# finv

F inverse cumulative distribution function

## Syntax

```X = finv(P,V1,V2) ```

## Description

`X = finv(P,V1,V2)` computes the inverse of the F cdf with numerator degrees of freedom `V1` and denominator degrees of freedom `V2` for the corresponding probabilities in `P`. `P`, `V1`, and `V2` can be vectors, matrices, or multidimensional arrays that all have the same size. A scalar input is expanded to a constant array with the same dimensions as the other inputs. `V1` and `V2` parameters must contain real positive values, and the values in `P` must lie on the interval [0 1].

The F inverse function is defined in terms of the F cdf as

`$x={F}^{-1}\left(p|{\nu }_{1},{\nu }_{2}\right)=\left\{x:F\left(x|{\nu }_{1},{\nu }_{2}\right)=p\right\}$`

where

`$p=F\left(x|{\nu }_{1},{\nu }_{2}\right)=\underset{0}{\overset{x}{\int }}\frac{\Gamma \left[\frac{\left({\nu }_{1}+{\nu }_{2}\right)}{2}\right]}{\Gamma \left(\frac{{\nu }_{1}}{2}\right)\Gamma \left(\frac{{\nu }_{2}}{2}\right)}{\left(\frac{{\nu }_{1}}{{\nu }_{2}}\right)}^{\frac{{\nu }_{1}}{2}}\frac{{t}^{\frac{{\nu }_{1}-2}{2}}}{{\left[1+\left(\frac{{\nu }_{1}}{{\nu }_{2}}\right)t\right]}^{\frac{{\nu }_{1}+{\nu }_{2}}{2}}}dt$`

## Examples

Find a value that should exceed 95% of the samples from an F distribution with 5 degrees of freedom in the numerator and 10 degrees of freedom in the denominator.

```x = finv(0.95,5,10) x = 3.3258 ```

You would observe values greater than 3.3258 only 5% of the time by chance.