Derivation of the Standard Normal Distribution as the Limiting Distribution of Student's t-Distribution
Theorem
If $T_n \sim t(n)$, then $$ T_n \ \overset{D}{\to} N(0,1) $$
- $N \left( \mu , \sigma^{2} \right)$ is the normal distribution with mean $\mu$ and variance $\sigma^{2}$.
- $t(r)$ is the t-distribution with $r$ degrees of freedom.
- $\overset{D}{\to}$ denotes convergence in distribution.
The Student t-distribution was originally born for statistical analysis when the sample is small. As the sample size grows, it becomes similar to the standard normal distribution, which in statistical terms is described as converging in distribution. Therefore, without any special procedure, simply having a large enough sample yields the standard normal distribution.
Derivation
Definition of the t-distribution: For degrees of freedom $\nu > 0$, the continuous probability distribution $t \left( \nu \right)$ with the following probability density function is called the t-distribution. $$ f(x) = {{ \Gamma \left( {{ \nu + 1 } \over { 2 }} \right) } \over { \sqrt{\nu \pi} \Gamma \left( {{ \nu } \over { 2 }} \right) }} \left( 1 + {{ x^{2} } \over { \nu }} \right)^{- {{ \nu + 1 } \over { 2 }}} \qquad ,x \in \mathbb{R} $$
Definition of the standard normal distribution: The normal distribution $N \left( 0,1^{2} \right)$ with the following probability density function is called the standard normal distribution. $$ f(z) = {{ 1 } \over { \sqrt{2 \pi} }} \exp \left[ - {{ z^{2} } \over { 2 }} \right] $$
$$ F_n(t) = \int_{-\infty}^{t} {{\Gamma ( (n+1)/2 ) } \over { \sqrt{\pi n} \Gamma (n/2) }} { {1} \over {(1 + y^{2} / n)^{(n+1)/2} } } dy $$ The cumulative distribution function of $T_n$ is given as above. By the continuity of $F_{n}$, $$ \lim_{n \to \infty} F_n (t) = \lim_{n \to \infty} \int_{-\infty}^{t} f_n (y) dy = \int_{-\infty}^{t} \lim_{n \to \infty} f_n (y) dy $$ Since $\Gamma (1/2) = \sqrt{\pi} $, we have $\displaystyle |f_n (y)| \le 2 f_1 (y) = { {1} \over {\pi} } { {2} \over {1 + y^2 } }$, and by the derivative of the arctangent function, $$ \displaystyle\lim_{n \to \infty} \int_{-\infty}^{t} f_n (y) dy< \int_{-\infty}^{t} 2 f_1 (y) dy = { {2} \over {\pi} } \tan ^{-1} t < \infty $$ Now it suffices to show exactly where $\displaystyle \lim_{n \to \infty} f_n (y)$ converges. First, let us split $f_n$ as follows. $$ \begin{align*} f_{n} (y) =& {{\Gamma ( (n+1)/2 ) } \over { \sqrt{\pi n} \Gamma (n/2) }} { {1} \over {(1 + y^{2} / n)^{(n+1)/2} } } \\ =& {{\Gamma ( (n+1)/2 ) } \over { \sqrt{ n/2} \Gamma (n/2) }} \cdot { {1} \over { \sqrt{2 \pi} (1 + y^{2} / n)^{(n+1)/2} } } \\ =& {{\Gamma ( (n+1)/2 ) } \over { \sqrt{ n/2} \Gamma (n/2) }} \cdot { {1} \over {(1 + y^{2} / n)^{1/2} } } \cdot { {1} \over {\sqrt{2 \pi }} } \left( 1 + { {y^{2}} \over {n} } \right) ^{-n/2} \end{align*} $$
Stirling’s approximation: $$ \lim_{n \to \infty} {{n!} \over {e^{n \ln n - n} \sqrt{ 2 \pi n} }} = 1 $$
Let us first find the limit of the first factor.
1 By Stirling’s approximation, for sufficiently large $n \in \mathbb{N}$, if we set $$ \Gamma (n) \approx e^{n \ln n - n } \sqrt{ 2 \pi n} = \left( {{ n } \over { e }} \right)^{n} \sqrt { 2 \pi n } $$ then for sufficiently large $n$, $$ \begin{align*} {{ {\Gamma ( (n+1)/2 ) } } \over { { \sqrt{ n / 2 } \Gamma (n/2) } }} \approx& \sqrt{ {{ 2 } \over { n }} } {{ \left( {{ n+1 } \over { 2e }} \right)^{{{ n+1 } \over { 2 }}} \sqrt{ 2 \pi (n+1)} } \over { \cdot \left( {{ n } \over { 2e }} \right)^{{{ n } \over { 2 }}} \sqrt{ 2 \pi n} }} \\ \approx& \sqrt{ {{ 2 } \over { n }} } \sqrt{ {{ n+1 } \over { 2e }} } \left( {{ n+1 } \over { n }} \right)^{n/2} \sqrt{ {{ n+1 } \over { n }} } \\ \approx& \sqrt{ {{ 1 } \over { e }}} \left( 1 + {{ 1 } \over { n }} \right)^{n/2} \end{align*} $$ so $$ \lim_{n \to \infty} {{ {\Gamma ( (n+1)/2 ) } } \over { { \sqrt{ n / 2 } \Gamma (n/2) } }} = 1 $$ and the second factor is $$ \lim_{n \to \infty} { {1} \over {(1 + y^{2} / n)^{1/2} } } =1 $$ and the third factor is $$ \lim_{n \to \infty} { {1} \over {\sqrt{2 \pi }} } \left( 1 + { {y^{2}} \over {n} } \right) ^{-n/2} = { {1} \over {\sqrt{2 \pi }} } e ^{- y^{2} / 2} $$ That is, $$ F_n(t) = \int_{-\infty}^{t} { {1} \over {\sqrt{2 \pi }} } e ^{- y^{2} / 2} dy $$ Therefore, $T_n$ converges in distribution to a random variable following the standard normal distribution.
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