=\operatorname {E} [Y^{-p}]={\frac {\Gamma (\beta -p)}{\Gamma (\beta )}},\;p<\beta .}$

Multiplying the corresponding moments gives

${\displaystyle \operatorname {E} [(X/Y)^{p}]=\operatorname {E} [X^{p}]\;\operatorname {E} [Y^{-p}]={\frac {\Gamma (\alpha +p)}{\Gamma (\alpha )}}{\frac {\Gamma (\beta -p)}{\Gamma (\beta )}},\;p<\beta .}$

Independently, it is known that the ratio of the two Gamma samples ${\displaystyle R=X/Y}$ follows the Beta Prime distribution:

${\displaystyle f_{\beta '}(r,\alpha ,\beta )=B(\alpha ,\beta )^{-1}r^{\alpha -1}(1+r)^{-(\alpha +\beta )}}$ whose moments are ${\displaystyle \operatorname {E} [R^{p}]={\frac {\mathrm {B} (\alpha +p,\beta -p)}{\mathrm {B} (\alpha ,\beta )}}}$

Substituting ${\displaystyle \mathrm {B} (\alpha ,\beta )={\frac {\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}}}$ we have ${\displaystyle \operatorname {E} [R^{p}]={\frac {\Gamma (\alpha +p)\Gamma (\beta -p)}{\Gamma (\alpha +\beta )}}{\Bigg /}{\frac {\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}}={\frac {\Gamma (\alpha +p)\Gamma (\beta -p)}{\Gamma (\alpha )\Gamma (\beta )}}}$ 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

${\displaystyle \operatorname {E} (X/Y)=\operatorname {E} (X)\operatorname {E} (1/Y)}$

which, in terms of probability distributions, is equivalent to

${\displaystyle \operatorname {E} (X/Y)=\int _{-\infty }^{\infty }xf_{x}(x)\,dx\times \int _{-\infty }^{\infty }y^{-1}f_{y}(y)\,dy}$

Note that ${\displaystyle \operatorname {E} (1/Y)\neq {\frac {1}{\operatorname {E} (Y)}}}$ i.e., ${\displaystyle \int _{-\infty }^{\infty }y^{-1}f_{y}(y)\,dy\neq {\frac {1}{\int _{-\infty }^{\infty }yf_{y}(y)\,dy}}}$

The variance of a ratio of independent variables is

${\displaystyle {\begin{aligned}\operatorname {Var} (X/Y)&=\operatorname {E} ([X/Y]^{2})-\operatorname {E^{2}} (X/Y)\\&=\operatorname {E} (X^{2})\operatorname {E} (1/Y^{2})-\operatorname {E} ^{2}(X)\operatorname {E} ^{2}(1/Y)\end{aligned}}}$

Normal ratio distributions

Correlated noncentral normal ratio

This was shown in Springer 1979 problem 4.28.

A transformation to the log domain was suggested by Katz(1978) (see binomial section below). Let the ratio be

${\displaystyle T\sim {\frac {\mu _{x}+\mathbb {N} (0,\sigma _{x}^{2})}{\mu _{y}+\mathbb {N} (0,\sigma _{y}^{2})}}={\frac {\mu _{x}+X}{\mu _{y}+Y}}={\frac {\mu _{x}}{\mu _{y}}}{\frac {1+{\frac {X}{\mu _{x}}}}{1+{\frac {Y}{\mu _{y}}}}}}$.

Take logs to get

${\displaystyle \log _{e}(T)=\log _{e}\left({\frac {\mu _{x}}{\mu _{y}}}\right)+\log _{e}\left(1+{\frac {X}{\mu _{x}}}\right)-\log _{e}\left(1+{\frac {Y}{\mu _{y}}}\right).}$

Since ${\displaystyle \log _{e}(1+\delta )=\delta -{\frac {\delta ^{2}}{2}}+{\frac {\delta ^{3}}{3}}+\cdots }$ then asymptotically

${\displaystyle \log _{e}(T)\approx \log _{e}\left({\frac {\mu _{x}}{\mu _{y}}}\right)+{\frac {X}{\mu _{x}}}-{\frac {Y}{\mu _{y}}}\sim \log _{e}\left({\frac {\mu _{x}}{\mu _{y}}}\right)+\mathbb {N} \left(0,{\frac {\sigma _{x}^{2}}{\mu _{x}^{2}}}+{\frac {\sigma _{y}^{2}}{\mu _{y}^{2}}}\right).}$

Alternatively, Geary (1930) suggested that

${\displaystyle t\approx {\frac {\mu _{y}T-\mu _{x}}{\sqrt {\sigma _{y}^{2}T^{2}-2\rho \sigma _{x}\sigma _{y}T+\sigma _{x}^{2}}}}}$

has approximately a standard Gaussian distribution:^[1] This transformation has been called the Geary–Hinkley transformation;^[7] the approximation is good if Y is unlikely to assume negative values, basically ${\displaystyle \mu _{y}>3\sigma _{y}}$.

Exact correlated noncentral normal ratio

This is developed by Dale (Springer 1979 problem 4.28) and Hinkley 1969. Geary showed how the correlated ratio ${\displaystyle z}$ could be transformed into a near-Gaussian form and developed an approximation for ${\displaystyle t}$ dependent on the probability of negative denominator values ${\displaystyle x+\mu _{x}<0}$ 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. Pham-Ghia has exhaustively discussed these methods. 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:

${\displaystyle z={\frac {x+\mu _{x}}{y+\mu _{y}}}}$

in which ${\displaystyle x,y}$ are zero-mean correlated normal variables with variances ${\displaystyle \sigma _{x}^{2},\sigma _{y}^{2}}$ and ${\displaystyle X,Y}$ have means ${\displaystyle \mu _{x},\mu _{y}.}$ Write ${\displaystyle x'=x-\rho y\sigma _{x}/\sigma _{y}}$ such that ${\displaystyle x',y}$ become uncorrelated and ${\displaystyle x'}$ has standard deviation

${\displaystyle \sigma _{x}'=\sigma _{x}{\sqrt {1-\rho ^{2}}}.}$

The ratio:

${\displaystyle z={\frac {x'+\rho y\sigma _{x}/\sigma _{y}+\mu _{x}}{y+\mu _{y}}}}$

is invariant under this transformation and retains the same pdf. The ${\displaystyle y}$ term in the numerator appears to be made separable by expanding:

${\displaystyle {x'+\rho y\sigma _{x}/\sigma _{y}+\mu _{x}}=x'+\mu _{x}-\rho \mu _{y}{\frac {\sigma _{x}}{\sigma _{y}}}+\rho (y+\mu _{y}){\frac {\sigma _{x}}{\sigma _{y}}}}$

to get

${\displaystyle z={\frac {x'+\mu _{x}'}{y+\mu _{y}}}+\rho {\frac {\sigma _{x}}{\sigma _{y}}}}$

in which ${\textstyle \mu '_{x}=\mu _{x}-\rho \mu _{y}{\frac {\sigma _{x}}{\sigma _{y}}}}$ and z has now become a ratio of uncorrelated non-central normal samples with an invariant z-offset (this is not formally proven, though appears to have been used by Geary),

Finally, to be explicit, the pdf of the ratio ${\displaystyle z}$ for correlated variables is found by inputting the modified parameters ${\displaystyle \sigma _{x}',\mu _{x}',\sigma _{y},\mu _{y}}$ and ${\displaystyle \rho '=0}$ into the Hinkley equation above which returns the pdf for the correlated ratio with a constant offset ${\displaystyle -\rho {\frac {\sigma _{x}}{\sigma _{y}}}}$ on ${\displaystyle z}$.

Contours of the correlated bivariate Gaussian distribution (not to scale) giving ratio x/y

pdf of the Gaussian ratio z and a simulation (points) for
${\displaystyle \sigma _{x}=\sigma _{y}=1,\mu _{x}=0,\mu _{y}=0.5,\rho =0.975}$

The figures above show an example of a positively correlated ratio with ${\displaystyle \sigma _{x}=\sigma _{y}=1,\mu _{x}=0,\mu _{y}=0.5,\rho =0.975}$ in which the shaded wedges represent the increment of area selected by given ratio ${\displaystyle x/y\in [r,r+\delta ]}$ 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 ${\displaystyle z=x/y\approx 1}$ the wedge has almost bypassed the main distribution mass altogether and this explains the local minimum in the theoretical pdf ${\displaystyle p_{Z}(x/y)}$. Conversely as ${\displaystyle x/y}$ 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.^[13] and has since been extended to the nonzero-mean and nonsymmetric case.^[14] In the correlated zero-mean case, the joint distribution of x, y is

${\displaystyle f_{x,y}(x,y)={\frac {1}{\pi ^{2}|\Sigma |}}\exp \left(-{\begin{bmatrix}x\\y\end{bmatrix}}^{H}\Sigma ^{-1}{\begin{bmatrix}x\\y\end{bmatrix}}\right)}$

where

${\displaystyle \Sigma ={\begin{bmatrix}\sigma _{x}^{2}&\rho \sigma _{x}\sigma _{y}\\\rho ^{*}\sigma _{x}\sigma _{y}&\sigma _{y}^{2}\end{bmatrix}},\;\;x=x_{r}+ix_{i},\;\;y=y_{r}+iy_{i}}$

${\displaystyle (\cdot )^{H}}$ is an Hermitian transpose and

${\displaystyle \rho =\rho _{r}+i\rho _{i}=\operatorname {E} {\bigg (}{\frac {xy^{*}}{\sigma _{x}\sigma _{y}}}{\bigg )}\;\;\in \;\left|\mathbb {C} \right|\leq 1}$

The PDF of ${\displaystyle Z=X/Y}$ is found to be

${\displaystyle {\begin{aligned}f_{z}(z_{r},z_{i})&={\frac {1-|\rho |^{2}}{\pi \sigma _{x}^{2}\sigma _{y}^{2}}}{\Biggr (}{\frac {|z|^{2}}{\sigma _{x}^{2}}}+{\frac {1}{\sigma _{y}^{2}}}-2{\frac {\rho _{r}z_{r}-\rho _{i}z_{i}}{\sigma _{x}\sigma _{y}}}{\Biggr )}^{-2}\\&={\frac {1-|\rho |^{2}}{\pi \sigma _{x}^{2}\sigma _{y}^{2}}}{\Biggr (}\;\;{\Biggr |}{\frac {z}{\sigma _{x}}}-{\frac {\rho ^{*}}{\sigma _{y}}}{\Biggr |}^{2}+{\frac {1-|\rho |^{2}}{\sigma _{y}^{2}}}{\Biggr )}^{-2}\end{aligned}}}$

In the usual event that ${\displaystyle \sigma _{x}=\sigma _{y}}$ we get

${\displaystyle f_{z}(z_{r},z_{i})={\frac {1-|\rho |^{2}}{\pi \left(\;\;|z-\rho ^{*}|^{2}+1-|\rho |^{2}\right)^{2}}}}$

Further closed-form results for the CDF are also given.

The ratio distribution of correlated complex variables, rho = 0.7 exp(i pi/4).

The graph shows the pdf of the ratio of two complex normal variables with a correlation coefficient of ${\displaystyle \rho =0.7\exp(i\pi /4)}$. The pdf peak occurs at roughly the complex conjugate of a scaled down ${\displaystyle \rho }$.

Ratio of log-normal

The ratio of independent or correlated log-normals is log-normal. This follows, because if ${\displaystyle X_{1}}$ and ${\displaystyle X_{2}}$ are log-normally distributed, then ${\displaystyle \ln(X_{1})}$ and ${\displaystyle \ln(X_{2})}$ 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.^{[note 1]}

This is important for many applications requiring the ratio of random variables that must be positive, where joint distribution of ${\displaystyle X_{1}}$ and ${\displaystyle X_{2}}$ 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 ${\displaystyle X_{i}}$ is the result of an accumulation of many small percentage changes and must be positive and approximately log-normally distributed.^[15]

Uniform ratio distribution

With two independent random variables following a uniform distribution, e.g.,

${\displaystyle p_{X}(x)={\begin{cases}1&0<x<1\\0&{\text{otherwise}}\end{cases}}}$

the ratio distribution becomes

${\displaystyle p_{Z}(z)={\begin{cases}1/2\qquad &0<z<1\\{\frac {1}{2z^{2}}}\qquad &z\geq 1\\0\qquad &{\text{otherwise}}\end{cases}}}$

Cauchy ratio distribution

If two independent random variables, X and Y each follow a Cauchy distribution with median equal to zero and shape factor ${\displaystyle a}$

${\displaystyle p_{X}(x|a)={\frac {a}{\pi (a^{2}+x^{2})}}}$

then the ratio distribution for the random variable ${\displaystyle Z=X/Y}$ is^[16]

${\displaystyle p_{Z}(z|a)={\frac {1}{\pi ^{2}(z^{2}-1)}}\ln(z^{2}).}$

This distribution does not depend on ${\displaystyle a}$ and the result stated by Springer^[8] (p158 Question 4.6) is not correct. The ratio distribution is similar to but not the same as the product distribution of the random variable ${\displaystyle W=XY}$:

${\displaystyle p_{W}(w|a)={\frac {a^{2}}{\pi ^{2}(w^{2}-a^{4})}}\ln \left({\frac {w^{2}}{a^{4}}}\right).}$^[8]

More generally, if two independent random variables X and Y each follow a Cauchy distribution with median equal to zero and shape factor ${\displaystyle a}$ and ${\displaystyle b}$ respectively, then:

1. The ratio distribution for the random variable ${\displaystyle Z=X/Y}$ is^[16]
   $${\displaystyle p_{Z}(z|a,b)={\frac {ab}{\pi ^{2}(b^{2}z^{2}-a^{2})}}\ln \left({\frac {b^{2}z^{2}}{a^{2}}}\right).}$$

   
2. The product distribution for the random variable ${\displaystyle W=XY}$ is^[16]
   $${\displaystyle p_{W}(w|a,b)={\frac {ab}{\pi ^{2}(w^{2}-a^{2}b^{2})}}\ln \left({\frac {w^{2}}{a^{2}b^{2}}}\right).}$$

   

The result for the ratio distribution can be obtained from the product distribution by replacing ${\displaystyle b}$ with ${\displaystyle {\frac {1}{b}}.}$

Ratio of standard normal to standard uniform

If X has a standard normal distribution and Y has a standard uniform distribution, then Z = X / Y has a distribution known as the slash distribution, with probability density function

${\displaystyle p_{Z}(z)={\begin{cases}\left[\varphi (0)-\varphi (z)\right]/z^{2}\quad &z\neq 0\\\varphi (0)/2\quad &z=0,\\\end{cases}}}$

where φ(z) is the probability density function of the standard normal distribution.^[17]

Chi-squared, Gamma, Beta distributions

Let G be a normal(0,1) distribution, Y and Z be chi-squared distributions with m and n degrees of freedom respectively, all independent, with ${\displaystyle f_{\chi }(x,k)={\frac {x^{{\frac {k}{2}}-1}e^{-x/2}}{2^{k/2}\Gamma (k/2)}}}$. Then

${\displaystyle {\frac {G}{\sqrt {Y/m}}}\sim t_{m}}$ the Student's t distribution

${\displaystyle {\frac {Y/m}{Z/n}}=F_{m,n}}$ i.e. Fisher's F-test distribution

${\displaystyle {\frac {Y}{Y+Z}}\sim \beta ({\tfrac {m}{2}},{\tfrac {n}{2}})}$ the beta distribution

${\displaystyle \;\;{\frac {Y}{Z}}\sim \beta '({\tfrac {m}{2}},{\tfrac {n}{2}})}$ the standard beta prime distribution

If ${\displaystyle V_{1}\sim {\chi '}_{k_{1}}^{2}(\lambda )}$, a noncentral chi-squared distribution, and ${\displaystyle V_{2}\sim {\chi '}_{k_{2}}^{2}(0)}$ and ${\displaystyle V_{1}}$ is independent of ${\displaystyle V_{2}}$ then

${\displaystyle {\frac {V_{1}/k_{1}}{V_{2}/k_{2}}}\sim F'_{k_{1},k_{2}}(\lambda )}$, a noncentral F-distribution.

${\displaystyle {\frac {m}{n}}F'_{m,n}=\beta '({\tfrac {m}{2}},{\tfrac {n}{2}}){\text{ or }}F'_{m,n}=\beta '({\tfrac {m}{2}},{\tfrac {n}{2}},1,{\tfrac {n}{m}})}$ defines ${\displaystyle F'_{m,n}}$, Fisher's F density distribution, the PDF of the ratio of two Chi-squares with m, n degrees of freedom.

The CDF of the Fisher density, found in F-tables is defined in the beta prime distribution article. If we enter an F-test table with m = 3, n = 4 and 5% probability in the right tail, the critical value is found to be 6.59. This coincides with the integral

${\displaystyle F_{3,4}(6.59)=\int _{6.59}^{\infty }\beta '(x;{\tfrac {m}{2}},{\tfrac {n}{2}},1,{\tfrac {n}{m}})dx=0.05}$

For gamma distributions U and V with arbitrary shape parameters α₁ and α₂ and their scale parameters both set to unity, that is, ${\displaystyle U\sim \Gamma (\alpha _{1},1),V\sim \Gamma (\alpha _{2},1)}$, where ${\displaystyle \Gamma (x;\alpha ,1)={\frac {x^{\alpha -1}e^{-x}}{\Gamma (\alpha )}}}$, then

${\displaystyle {\frac {U}{U+V}}\sim \beta (\alpha _{1},\alpha _{2}),\qquad {\text{ expectation }}={\frac {\alpha _{1}}{\alpha _{1}+\alpha _{2}}}}$

${\displaystyle {\frac {U}{V}}\sim \beta '(\alpha _{1},\alpha _{2}),\qquad \qquad {\text{ expectation }}={\frac {\alpha _{1}}{\alpha _{2}-1}},\;\alpha _{2}>1}$

${\displaystyle {\frac {V}{U}}\sim \beta '(\alpha _{2},\alpha _{1}),\qquad \qquad {\text{ expectation }}={\frac {\alpha _{2}}{\alpha _{1}-1}},\;\alpha _{1}>1}$

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

If ${\displaystyle U\sim \Gamma (\alpha _{1},\theta _{1}),\;V\sim \Gamma (\alpha _{2},\theta _{2})}$, then by rescaling the ${\displaystyle \theta }$ parameter to unity we have

${\displaystyle {\frac {\frac {U}{\theta _{1}}}{{\frac {U}{\theta _{1}}}+{\frac {V}{\theta _{2}}}}}={\frac {\theta _{2}U}{\theta _{2}U+\theta _{1}V}}\sim \beta (\alpha _{1},\alpha _{2})}$

${\displaystyle {\frac {\frac {U}{\theta _{1}}}{\frac {V}{\theta _{2}}}}={\frac {\theta _{2}}{\theta _{1}}}{\frac {U}{V}}\sim \beta '(\alpha _{1},\alpha _{2})}$

Thus

${\displaystyle {\frac {U}{V}}\sim \beta '(\alpha _{1},\alpha _{2},1,{\frac {\theta _{1}}{\theta _{2}}})\quad {\text{ and }}\operatorname {E} \left[{\frac {U}{V}}\right]={\frac {\theta _{1}}{\theta _{2}}}{\frac {\alpha _{1}}{\alpha _{2}-1}}}$

in which ${\displaystyle \beta '(\alpha ,\beta ,p,q)}$ represents the generalised beta prime distribution.

In the foregoing it is apparent that if ${\displaystyle X\sim \beta '(\alpha _{1},\alpha _{2},1,1)\equiv \beta '(\alpha _{1},\alpha _{2})}$ then ${\displaystyle \theta X\sim \beta '(\alpha _{1},\alpha _{2},1,\theta )}$. More explicitly, since

${\displaystyle \beta '(x;\alpha _{1},\alpha _{2},1,R)={\frac {1}{R}}\beta '({\frac {x}{R}};\alpha _{1},\alpha _{2})}$

if ${\displaystyle U\sim \Gamma (\alpha _{1},\theta _{1}),V\sim \Gamma (\alpha _{2},\theta _{2})}$ then

${\displaystyle {\frac {U}{V}}\sim {\frac {1}{R}}\beta '({\frac {x}{R}};\alpha _{1},\alpha _{2})={\frac {\left({\frac {x}{R}}\right)^{\alpha _{1}-1}}{\left(1+{\frac {x}{R}}\right)^{\alpha _{1}+\alpha _{2}}}}\cdot {\frac {1}{\;R\;B(\alpha _{1},\alpha _{2})}},\;\;x\geq 0}$

where

${\displaystyle R={\frac {\theta _{1}}{\theta _{2}}},\;\;\;B(\alpha _{1},\alpha _{2})={\frac {\Gamma (\alpha _{1})\Gamma (\alpha _{2})}{\Gamma (\alpha _{1}+\alpha _{2})}}}$

Rayleigh Distributions

If X, Y are independent samples from the Rayleigh distribution ${\displaystyle f_{r}(r)=(r/\sigma ^{2})e^{-r^{2}/2\sigma ^{2}},\;\;r\geq 0}$, the ratio Z = X/Y follows the distribution^[18]

${\displaystyle f_{z}(z)={\frac {2z}{(1+z^{2})^{2}}},\;\;z\geq 0}$

and has cdf

${\displaystyle F_{z}(z)=1-{\frac {1}{1+z^{2}}}={\frac {z^{2}}{1+z^{2}}},\;\;\;z\geq 0}$

The Rayleigh distribution has scaling as its only parameter. The distribution of ${\displaystyle Z=\alpha X/Y}$ follows

${\displaystyle f_{z}(z,\alpha )={\frac {2\alpha z}{(\alpha +z^{2})^{2}}},\;\;z>0}$

and has cdf

${\displaystyle F_{z}(z,\alpha )={\frac {z^{2}}{\alpha +z^{2}}},\;\;\;z\geq 0}$

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

The generalized gamma distribution is

${\displaystyle f(x;a,d,r)={\frac {r}{\Gamma (d/r)a^{d}}}x^{d-1}e^{-(x/a)^{r}}\;x\geq 0;\;\;a,\;d,\;r>0}$

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.

If ${\displaystyle U\sim f(x;a_{1},d_{1},r),\;\;V\sim f(x;a_{2},d_{2},r){\text{ are independent, and }}W=U/V}$

then^[19] ${\textstyle g(w)={\frac {r\left({\frac {a_{1}}{a_{2}}}\right)^{d_{2}}}{B\left({\frac {d_{1}}{r}},{\frac {d_{2}}{r}}\right)}}{\frac {w^{-d_{2}-1}}{\left(1+\left({\frac {a_{2}}{a_{1}}}\right)^{-r}w^{-r}\right)^{\frac {d_{1}+d_{2}}{r}}}},\;\;w>0}$

where ${\displaystyle B(u,v)={\frac {\Gamma (u)\Gamma (v)}{\Gamma (u+v)}}}$