An F-distribution

Problem

  • Setup:
    • a linear regression model \(y=Xβ+ε\) with a random sample
    • \(ε_i|x_i \sim N(0,σ^2)\)
    • \(R\) is an \(J×k\) matrix of parameters

Show that (conditionally on \(X\) )

\[ \frac{{\left( Rb-Rβ \right)}'{\left( R{\left( X'X \right)}^{-1}R' \right)}^{-1}\left( Rb-Rβ \right)}{Js^2} \sim F_{J,n-k}\]

Solution

Conditionally on \(X\) , \(b \sim N\left( β,σ^2{\left( X'X \right)}^{-1} \right)\) . \(Rb\) contains \(J\) a linear combination of normally distributed random variables so \(Rb\) must be (multivariate) normal. We have

\[E\left( Rb \right)=RE\left( b \right)=Rβ\]

and

\[Var\left( Rb \right)=RVar\left( b \right)R'=σ^2R{\left( X'X \right)}^{-1}R'\]

We have

\[Rb \sim N\left( Rβ,Var\left( Rb \right) \right)\]

or

\[Rb \sim N\left( Rβ,RVar\left( b \right)R' \right)\]

or

\[Rb \sim N\left( Rβ,σ^2R{\left( X'X \right)}^{-1}R' \right)\]

\(R{\left( X'X \right)}^{-1}R'\) is positive definite and (see Standardizing a multivariate normal random vector)

\[{\left( RVar\left( b \right)R' \right)}^{-1/2}(Rb-Rβ) \sim N(0,I_J)\]

Now if \(z \sim N\left( 0,I_J \right)\) then each \(z_j\) is \(N\left( 0,1 \right)\) and \(z_j^2 \sim χ_2^1\) and \(z_1^2+ \ldots +z_J^2=z'z \sim χ_2^J\) . Therefore,

\[{\left( {\left( RVar\left( b \right)R' \right)}^{-1/2}\left( Rb-Rβ \right) \right)}'\left( {\left( RVar\left( b \right)R' \right)}^{-1/2}(Rb-Rβ) \right) \sim χ_2^J\]

\(RVar\left( b \right)R'\) is symmetric so the left-hand side is

\[\left( Rb-Rβ \right){\left( RVar\left( b \right)R' \right)}^{-1/2}{\left( RVar\left( b \right)R' \right)}^{-1/2}(Rb-Rβ)=\]

\[{\left( Rb-Rβ \right)}'{\left( RVar\left( b \right)R' \right)}^{-1}\left( Rb-Rβ \right) \sim χ_J^2\]

Since

\[ \frac{\left( n-k \right)s^2}{σ^2} \sim χ_{n-k}^2\]

we have

\[ \frac{{\left( Rb-Rβ \right)}'{\left( RVar\left( b \right)R' \right)}^{-1}\left( Rb-Rβ \right)/J}{ \frac{\left( n-k \right)s^2}{σ^2}/(n-k)} \sim F_{J,n-k}\]

Simplifying the left-hand side

\[ \frac{{\left( Rb-Rβ \right)}'{\left( σ^2R{\left( X'X \right)}^{-1}R' \right)}^{-1}\left( Rb-Rβ \right)}{J} \frac{σ^2}{s^2}=\]

\[ \frac{{\left( Rb-Rβ \right)}'{\left( R{\left( X'X \right)}^{-1}R' \right)}^{-1}\left( Rb-Rβ \right)}{Js^2}\]