What is the distribution of the ratio of two normals?

A related question is A/B testing ratio of sums The following is from a part of an answer to that question.

(You state that both variables are positive. And thus you do not have real normal distributions, which can take the value zero. You will have something that resembles the Cauchy distribution, but it does not have the undefined mean.)

For two (asymptomatically) Gaussian distributed variables (that are potentially correlated) you can use the Delta method or use an exact expression to express the ratio distribution.

The use of the Delta method for the estimation of ratio's is described here. The result of this application of the Delta method actually coincides with an approximation of Hinkley's result, an exact expression for the ratio of two correlated normal distributed variables (Hinkley D.V., 1969, On the Ratio of Two Correlated Normal Random Variables, Biometrica vol. 56 no. 3).

For $Z = \frac{X}{Y}$ with $$ \begin{bmatrix}X\\Y\end{bmatrix} \sim N\left(\begin{bmatrix} \mu_x \\ \mu_y \end{bmatrix} , \begin{bmatrix} \sigma_x^2 & \rho \sigma_x \sigma_y \\ \rho \sigma_x \sigma_y & \sigma_y^2 \end{bmatrix} \right) $$ The exact result is: $$ f(z) = \frac{b(z)d(z)}{a(z)^3} \frac{1}{\sqrt{2\pi} \sigma_X\sigma_Y} \left[ \Phi \left( \frac{b(z)}{\sqrt{1-\rho^2}a(z)} \right) - \Phi \left( - \frac{b(z)}{\sqrt{1-\rho^2}a(z)} \right) \right] + \frac{\sqrt{1-\rho^2}}{\pi \sigma_X \sigma_Y a(z)^2} \exp \left( -\frac{c}{2(1-\rho^2)}\right) $$ with $$ \begin{array}{} a(z) &=& \left( \frac{z^2}{\sigma_X^2} - \frac{2 \rho z}{\sigma_X \sigma_Y} + \frac{1}{\sigma_Y^2} \right) ^{\frac{1}{2}} \\ b(z) &=& \frac{\mu_X z}{ \sigma_X^2} - \frac{\rho (\mu_X+ \mu_Y z)}{ \sigma_X \sigma_Y} + \frac{\mu_Y}{\sigma_Y^2} \\ c &=& \frac{\mu_X^2}{\sigma_Y^2} - \frac{2 \rho \mu_X \mu_Y + }{\sigma_X \sigma_Y} + \frac{\mu_Y^2}{\sigma_Y^2}\\ d(z) &=& \text{exp} \left( \frac {b(z)^2 - c a(z) ^2}{2(1-\rho^2)a(z)^2}\right) \end{array}$$ And an approximation based on an assymptotic behaviour is: (for $\mu_Y/\sigma_Y \to \infty$): $$ F(z) \to \Phi\left( \frac{z - \mu_X/\mu_Y}{\sigma_X \sigma_Y a(z)/\mu_Y} \right) $$ You end up with the Delta method result when you insert the approximation $a(z) = a(\mu_X/\mu_Y)$ $$a(z) \sigma_X \sigma_Y /\mu_Y \approx a(\mu_X/\mu_Y) \sigma_X \sigma_Y /\mu_Y = \left( \frac{\mu_X^2\sigma_Y^2}{\mu_Y^4} - \frac{2 \mu_X \rho \sigma_X \sigma_Y}{\mu_Y^3} + \frac{\sigma_X^2}{\mu_Y^2} \right) ^{\frac{1}{2}}$$

What is the distribution of the ratio of two normals?

A related question is A/B testing ratio of sums The following is from a part of an answer to that question.

(You state that both variables are positive. And thus you do not have real normal distributions, which can take the value zero. You will have something that resembles the Cauchy distribution, but it does not have the undefined mean.)

For two (asymptomatically) Gaussian distributed variables (that are potentially correlated) you can use the Delta method or use an exact expression to express the ratio distribution.

The use of the Delta method for the estimation of ratio's is described here. The result of this application of the Delta method actually coincides with an approximation of Hinkley's result, an exact expression for the ratio of two correlated normal distributed variables (Hinkley D.V., 1969, On the Ratio of Two Correlated Normal Random Variables, Biometrica vol. 56 no. 3).

For $Z = \frac{X}{Y}$ with $$ \begin{bmatrix}X\\Y\end{bmatrix} \sim N\left(\begin{bmatrix} \mu_x \\ \mu_y \end{bmatrix} , \begin{bmatrix} \sigma_x^2 & \rho \sigma_x \sigma_y \\ \rho \sigma_x \sigma_y & \sigma_y^2 \end{bmatrix} \right) $$ The exact result is: $$ f(z) = \frac{b(z)d(z)}{a(z)^3} \frac{1}{\sqrt{2\pi} \sigma_X\sigma_Y} \left[ \Phi \left( \frac{b(z)}{\sqrt{1-\rho^2}a(z)} \right) - \Phi \left( - \frac{b(z)}{\sqrt{1-\rho^2}a(z)} \right) \right] + \frac{\sqrt{1-\rho^2}}{\pi \sigma_X \sigma_Y a(z)^2} \exp \left( -\frac{c}{2(1-\rho^2)}\right) $$ with $$ \begin{array}{} a(z) &=& \left( \frac{z^2}{\sigma_X^2} - \frac{2 \rho z}{\sigma_X \sigma_Y} + \frac{1}{\sigma_Y^2} \right) ^{\frac{1}{2}} \\ b(z) &=& \frac{\mu_X z}{ \sigma_X^2} - \frac{\rho (\mu_X+ \mu_Y z)}{ \sigma_X \sigma_Y} + \frac{\mu_Y}{\sigma_Y^2} \\ c &=& \frac{\mu_X^2}{\sigma_Y^2} - \frac{2 \rho \mu_X \mu_Y + }{\sigma_X \sigma_Y} + \frac{\mu_Y^2}{\sigma_Y^2}\\ d(z) &=& \text{exp} \left( \frac {b(z)^2 - c a(z) ^2}{2(1-\rho^2)a(z)^2}\right) \end{array}$$ And an approximation based on an assymptotic behaviour is: (for $\mu_Y/\sigma_Y \to \infty$): $$ F(z) \to \Phi\left( \frac{z - \mu_X/\mu_Y}{\sigma_X \sigma_Y a(z)/\mu_Y} \right) $$ You end up with the Delta method result when you insert the approximation $a(z) = a(\mu_X/\mu_Y)$ $$a(z) \sigma_X \sigma_Y /\mu_Y \approx a(\mu_X/\mu_Y) \sigma_X \sigma_Y /\mu_Y = \left( \frac{\mu_X^2\sigma_Y^2}{\mu_Y^4} - \frac{2 \mu_X \rho \sigma_X \sigma_Y}{\mu_Y^3} + \frac{\sigma_X^2}{\mu_Y^2} \right) ^{\frac{1}{2}}$$

(Sidenote: The above quote has been updated with some references to earlier descriptions and simpler descriptions. An earlier description of the exact expression was given by George Marsaglia 1965 in the JASA Vol. 60, No. 309. A simple modern description is given in 2006 in Jstatsoft Volume 16 Issue 4)

What is the distribution of the ratio of two normals?

A related question is A/B testing ratio of sums The following from a part of an answer to that question.

(And since you state that both variables are positive, and thus not real normal distributions, which can take the value zero, you will have something that might resemble the Cauchy distribution, but does not have undefined mean)

For two (asymptomatically) normal distributed that are potentially correlated you can use the Delta method or use an exact expression to express the ratio distribution.

The use of the Delta method for the estimation of ratio's is described here. The result of this application of the Delta method actually coincides with an approximation of Hinkley's result, an exact expression for the ratio of two correlated normal distributed variables (Hinkley D.V., 1969, On the Ratio of Two Correlated Normal Random Variables, Biometrica vol. 56 no. 3).

For $Z = \frac{X}{Y}$ with $$ \begin{bmatrix}X\\Y\end{bmatrix} \sim N\left(\begin{bmatrix} \mu_x \\ \mu_y \end{bmatrix} , \begin{bmatrix} \sigma_x^2 & \rho \sigma_x \sigma_y \\ \rho \sigma_x \sigma_y & \sigma_y^2 \end{bmatrix} \right) $$ The exact result is: $$ f(z) = \frac{b(z)d(z)}{a(z)^3} \frac{1}{\sqrt{2\pi} \sigma_X\sigma_Y} \left[ \Phi \left( \frac{b(z)}{\sqrt{1-\rho^2}a(z)} \right) - \Phi \left( - \frac{b(z)}{\sqrt{1-\rho^2}a(z)} \right) \right] + \frac{\sqrt{1-\rho^2}}{\pi \sigma_X \sigma_Y a(z)^2} \exp \left( -\frac{c}{2(1-\rho^2)}\right) $$ with $$ \begin{array}{} a(z) &=& \left( \frac{z^2}{\sigma_X^2} - \frac{2 \rho z}{\sigma_X \sigma_Y} + \frac{1}{\sigma_Y^2} \right) ^{\frac{1}{2}} \\ b(z) &=& \frac{\mu_X z}{ \sigma_X^2} - \frac{\rho (\mu_X+ \mu_Y z)}{ \sigma_X \sigma_Y} + \frac{\mu_Y}{\sigma_Y^2} \\ c &=& \frac{\mu_X^2}{\sigma_Y^2} - \frac{2 \rho \mu_X \mu_Y + }{\sigma_X \sigma_Y} + \frac{\mu_Y^2}{\sigma_Y^2}\\ d(z) &=& \text{exp} \left( \frac {b(z)^2 - c a(z) ^2}{2(1-\rho^2)a(z)^2}\right) \end{array}$$ And an approximation based on an assymptotic behaviour is: (for $\mu_Y/\sigma_Y \to \infty$): $$ F(z) \to \Phi\left( \frac{z - \mu_X/\mu_Y}{\sigma_X \sigma_Y a(z)/\mu_Y} \right) $$ You end up with the Delta method result when you insert the approximation $a(z) = a(\mu_X/\mu_Y)$ $$a(z) \sigma_X \sigma_Y /\mu_Y \approx a(\mu_X/\mu_Y) \sigma_X \sigma_Y /\mu_Y = \left( \frac{\mu_X^2\sigma_Y^2}{\mu_Y^4} - \frac{2 \mu_X \rho \sigma_X \sigma_Y}{\mu_Y^3} + \frac{\sigma_X^2}{\mu_Y^2} \right) ^{\frac{1}{2}}$$

The values for $\mu_X,\mu_Y,\sigma_X,\sigma_Y,\rho$ can be estimated from your observations which allow you to estimate the variance and mean of the distribution for single users and related to this the variance and mean for the sample distribution of the sum of several users.

What is the distribution of the ratio of two normals?

A related question is A/B testing ratio of sums The following is from a part of an answer to that question.

(You state that both variables are positive. And thus you do not have real normal distributions, which can take the value zero. You will have something that resembles the Cauchy distribution, but it does not have the undefined mean.)

For two (asymptomatically) Gaussian distributed variables (that are potentially correlated) you can use the Delta method or use an exact expression to express the ratio distribution.

The use of the Delta method for the estimation of ratio's is described here. The result of this application of the Delta method actually coincides with an approximation of Hinkley's result, an exact expression for the ratio of two correlated normal distributed variables (Hinkley D.V., 1969, On the Ratio of Two Correlated Normal Random Variables, Biometrica vol. 56 no. 3).

For $Z = \frac{X}{Y}$ with $$ \begin{bmatrix}X\\Y\end{bmatrix} \sim N\left(\begin{bmatrix} \mu_x \\ \mu_y \end{bmatrix} , \begin{bmatrix} \sigma_x^2 & \rho \sigma_x \sigma_y \\ \rho \sigma_x \sigma_y & \sigma_y^2 \end{bmatrix} \right) $$ The exact result is: $$ f(z) = \frac{b(z)d(z)}{a(z)^3} \frac{1}{\sqrt{2\pi} \sigma_X\sigma_Y} \left[ \Phi \left( \frac{b(z)}{\sqrt{1-\rho^2}a(z)} \right) - \Phi \left( - \frac{b(z)}{\sqrt{1-\rho^2}a(z)} \right) \right] + \frac{\sqrt{1-\rho^2}}{\pi \sigma_X \sigma_Y a(z)^2} \exp \left( -\frac{c}{2(1-\rho^2)}\right) $$ with $$ \begin{array}{} a(z) &=& \left( \frac{z^2}{\sigma_X^2} - \frac{2 \rho z}{\sigma_X \sigma_Y} + \frac{1}{\sigma_Y^2} \right) ^{\frac{1}{2}} \\ b(z) &=& \frac{\mu_X z}{ \sigma_X^2} - \frac{\rho (\mu_X+ \mu_Y z)}{ \sigma_X \sigma_Y} + \frac{\mu_Y}{\sigma_Y^2} \\ c &=& \frac{\mu_X^2}{\sigma_Y^2} - \frac{2 \rho \mu_X \mu_Y + }{\sigma_X \sigma_Y} + \frac{\mu_Y^2}{\sigma_Y^2}\\ d(z) &=& \text{exp} \left( \frac {b(z)^2 - c a(z) ^2}{2(1-\rho^2)a(z)^2}\right) \end{array}$$ And an approximation based on an assymptotic behaviour is: (for $\mu_Y/\sigma_Y \to \infty$): $$ F(z) \to \Phi\left( \frac{z - \mu_X/\mu_Y}{\sigma_X \sigma_Y a(z)/\mu_Y} \right) $$ You end up with the Delta method result when you insert the approximation $a(z) = a(\mu_X/\mu_Y)$ $$a(z) \sigma_X \sigma_Y /\mu_Y \approx a(\mu_X/\mu_Y) \sigma_X \sigma_Y /\mu_Y = \left( \frac{\mu_X^2\sigma_Y^2}{\mu_Y^4} - \frac{2 \mu_X \rho \sigma_X \sigma_Y}{\mu_Y^3} + \frac{\sigma_X^2}{\mu_Y^2} \right) ^{\frac{1}{2}}$$