Chi-square Distribution
โํ๋ฅ ๊ณผ ํต๊ณ(MATH230)โ ์์ ์์ ๋ฐฐ์ด ๊ฒ๊ณผ ๊ณต๋ถํ ๊ฒ์ ์ ๋ฆฌํ ํฌ์คํธ์ ๋๋ค. ์ ์ฒด ํฌ์คํธ๋ Probability and Statistics์์ ํ์ธํ์ค ์ ์์ต๋๋ค ๐ฒ
์๋ฆฌ์ฆ: Continuous Probability Distributions
์ ํ ๊ฐ๋ ์ผ๋ก Gamma Distribution์ ๋ํด ์๊ณ ์์ด์ผ ํ๋ค.
\[f(x; \alpha, \beta) = \begin{cases} C_{\alpha, \beta} \cdot x^{\alpha-1} e^{-\frac{x}{\beta}} & \text{for } x > 0 \\ \quad 0 & \text{else} \end{cases}\] \[C_{\alpha, \beta} = \frac{1}{\Gamma(\alpha) \cdot \beta^{\alpha}}\]Chi-square Distribution
Definition. Chi-square Distribution
A RV $X$ is called a <Chi-square RV> with $n$ degrees of freedom, denoted as $X \sim \chi^2(n)$,
if it has a Gamma distribution with $\alpha = n/2$ and $\beta=2$.
That is, its pdf is given by
\[f(x; n/2, 2) = \frac{1}{\Gamma(n/2) \cdot 2^{n/2}} \cdot x^{n/2 - 1} \cdot e^{-x/2}\] \[\chi^2(n) = \text{Gamma}\left(\frac{n}{2}, 2\right)\]Remark.
1. If $Z \sim N(0, 1)$, then $Z^2 \sim \chi^2(1)$.
proof.
For $Z \sim N(0, 1)$, let $Y = Z^2$.
Letโs see cdf $P(Y \le y)$,
\[\begin{aligned} F(y) &= P(Y \le y) = P(Z^2 \le y) \\ &= P(-\sqrt{y} \le Z \le \sqrt{y}) \end{aligned}\]๊ทธ๋ผ ์ด์ ์ ๊ท๋ถํฌ $Z$์์์ ํ๋ฅ ์ ๊ตฌํ๋ ๊ฒ์ด๋ฏ๋ก ์ ๋ถ์์ ๊ตฌ์ฑํ๋ฉด,
\[\begin{aligned} \int^{\sqrt{y}}_{-\sqrt{y}} \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} dz &= 2 \int^{\sqrt{y}}_{0} \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} dz \end{aligned}\]์์ ๊ณผ์ ์์๋ ์ ๊ท๋ถํฌ์ ์ฐํจ์ ํน์ฑ์ ์ฌ์ฉํ ๊ฒ์ด๋ค. ์์ ์์์ $z = \sqrt{x}$๋ก ์นํ์ ๋ถ์ ์งํํด๋ณด์.
\[z = \sqrt{x} \iff dz = \frac{1}{2\sqrt{x}} dx\]๊ทธ๋ฆฌ๊ณ ์ ๋ถ์์ ๋์ ํ๋ฉด,
\[\begin{aligned} 2 \int^{\sqrt{y}}_{0} \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} dz &= \frac{1}{\sqrt{2\pi}} \cancel{2} \int^y_0 \frac{1}{\cancel{2}\sqrt{x}} e^{-\frac{x}{2}} dx \\ &= \frac{1}{\sqrt{2\pi}} \int^y_0 x^{-\frac{1}{2}} e^{-\frac{x}{2}} dx \end{aligned}\]์ฆ, $Y = Z^2$์ cdf๋
\[F(y) = \frac{1}{\sqrt{2\pi}} \int^y_0 x^{\frac{1}{2} - 1} e^{-\frac{x}{2}} dx\]์ด๋ค. ์ด์ pdf๋ฅผ ๊ตฌํ๊ธฐ ์ํด ์๋ณ์ ๋ฏธ๋ถํ๋ฉด,
\[\begin{aligned} f(y) = \frac{d}{dy} F(y) = \frac{1}{\sqrt{2\pi}} y^{\frac{1}{2} - 1} e^{-\frac{y}{2}} \end{aligned}\]์ด๋, ๊ฐ๋งํจ์ $\Gamma(1/2)$๋ $\sqrt{\pi}$์ ๊ฐ์ ๊ฐ๋๋ค. ๋ฐ๋ผ์,
\[\begin{aligned} f(y) &= \frac{1}{\sqrt{2\pi}} y^{\frac{1}{2} - 1} e^{-\frac{y}{2}} \\ &= \frac{1}{\Gamma(1/2) \cdot 2^{\frac{1}{2}}} \cdot y^{\frac{1}{2} - 1} e^{-\frac{y}{2}} \end{aligned}\]์ด๊ฒ์ ๊ณง, ๊ฐ๋ง ๋ถํฌ $\text{Gamma}(1/2, 2)$์ pdf์ ๊ฐ๋ค! ๋ฐ๋ผ์,
\[\left(Z(0, 1)\right)^2 \overset{D}{=} \text{Gamma}(1/2, 2) \overset{D}{=} \chi^2(1)\]2. If $X \sim \chi^2(n)$, then
- $E[X] = n$
- $\text{Var}(X) = 2n$
๋งบ์๋ง
์ด์ด์ง๋ ํฌ์คํธ์์๋ <Weibull Distribution>์ ํตํด <๊ฒฐํจ๋ฅ ; Failure rate>์ <์ ๋ขฐ๋; Reliability>์ ๋ชจ๋ธ๋งํ๋ค. ์ด ๋ถ๋ถ์ ์ ๊ท ์์ ์์๋ ์๊ฐ๋ง ํ๊ณ ๋์ด๊ฐ ๋ถ๋ถ์ด๊ธฐ ๋๋ฌธ์ ๊ด์ฌ์ด ์๊ฑฐ๋ ๊ผญ ํ์ํ๊ฒ ์๋๋ผ๋ฉด ๊ฑด๋ ๋ฐ์ด๋ ๊ด์ฐฎ๋ค.