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I encountered the following three distributions of a continuous random variable, with the corresponding pdf's;

  • the normal(Gaussian) distribution $$f_{\mu,\sigma}(x)=\frac1{\sigma\sqrt{2\pi}}e^{-\frac12\left(\frac{x-\mu}{\sigma}\right)^2}$$
  • the Cauchy distribution

$$f_{x_0,\gamma}(x)=\frac1{\pi\gamma\left[1+\left(\frac{x-x_0}\gamma\right)^2\right]} =\frac1{\pi\gamma}\left[\frac{\gamma^2}{(x-x_0)^2+\gamma}\right]$$

  • the student t-distribution $$f_\nu(t)=\frac{\Gamma(\frac{\nu+1}2)}{\sqrt{\nu\pi}\; \Gamma(\frac\nu2)}\left(1+\frac{t^2}\nu\right)^{-\frac{\nu+1}2}$$

The formulas are all from wikipedia. And it says the first two distributions are special cases of the third one. Setting $\nu=1$ to the third pdf, we get the second pdf with $x_0=0$ and $\gamma=1$ very easily. The only nontrivial part of which is to evaluate $\Gamma\left(\frac12\right)=\sqrt{\pi}$, but it seems feasible anyway.

But how come are the first and the third related? Wikipedia says that the third 'approaches' to the first($\mu=0$, $\sigma=1$) as $\nu\to\infty$. But how? Can anyone give me a possible answer?

govin
  • 345

2 Answers2

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Hints:

  1. Use the definition of $\Gamma(x) = (x-1)!$ and Stirling's approximation for factorial
  2. Use the definition of $e^{x} = \lim_{n \to \infty}(1+\frac{x}{n})^n$ Also, if limit exists, then the limit of the product is equal to the product of limits
Alex
  • 19,262
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A more explicit version of @Alex's answer:

For large $\nu$, linearizing the log-Gamma function gives

$$\tfrac{\Gamma\left(\tfrac{\nu+1}{2}\right)}{\Gamma\left(\tfrac{\nu}{2}\right)}\sim\sqrt{\tfrac{\Gamma\left(\tfrac{\nu+2}{2}\right)}{\Gamma\left(\tfrac{\nu}{2}\right)}}=\sqrt{\tfrac{\nu}{2}},$$

and the other factor

$$\left(1+\tfrac{t^2}{\nu}\right)^{-\tfrac{\nu+1}{2}}\sim\left(1+\tfrac{t^2}{\nu}\right)^{-\tfrac{\nu}{2}}\sim\exp\tfrac{-t^2}{2},$$

so $\lim_{\nu\to\infty}f_\nu(t)=\tfrac{1}{\sqrt{2\pi}}\exp\tfrac{-t^2}{2}$.

J.G.
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