HP17bII+ Programming t-distribution
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07-25-2015, 10:12 AM
Post: #9
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RE: HP17bII+ Programming t-distribution
(07-23-2015 09:52 PM)Don Shepherd Wrote: according to the formula given at Wikipedia \(\frac{\Gamma \left(\frac{\nu+1}{2} \right)} {\sqrt{\nu\pi}\,\Gamma \left(\frac{\nu}{2} \right)} \left(1+\frac{x^2}{\nu} \right)^{-\frac{\nu+1}{2}}\) We can use the following identities:
For \(\nu\) even: \(\begin{align} \Gamma(\frac{\nu+1}{2})&=\frac{\nu-1}{2}\cdot\frac{\nu-3}{2}\cdots\frac{1}{2}\cdot\Gamma\left(\tfrac{1}{2}\right) \\ &=\frac{\nu-1}{2}\cdot\frac{\nu-2}{\nu-2}\cdot\frac{\nu-3}{2}\cdot\frac{\nu-4}{\nu-4}\cdots\frac{2}{2}\cdot\frac{1}{2}\cdot\Gamma\left(\tfrac{1}{2}\right) \\ &=\frac{\nu-1}{2}\cdot\frac{\nu-2}{2\cdot(\frac{\nu}{2}-1)}\cdot\frac{\nu-3}{2}\cdot\frac{\nu-4}{2\cdot(\frac{\nu}{2}-2)}\cdots\frac{2}{2\cdot1}\cdot\frac{1}{2}\cdot\Gamma\left(\tfrac{1}{2}\right) \\ &=\frac{(\nu-1)!}{2^{\nu-1}\,(\frac{\nu}{2}-1)!}\,\sqrt{\pi} \\ \end{align}\) \(\Gamma(\frac{\nu}{2})=(\frac{\nu}{2}-1)!\) \(\begin{align} \frac{\Gamma(\frac{\nu+1}{2})} {\sqrt{\pi}\,\Gamma(\frac{\nu}{2})}&=\frac{(\nu-1)!}{\sqrt{\pi}\,2^{n-1}\,(\frac{\nu}{2}-1)!\,(\frac{\nu}{2}-1)!}\,\sqrt{\pi} \\ &=\frac{(\nu-1)!}{2^{n-1}\,((\frac{\nu}{2}-1)!)^2} \\ \end{align}\) For \(\nu\) odd: \(\Gamma(\tfrac{\nu+1}{2})=(\tfrac{\nu+1}{2}-1)!=\frac{\nu-1}{2}!\) \(\begin{align} \Gamma(\frac{\nu}{2})&=\frac{\nu-2}{2}\cdot\frac{\nu-4}{2}\cdots\frac{1}{2}\cdot\Gamma\left(\tfrac{1}{2}\right) \\ &=\frac{\nu-1}{\nu-1}\cdot\frac{\nu-2}{2}\cdot\frac{\nu-3}{\nu-3}\cdot\frac{\nu-4}{2}\cdots\frac{2}{2}\cdot\frac{1}{2}\cdot\Gamma\left(\tfrac{1}{2}\right) \\ &=\frac{\nu-1}{2\cdot\frac{\nu-1}{2}}\cdot\frac{\nu-2}{2}\cdot\frac{\nu-3}{2\cdot(\frac{\nu-1}{2}-1)}\cdot\frac{\nu-4}{2}\cdots\frac{2}{2\cdot1}\cdot\frac{1}{2}\cdot\Gamma\left(\tfrac{1}{2}\right) \\ &=\frac{(\nu-1)!}{2^{\nu-1}\,\frac{\nu-1}{2}!}\,\sqrt{\pi} \\ \end{align}\) \(\begin{align} \frac{\Gamma(\frac{\nu+1}{2})} {\sqrt{\pi}\,\Gamma(\frac{\nu}{2})}&=\frac{\frac{\nu-1}{2}!\,2^{\nu-1}\,\frac{\nu-1}{2}!}{\sqrt{\pi}\,(\nu-1)!\,\sqrt{\pi}} \\ &=\frac{2^{\nu-1}\,(\frac{\nu-1}{2}!)^2}{(\nu-1)!\,\pi} \\ \end{align}\) Kind regards Thomas |
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