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I was looking at this question where it is shown that a Student's t-distribution converges to a standard normal distribution as the degrees of freedom tend to infinity.We start with the Student's t-distribution: $$f_T(t) = \frac{\Gamma\left(\frac{k+1}{2}\right)}{\sqrt{k\pi}\Gamma\left(\frac{k}{2}\right)}\left(1+\frac{t^2}{k}\right)^{-\frac{k+1}{2}}$$ for $t\in \mathbb{R}$ and where $k$ represent the degrees of freedom. Then let $k \to \infty$ so \begin{align} \lim_{k \to \infty} f_T(t) &= \lim_{k \to \infty} \frac{\Gamma\left(\frac{k+1}{2}\right)}{\sqrt{k\pi}\Gamma\left(\frac{k}{2}\right)}\left(1+\frac{t^2}{k}\right)^{-\frac{k+1}{2}}\\ &= \lim_{k \to \infty}\frac{\Gamma\left(\frac{k+1}{2}\right)}{\sqrt{k\pi}\Gamma\left(\frac{k}{2}\right)}\cdot \lim_{k \to \infty}\left(1+\frac{t^2}{k}\right)^{-\frac{k+1}{2}} \end{align} and then the answer suggests that using Stirlings approximation gets us that $$\lim_{k \to \infty}\frac{\Gamma\left(\frac{k+1}{2}\right)}{\sqrt{k\pi}\Gamma\left(\frac{k}{2}\right)}=\frac{1}{\sqrt{\pi}}\lim_{k \to \infty} \frac{\sqrt{k/2}}{\sqrt{k}} \tag{1}$$

I tried using the fact that, for big $k$ we have that $$\Gamma(k) \approx \sqrt{\frac{2 \pi}{k}}\left(\frac{k}{e} \right)^k $$ but simply couldn't make the algebra work.

How can we see that $(1)$ is true? Any help is appreciated.

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2 Answers 2

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Here is an alternative argument without going into the details of the pdf:

Let $T_n$ have a $t$ distribution with $n$ degrees of freedom. Then by definition $$T_n=\frac{X}{\sqrt{Y_n/n}}$$

, where $X\sim N(0,1)$ is independent of $Y_n\sim\chi^2_n$.

Since $Y_n$ is distributed as $\sum\limits_{i=1}^n X_i^2$ where $X_i$'s are i.i.d standard normal, by law of large numbers $$\frac{Y_n}{n}\stackrel{P}\longrightarrow E\left(X_1^2\right)=1$$

Therefore, $$\sqrt{Y_n/n}\stackrel{P}\longrightarrow 1\tag{1}$$

And of course, $$X\stackrel{L}\longrightarrow N(0,1)\tag{2}$$

Applying Slutsky's theorem on $(1)$ and $(2)$, it follows that $$T_n\stackrel{L}\longrightarrow N(0,1)$$

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  • $\begingroup$ Thank you very much for you answer, it is a nice alternative! $\endgroup$
    – Bergson
    May 26, 2019 at 18:58
  • $\begingroup$ wow, that is a very nice explanation. the simpler the better, and this is as simple as it gets. $\endgroup$ Jun 15, 2021 at 17:04
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You have all the correct ingredients. Let us work with the expression $f(k)=\sqrt{\dfrac{2\pi}{k}}\left(\dfrac{k}{e}\right)^k.$ Then $$\dfrac{f\left(\dfrac{k+1}{2}\right)}{f\left(\dfrac{k}{2}\right)}=\dfrac{\sqrt{\dfrac{2}{k+1}}\left(\dfrac{k+1}{2e}\right)^{\dfrac{k+1}{2}}}{\sqrt{\dfrac{2}{k}}\left(\dfrac{k}{2e}\right)^{\dfrac{k}{2}}}=\dfrac{\sqrt{k}}{\sqrt{2e}}\times\left(\dfrac{k+1}{k}\right)^{k/2}.$$ Now use the fact that $\underset{k\to\infty}{\lim}\left(1+\dfrac{1}{k}\right)^{k/2}=e^{1/2}=\sqrt{e}.$ This means $\underset{k\to\infty}{\lim}\dfrac{\Gamma\left(\frac{k+1}{2}\right)}{\sqrt{k\pi}\Gamma\left(\frac{k}{2}\right)}=\dfrac{1}{\sqrt{\pi}}\underset{k\to\infty}{\lim}\dfrac{\sqrt{k/2}}{\sqrt{k}}.$

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  • $\begingroup$ Thanks a lot! (there is a minor typo when you first define $f(k)$ but it doesn't really affect your answer) $\endgroup$
    – Bergson
    May 26, 2019 at 18:57
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    $\begingroup$ @ThomasBladt thanks for pointing out. Corrected. $\endgroup$ May 26, 2019 at 19:24

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