User:Beakerboy/sandbox1

The studentized range is defined as $q={\frac {{\sqrt {n}}r}{{\sqrt {2}}S}}$. To create the studentized range PDF, we will need to integrate over all combinations of a range distribution, and the $\chi$ distribution which would give a value of q. The range distribution is:

$f(r;k)=k(k-1)\int _{-\infty }^{\infty }f(u-r)f(u)\left[F(u)-F(u-r)\right]^{k-2}\,{\text{d}}u$

The f and F are the averages of n standard normal variables, which is $f(z)={\sqrt {n}}\phi ({\sqrt {n}}z)$, and $F(z)=\phi ({\sqrt {n}}z)$. This gives us:

$f(r;k)=k(k-1)\int _{-\infty }^{\infty }{\sqrt {n}}\phi ({\sqrt {n}}(u-r)){\sqrt {n}}\phi ({\sqrt {n}}u)\left[\Phi ({\sqrt {n}}u)-\Phi ({\sqrt {n}}(u-r))\right]^{k-2}\,{\text{d}}u$

if $t={\sqrt {n}}u$ then ${\text{d}}t={\sqrt {n}}{\text{d}}u$

$f(r;k)=k(k-1){\sqrt {n}}\int _{-\infty }^{\infty }\phi (t-{\sqrt {n}}r)\phi (t)\left[\Phi (t)-\Phi (t-{\sqrt {n}}r)\right]^{k-2}{\text{d}}t$

Combining this with the $\chi$ distribution gives:

$f(q;k,\nu )={\frac {1}{2^{\frac {\nu }{2}}\Gamma \left({\frac {\nu }{2}}\right)}}\int _{0}^{\infty }x^{{\frac {\nu }{2}}-1}e^{-{\frac {x}{2}}}k(k-1){\sqrt {n}}\int _{-\infty }^{\infty }\phi (t-{\sqrt {n}}r)\phi (t)\left[\Phi (t)-\Phi (t-{\sqrt {n}}r)\right]^{k-2}{\text{d}}t{\text{d}}x$

or

$f(q;k,\nu )={\frac {1}{{\sqrt {2}}^{\nu }\Gamma \left({\frac {\nu }{2}}\right)}}\int _{0}^{\infty }{\sqrt {x}}^{\nu -2}e^{-{\frac {x}{2}}}k(k-1){\sqrt {n}}\int _{-\infty }^{\infty }\phi (t-{\sqrt {n}}r)\phi (t)\left[\Phi (t)-\Phi (t-{\sqrt {n}}r)\right]^{k-2}{\text{d}}t{\text{d}}x$

Since x in the $\chi ^{2}$distribution is defined as $x=\nu S^{2}$ we can set $y={\frac {\sqrt {2}}{\sqrt {\nu }}}{\sqrt {x}}$ to make ${\sqrt {n}}r=yq$ and ${\text{d}}y={\frac {{\text{d}}x}{\sqrt {2\nu x}}}$.

$f(q;k,\nu )={\frac {\sqrt {\nu }}{{\sqrt {2}}^{\nu -1}\Gamma \left({\frac {\nu }{2}}\right)}}\int _{0}^{\infty }{\sqrt {x}}^{\nu -1}e^{-{\frac {x}{2}}}k(k-1){\sqrt {n}}\int _{-\infty }^{\infty }\phi (t-{\sqrt {n}}r)\phi (t)\left[\Phi (t)-\Phi (t-{\sqrt {n}}r)\right]^{k-2}{\text{d}}t{\frac {{\text{d}}x}{\sqrt {\nu 2x}}}$

$f(q;k,\nu )={\frac {{\sqrt {\nu }}^{\nu }}{2^{\nu -1}\Gamma \left({\frac {\nu }{2}}\right)}}\int _{0}^{\infty }y^{\nu -1}e^{-{\frac {\nu y^{2}}{4}}}k(k-1){\sqrt {n}}\int _{-\infty }^{\infty }\phi (t-yq)\phi (t)\left[\Phi (t)-\Phi (t-yq)\right]^{k-2}{\text{d}}t{\text{d}}y$