User:Beakerboy/sandbox2

Converting one form of the CDF to the other:
 * $$F(q;k,\nu)=\frac{k\nu^{\frac{\nu}{2}}}{\Gamma\left(\frac{\nu}{2}\right)2^{\frac{\nu}{2}-1}}\int_0^\infty x^{\nu-1}e^{\frac{-\nu x^2}{2}}\left[\int_{-\infty}^\infty \phi(u) (\Phi(u)-\Phi(u-qx))^{k-1} \, \text{d}u\right]\text{d}x$$


 * $$f(q;k,\nu)=\frac{\text{d}F(q;k,\nu)}{\text{d}q}=\frac{k\nu^{\frac{\nu}{2}}}{\Gamma\left(\frac{\nu}{2}\right)2^{\frac{\nu}{2}-1}}\int_0^\infty x^{\nu-1}e^{\frac{-\nu x^2}{2}}\left[\int_{-\infty}^\infty \phi(u) \frac{\text{d}(\Phi(u)-\Phi(u-qx))^{k-1}}{\text{d}q} \, \text{d}u\right]\text{d}x$$


 * $$f(q;k,\nu)=\frac{k(k-1)\nu^{\frac{\nu}{2}}}{\Gamma\left(\frac{\nu}{2}\right)2^{\frac{\nu}{2}-1}}\int_0^\infty x^{\nu}e^{\frac{-\nu x^2}{2}}\left[\int_{-\infty}^\infty \phi(u) \phi(u-qx)(\Phi(u)-\Phi(u-qx))^{k-2} \, \text{d}u\right]\text{d}x$$


 * $$f(q;k,\nu)=\frac{k(k-1)\nu^{\frac{\nu}{2}}}{\Gamma\left(\frac{\nu}{2}\right)2^{\frac{\nu}{2}-1}}\int_0^\infty x^{\nu}e^{\frac{-\nu x^2}{2}}\left[\int_{-\infty}^\infty \phi(u) \phi(u+qx)(\Phi(u+qx)-\Phi(u))^{k-2} \, \text{d}u\right]\text{d}x$$


 * $$F(q;k,\nu)=\frac{k(k-1)\nu^{\frac{\nu}{2}}}{\Gamma\left(\frac{\nu}{2}\right)2^{\frac{\nu}{2}-1}}\int_0^\infty x^{\nu}e^{\frac{-\nu x^2}{2}}\left[\int_{-\infty}^\infty \phi(u) \left(\int_{-\infty}^q\phi(u+qx)(\Phi(u+qx)-\Phi(u))^{k-2} \text{d}q\right) \, \text{d}u\right]\text{d}x$$


 * $$1-F(q;k,\nu)=\frac{k(k-1)\nu^{\frac{\nu}{2}}}{\Gamma\left(\frac{\nu}{2}\right)2^{\frac{\nu}{2}-1}}\int_0^\infty x^{\nu}e^{\frac{-\nu x^2}{2}}\left[\int_{-\infty}^\infty \phi(u) \left(\int_q^\infty\phi(u+qx)(\Phi(u+qx)-\Phi(u))^{k-2} \text{d}q\right)\text{d}u\right]\text{d}x$$


 * $$1-F(q;k,\nu)=\frac{\nu^{\frac{\nu}{2}}}{\Gamma\left(\frac{\nu}{2}\right)2^{\frac{\nu}{2}-1}}\int_0^\infty x^{\nu-1}e^{\frac{-\nu x^2}{2}}\left[1-k\int_{-\infty}^\infty \phi(u)(\Phi(u+qx)-\Phi(u))^{k-1}\text{d}u\right]\text{d}x$$

if $$t=qx$$ then $$\frac{\text{d}t}{t} = \frac{\text{d}x}{x}$$


 * $$1-F(q;k,\nu)=\frac{\nu^{\frac{\nu}{2}}}{\Gamma\left(\frac{\nu}{2}\right)2^{\frac{\nu}{2}-1}}\int_0^\infty \left(\left(\frac{t}{q}\right)e^{\frac{-\left(\frac{t}{q}\right)^2}{2}}\right)^\nu\left[1-k\int_{-\infty}^\infty \phi(u) (\Phi(u+t)-\Phi(u))^{k-1}\text{d}u\right]\frac{\text{d}t}{t}$$


 * $$1-F(q;k,\nu)=\frac{\sqrt{2\pi}^\nu \nu^{\frac{\nu}{2}}}{\Gamma\left(\frac{\nu}{2}\right)2^{\frac{\nu}{2}-1}}\int_0^\infty \left(\left(\frac{t}{q}\right)\phi\left(\frac{t}{q}\right)\right)^\nu\left[1-k\int_{-\infty}^\infty \phi(u) (\Phi(u+t)-\Phi(u))^{k-1}\text{d}u\right]\frac{\text{d}t}{t}$$


 * $$1-F(q;k,\nu)=\frac{2\left(\frac{\sqrt{\pi\nu}}{4}\right)^\nu}{\Gamma\left(\frac{\nu}{2}\right)}\int_0^\infty \left(4\left(\frac{t}{q}\right)\phi\left(\frac{t}{q}\right)\right)^\nu \left[1-k\int_{-\infty}^\infty \phi(u) (\Phi(u+t)-\Phi(u))^{k-1}\text{d}u\right]\frac{\text{d}t}{t}$$

if $$c_v=\frac{2\left(\frac{\sqrt{\pi\nu}}{4}\right)^\nu}{\Gamma\left(\frac{\nu}{2}\right)}$$ and $$P_n(t)=k\int_{-\infty}^\infty \phi(u) (\Phi(u+t)-\Phi(u))^{k-1}\text{d}u$$


 * $$1-F(q;k,\nu)=c_v\int_0^\infty \left(4\left(\frac{t}{q}\right)\phi\left(\frac{t}{q}\right)\right)^\nu(1- P_n(t))\frac{\text{d}t}{t}$$