Note: This page has been translated by MathWorks. Please click here

To view all translated materals including this page, select Japan from the country navigator on the bottom of this page.

To view all translated materals including this page, select Japan from the country navigator on the bottom of this page.

*F* inverse cumulative distribution function

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

`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

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

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.

Was this topic helpful?