Limit of tt-distribution as nn goes to infinity

I found in my intro to stats textbook that t-distribution approaches the standard normal as n goes to infinity. The textbook gives the density for t-distribution as follows, f(t)=Γ(n+12)nπΓ(n2)(1+t2n)n+12

I think it might be possible to show that this density converges (uniformly) to the density of normal as n goes to infinity. Given lim, it would be great if we can show
\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}\to \sqrt{\frac{n}{2}} as n\to \infty, yet I am stuck here. Can someone point out how to proceed or an alternative way to show that t-distribution converges to normal as n\to \infty.

Answer

Stirling’s approximation gives \Gamma(z) = \sqrt{\frac{2\pi}{z}}\,{\left(\frac{z}{e}\right)}^z \left(1 + O\left(\tfrac{1}{z}\right)\right) so

\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n}{2})} = \dfrac{\sqrt{\frac{2\pi}{\frac{n+1}{2}}}\,{\left(\frac{\frac{n+1}{2}}{e}\right)}^{\frac{n+1}{2}}}{\sqrt{\frac{2\pi}{\frac{n}{2}}}\,{\left(\frac{\frac{n}{2}}{e}\right)}^{\frac{n}{2}}}\left(1 + O\left(\tfrac{1}{n}\right)\right)\\= {\sqrt{\frac{\frac{n+1}{2}}{e}}}\left(1+\frac1n\right)^{\frac{n}{2}}\left(1 + O\left(\tfrac{1}{n}\right)\right) \\= \sqrt{\frac{n}{2}} \left(1 + O\left(\tfrac{1}{n}\right)\right)\\ \to \sqrt{\frac{n}{2}} and you may have a slight typo in your question

In fact when considering limits as n\to \infty, you should not have n in the solution; instead you can say the ratio tends to 1 and it turns out here that the difference tends to 0. Another point is that \sqrt{\frac{n}{2}-\frac14} is a better approximation, in that not only does the difference tend to 0, but so too does the difference of the squares.

Attribution
Source : Link , Question Author : Yilei Huang , Answer Author : Henry

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