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$$\mathrm{d}t(x_i; \nu) = \frac{\Gamma \bigg(\frac{\nu+1}{2}\bigg) }{\Gamma (\frac{\nu}{2}) \sqrt{\pi\nu}} \Bigg( 1+\frac{x_i^2}{\nu} \Bigg)^{-\frac{\nu+1}{2} } \enspace, i=1,2$$

How to derive a closed-form solution of the above by simplifying out of the gamma function?

Is the gamma function that is being used above (for $z>0$) the same as $$\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}\mathrm{d}x ?$$

develarist
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  • Where have you come across this expression ? If you add appropriate context it will be helpful – Infinity_hunter Nov 24 '20 at 11:49
  • it appears in the formula for the Student $t$-copula's density: https://math.stackexchange.com/questions/3840404/analytical-closed-form-solution-for-the-entropy-of-student-t-copula – develarist Nov 24 '20 at 11:54

1 Answers1

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The symbol $\Gamma$ in the factor $\frac{\Gamma \bigg(\frac{\nu+1}{2}\bigg) }{\Gamma (\frac{\nu}{2}) \sqrt{\pi\nu}} $ does denote the gamma function.

The expression for the pdf of the univariate Student's t-distributution $dt(x_i; \nu)$ is already in closed-form in the sense that the gamma function with an arbitrary real argument cannot be expressed in terms of more elementary functions other than in integral, infinite series, or infinite product representations (many of which are shown in the link). I don't believe that would qualify as "simplifying out of the gamma function".

If the argument is a positive integer, then we have the immense simplification $\Gamma(n) = (n-1)!$. However, for a degree-of-freedom parameter $\nu \not \in \mathbb{Z}$, both arguments $\nu/2$ and $(\nu+1)/2$ are not integers. Even if $\nu$ is an integer one or the other of the arguments is not an integer.

An alternative representation that "appears" less complicated uses the beta function which is related to the gamma function through

$$\int_0^1 t^{x-1}(1-t)^{y-1} \, dt := B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$

Hence,

$$\frac{\Gamma \bigg(\frac{\nu+1}{2}\bigg) }{\Gamma (\frac{\nu}{2})\sqrt{\pi}}= \frac{\Gamma \bigg(\frac{\nu+1}{2}\bigg) }{\Gamma (\frac{\nu}{2})\Gamma (\frac{1}{2})}= \frac{1}{B \left(\frac{1}{2}, \frac{\nu}{2} \right)},$$

and $\displaystyle dt(x_i;\nu) = \frac{1}{\sqrt{\nu}B \left(\frac{1}{2}, \frac{\nu}{2} \right)} \Bigg( 1+\frac{x_i^2}{\nu} \Bigg)^{-\frac{\nu+1}{2} }.$

One may ask why the gamma function appears in this pdf to begin with. A random variable $X$ that is Student's t-distributed with $\nu$ degrees of freedom is obtained as $X = Z/S$ where $Z \sim \mathcal{N}(0,1)$ is a standard normal random variable and $S = \sqrt{Y/\nu}$ where $Y \sim \chi^2(\nu)$ has a chi-square distribution.

The cdf for $Y$ is

$$F_Y(y) = C \int_0^y u^{(\nu-2)/2}e^{-u/2} \, du,$$

where the normalizing factor $C$ is chosen so that $F_Y(y) \to 1$ as $y \to \infty$, that is

$$\frac{1}{C}= \int_0^\infty u^{(\nu-2)/2}e^{-u/2} \, du = \int_0^\infty 2^{\nu/2 -1}t^{\nu/2 -1 }e^{-t} \, 2dt = 2^{\nu/2} \Gamma(\nu/2).$$

After many more steps of transformation the pdf $dt(x;\nu)$ is obtained, retaining the factor $\Gamma(\nu/2)$ and introducing $\Gamma((\nu+1)/2)$.

RRL
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  • thanks for discussing. switching from gamma to beta leads to how to simplify out of beta – develarist Nov 25 '20 at 03:11
  • @develarist: You're welcome. – RRL Nov 25 '20 at 05:28
  • There is no way to simplify $\displaystyle dt(x_i;\nu) = \frac{1}{\sqrt{\nu}B \left(\frac{1}{2}, \frac{\nu}{2} \right)} \Bigg( 1+\frac{x_i^2}{\nu} \Bigg)^{-\frac{\nu+1}{2} }$ out of the beta function either? – develarist Nov 25 '20 at 05:33
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    Not in terms of known elementary functions (trigonometric, rational, etc.) -- just series, product or integral representations as shown in the Wikipedia link. You face the same situation with something as familiar as the normal CDF which involves the error function $erf (z)= \frac{2}{\sqrt{\pi}}\int_0^z e^{-x^2} , dx$. – RRL Nov 25 '20 at 05:45
  • What happened to the $\pi$ in $\frac{\Gamma \bigg(\frac{\nu+1}{2}\bigg) }{\Gamma (\frac{\nu}{2})\sqrt{\pi}}= \frac{\Gamma \bigg(\frac{\nu+1}{2}\bigg) }{\Gamma (\frac{\nu}{2})\Gamma (\frac{1}{2})}$? – develarist Nov 26 '20 at 11:34
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    $\Gamma \left( \frac{1}{2} \right) = \int_0^\infty x^{-1/2}e^{-x} , dx = 2\int_0^\infty e^{-u^2} , du = \sqrt{\pi}$ – RRL Nov 26 '20 at 16:55