📂Probability Distribution

Deriving Standard Normal Distribution as a Limiting Distribution of Student's t-Distribution

Theorem

If $T_n \sim t(n)$ then $$T_n \ \overset{D}{\to} N(0,1)$$

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• $N \left( \mu , \sigma^{2} \right)$ is a normal distribution with mean $\mu$ and variance $\sigma^{2}$.
• $t(r)$ is a t-distribution with degrees of freedom $r$.
• $\overset{D}{\to}$ respectively imply distribution convergence.

Originally, the Student t-distribution was created for statistical analysis when the sample size is small. As the sample size increases, it becomes similar to the standard normal distribution, which in statistical terms is said to converge. Thus, without any special process, simply having a large sample induces the standard normal distribution.

Derivation

Definition of the t-distribution: For degrees of freedom $\nu > 0$, the following probability density function defines a continuous probability distribution $t \left( \nu \right)$ as 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 following probability density function defines a normal distribution $N \left( 0,1^{2} \right)$ as 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. Due to 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 $\Gamma (1/2) = \sqrt{\pi}$, $\displaystyle |f_n (y)| \le 2 f_1 (y) = { {1} \over {\pi} } { {2} \over {1 + y^2 } }$ is true and according to the differentiation 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 is necessary to show where $\displaystyle \lim_{n \to \infty} f_n (y)$ specifically converges. First, let’s 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’s calculate the limit of the first term.

¹ By Stirling’s approximation, for sufficiently large $n \in \mathbb{N}$, $$\Gamma (n) \approx e^{n \ln n - n } \sqrt{ 2 \pi n} = \left( {{ n } \over { e }} \right)^{n} \sqrt { 2 \pi n }$$ assuming for sufficiently large $n$, $$\begin{align*} {{ {\Gamma ( (n+1)/2 ) } } \over { { \sqrt{ n / 2 } \Gamma (n/2) } }} \approx& \sqrt{ {{ 2 } \over { n }} } {{ \left( {{ n+1 } \over { 2e }} \right)^{{{ n+1 } \over { 2 }}} \sqrt{ 2 \pi (n+1)} } \over { \cdot \left( {{ n } \over { 2e }} \right)^{{{ n } \over { 2 }}} \sqrt{ 2 \pi n} }} \\ \approx& \sqrt{ {{ 2 } \over { n }} } \sqrt{ {{ n+1 } \over { 2e }} } \left( {{ n+1 } \over { n }} \right)^{n/2} \sqrt{ {{ n+1 } \over { n }} } \\ \approx& \sqrt{ {{ 1 } \over { e }}} \left( 1 + {{ 1 } \over { n }} \right)^{n/2} \end{align*}$$ therefore, $$\lim_{n \to \infty} {{ {\Gamma ( (n+1)/2 ) } } \over { { \sqrt{ n / 2 } \Gamma (n/2) } }} = 1$$ and the second term is $$\lim_{n \to \infty} { {1} \over {(1 + y^{2} / n)^{1/2} } } =1$$ and the third term is $$\lim_{n \to \infty} { {1} \over {\sqrt{2 \pi }} } \left( 1 + { {y^{2}} \over {n} } \right) ^{-n/2} = { {1} \over {\sqrt{2 \pi }} } e ^{- y^{2} / 2}$$ thus, $$F_n(t) = \int_{-\infty}^{t} { {1} \over {\sqrt{2 \pi }} } e ^{- y^{2} / 2} dy$$ Hence, $T_n$ converges in distribution to a random variable that follows the standard normal distribution.

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