시간이 지남에 따라 일어나는 어떤 특정한 사건 x x x X ∼ POI ( λ ) X\sim \text{POI}(\lambda) X ∼ POI ( λ )
1. 확률밀도함수
X ∼ POI ( λ ) X\sim \text{POI}(\lambda) X ∼ POI ( λ )
P ( X = x ) = λ x e − λ x ! x = 0 , 1 , 2 , ⋯ P(X = x) = \dfrac{\lambda^x e^{-\lambda}}{x!}\quad x = 0, 1, 2, \cdots P ( X = x ) = x ! λ x e − λ x = 0 , 1 , 2 , ⋯ cf. 포아송분포에 의한 이항분포의 근사 X X X B ( n , p ) B(n,\ p) B ( n , p ) n n n p p p n p = λ np=\lambda n p = λ x = 0 , 1 , 2 , … x=0,1,2,\ldots x = 0 , 1 , 2 , …
lim n → ∞ ( n x ) p x ( 1 − p ) n − x = exp ( − λ ) λ x x ! \lim_{n\rightarrow\infty}\binom{n}{x}p^x(1-p)^{n-x}=\dfrac{\exp(-\lambda)\lambda^x}{x!} n → ∞ lim ( x n ) p x ( 1 − p ) n − x = x ! exp ( − λ ) λ x 예 n > 100 , p < 0.01 n>100,\ p<0.01 n > 1 0 0 , p < 0 . 0 1 λ \lambda λ n p np n p n = 50 , p = 0.02 , n = 100 , p = 0.01 , n = 200 , p = 0.005 , λ = 1 n=50,\ p=0.02,\ n=100,\ p=0.01,\ n=200,\ p=0.005,\ \lambda=1 n = 5 0 , p = 0 . 0 2 , n = 1 0 0 , p = 0 . 0 1 , n = 2 0 0 , p = 0 . 0 0 5 , λ = 1
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2. 기댓값과 분산
X ∼ POI ( λ ) X\sim \text{POI}(\lambda) X ∼ POI ( λ )
1) 기댓값
E ( X ) = ∑ x = 0 ∞ x e − λ λ x x ! = λ e − λ ∑ x = 1 ∞ λ x − 1 ( x − 1 ) ! ( ∵ e x = ∑ n = 0 ∞ x n n ! ) = λ \begin{aligned} E(X) &=\sum_{x=0}^{\infty}x\frac{e^{-\lambda}\lambda^x}{x!} \\ &=\lambda e^{-\lambda}\sum_{x=1}^{\infty}\frac{\lambda^{x-1}}{(x-1)!}\quad (\because e^x=\sum_{n=0}^{\infty}\dfrac{x^n}{n!})\\ &=\lambda \end{aligned} E ( X ) = x = 0 ∑ ∞ x x ! e − λ λ x = λ e − λ x = 1 ∑ ∞ ( x − 1 ) ! λ x − 1 ( ∵ e x = n = 0 ∑ ∞ n ! x n ) = λ \\[40pt]
2) 분산
E ( X 2 ) = E [ X ( X − 1 ) ] + E ( X ) = ∑ x = 0 ∞ x ( x − 1 ) e − λ λ x x ! + λ = λ 2 e − λ ∑ x = 2 ∞ λ x − 2 ( x − 2 ) ! + λ = λ 2 e − λ ∑ k = 0 ∞ λ k k ! + λ = λ 2 + λ ∴ Var ( X ) = λ 2 + λ − λ 2 = λ \begin{aligned} E(X^2) &=E[X(X-1)]+E(X)\\[10pt] &=\sum_{x=0}^{\infty}x(x-1)\frac{e^{-\lambda}\lambda^x}{x!}+\lambda \\[15pt] &=\lambda^2e^{-\lambda}\sum_{x=2}^{\infty}\frac{\lambda^{x-2}}{(x-2)!}+\lambda \\[15pt] &=\lambda^2e^{-\lambda}\sum_{k=0}^{\infty}\frac{\lambda^{k}}{k!}+\lambda\\[15pt] &= \lambda^2+\lambda \\[20pt] \therefore \text{Var}(X) &=\lambda^2+\lambda-\lambda^2 \\[5pt] &=\lambda \end{aligned} E ( X 2 ) ∴ Var ( X ) = E [ X ( X − 1 ) ] + E ( X ) = x = 0 ∑ ∞ x ( x − 1 ) x ! e − λ λ x + λ = λ 2 e − λ x = 2 ∑ ∞ ( x − 2 ) ! λ x − 2 + λ = λ 2 e − λ k = 0 ∑ ∞ k ! λ k + λ = λ 2 + λ = λ 2 + λ − λ 2 = λ \\[60pt]
3. 적률생성함수
X ∼ POI ( λ ) X\sim \text{POI}(\lambda) X ∼ POI ( λ )
M X ( t ) = E ( e t x ) = ∑ x = 0 n e t x e − λ λ x x ! = e − λ ∑ x = 0 n ( λ e t ) x x ! = e − λ e λ e t ( ∵ e x = ∑ n = 0 ∞ x n n ! ) = e λ ( e t − 1 ) \begin{aligned} M_X(t) &=E(e^{tx}) \\[10pt] &=\sum_{x=0}^{n}e^{tx}\dfrac{e^{-\lambda}\lambda^x}{x!} \\[10pt] &=e^{-\lambda}\sum_{x=0}^{n}\dfrac{(\lambda e^{t})^x}{x!} \\[10pt] &=e^{-\lambda}e^{\lambda e^{t}}\quad (\because e^x=\sum_{n=0}^{\infty}\dfrac{x^n}{n!})\\[10pt] &=e^{\lambda(e^t-1)} \end{aligned} M X ( t ) = E ( e t x ) = x = 0 ∑ n e t x x ! e − λ λ x = e − λ x = 0 ∑ n x ! ( λ e t ) x = e − λ e λ e t ( ∵ e x = n = 0 ∑ ∞ n ! x n ) = e λ ( e t − 1 ) 따라서 적률생성함수를 활용한 기댓값과 분산은
기댓값M X ( 1 ) ( t ) = ( d d t ) ( e λ ( e t − 1 ) ) = e λ ( e t − 1 ) ⋅ d d t ( λ ( e t − 1 ) ) = e λ ( e t − 1 ) ⋅ λ e t ∴ E ( X ) = M X ( 1 ) ( 0 ) = e λ ( e t − 1 ) ⋅ λ e t = λ \begin{aligned} M^{(1)}_X(t) &=\left(\dfrac{d}{dt}\right)\left(e^{\lambda(e^t-1)}\right)=e^{\lambda (e^t - 1)} \cdot \frac{d}{dt} \left( \lambda (e^t - 1) \right)= e^{\lambda (e^t - 1)} \cdot \lambda e^t \\[30pt] \therefore E(X) &=M^{(1)}_X(0)=e^{\lambda (e^t - 1)} \cdot \lambda e^t=\lambda \end{aligned} M X ( 1 ) ( t ) ∴ E ( X ) = ( d t d ) ( e λ ( e t − 1 ) ) = e λ ( e t − 1 ) ⋅ d t d ( λ ( e t − 1 ) ) = e λ ( e t − 1 ) ⋅ λ e t = M X ( 1 ) ( 0 ) = e λ ( e t − 1 ) ⋅ λ e t = λ
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M X ( 2 ) ( t ) = ( d / d t ) ( e λ ( e t − 1 ) ⋅ λ e t ) = ( d d t e λ ( e t − 1 ) ) ⋅ ( λ e t ) + e λ ( e t − 1 ) ⋅ d d t ( λ e t ) = { e λ ( e t − 1 ) ⋅ ( λ e t ) ⋅ ( λ e t ) } + e λ ( e t − 1 ) ⋅ λ e t = e λ ( e t − 1 ) ⋅ λ 2 e 2 t + e λ ( e t − 1 ) ⋅ λ e t = e λ ( e t − 1 ) ⋅ λ e t ( λ e t + 1 ) . M X ( 2 ) ( 0 ) = e λ ( e t − 1 ) ⋅ λ e t ( λ e t + 1 ) = λ ( λ + 1 ) ∴ Var ( X ) = λ ( λ + 1 ) − λ 2 = λ \begin{aligned} M^{(2)}_X(t) &=(d/dt)(e^{\lambda (e^t - 1)} \cdot \lambda e^t) \\[10pt] &= \left( \frac{d}{dt} e^{\lambda (e^t - 1)} \right) \cdot (\lambda e^t) + e^{\lambda (e^t - 1)} \cdot \frac{d}{dt} (\lambda e^t) \\[15pt] &= \{e^{\lambda (e^t - 1)} \cdot (\lambda e^t) \cdot (\lambda e^t)\} + e^{\lambda (e^t - 1)} \cdot \lambda e^t \\[10pt] &= e^{\lambda (e^t - 1)} \cdot \lambda^2 e^{2t} + e^{\lambda (e^t - 1)} \cdot \lambda e^t \\[10pt] &=e^{\lambda (e^t - 1)} \cdot \lambda e^t (\lambda e^t + 1). \\[15pt] M^{(2)}_X(0)&=e^{\lambda (e^t - 1)} \cdot \lambda e^t (\lambda e^t + 1)=\lambda (\lambda + 1) \\[15pt] \therefore \text{Var}(X)&=\lambda (\lambda + 1)-\lambda^2=\lambda \end{aligned} M X ( 2 ) ( t ) M X ( 2 ) ( 0 ) ∴ Var ( X ) = ( d / d t ) ( e λ ( e t − 1 ) ⋅ λ e t ) = ( d t d e λ ( e t − 1 ) ) ⋅ ( λ e t ) + e λ ( e t − 1 ) ⋅ d t d ( λ e t ) = { e λ ( e t − 1 ) ⋅ ( λ e t ) ⋅ ( λ e t ) } + e λ ( e t − 1 ) ⋅ λ e t = e λ ( e t − 1 ) ⋅ λ 2 e 2 t + e λ ( e t − 1 ) ⋅ λ e t = e λ ( e t − 1 ) ⋅ λ e t ( λ e t + 1 ) . = e λ ( e t − 1 ) ⋅ λ e t ( λ e t + 1 ) = λ ( λ + 1 ) = λ ( λ + 1 ) − λ 2 = λ \\[60pt]
4. 표본분포의 근사
X ∼ POI ( λ ) X\sim \text{POI}(\lambda) X ∼ POI ( λ ) X 1 , X 2 , … , X n X_1,X_2,\ldots,X_n X 1 , X 2 , … , X n
표본평균
E ( X ˉ n ) = E ( 1 n ∑ i = 1 n X i ) = 1 n { ∑ i = 1 n E ( X i ) } = 1 n ⋅ n ⋅ E ( X ) = λ \begin{aligned} E(\bar{X}_n)&=E(\dfrac{1}{n}\sum_{i=1}^{n}X_i) \\[15pt] &=\dfrac{1}{n}\left\{\sum_{i=1}^{n}E(X_i)\right\} \\[15pt] &=\dfrac{1}{n}\cdot n\cdot E(X)=\lambda \end{aligned} E ( X ˉ n ) = E ( n 1 i = 1 ∑ n X i ) = n 1 { i = 1 ∑ n E ( X i ) } = n 1 ⋅ n ⋅ E ( X ) = λ \\[20pt]
Var ( X ˉ n ) = Var ( 1 n ∑ i = 1 n X i ) = 1 n 2 Var ( ∑ i = 1 n X i ) = 1 n 2 ∑ i = 1 n Var ( X i ) = n λ n 2 = λ n \begin{aligned} \text{Var}(\bar{X}_n) &=\text{Var}(\dfrac{1}{n}\sum_{i=1}^{n}X_i) \\[15pt] &=\dfrac{1}{n^2}\text{Var}(\sum_{i=1}^{n}X_i) \\[15pt] &=\dfrac{1}{n^2}\sum_{i=1}^{n}\text{Var}(X_i) \\[15pt] &=\dfrac{n\lambda}{n^2} \\[10pt] &=\dfrac{\lambda}{n} \end{aligned} Var ( X ˉ n ) = Var ( n 1 i = 1 ∑ n X i ) = n 2 1 Var ( i = 1 ∑ n X i ) = n 2 1 i = 1 ∑ n Var ( X i ) = n 2 n λ = n λ \\[40pt]
표본분산
E ( S n 2 ) = E [ 1 n − 1 ∑ i = 1 n ( X i − X ˉ n ) 2 ] = E [ 1 n − 1 ∑ i = 1 n ( X i − λ + λ − X ˉ n ) 2 ] = E [ 1 n − 1 ∑ i = 1 n [ ( X i − λ ) 2 + ( λ − X ˉ n ) 2 + 2 ( X i − λ ) ( λ − X ˉ n ) ] ] = 1 n − 1 E [ ∑ i = 1 n ( X i − λ ) 2 − n ( X ˉ n − λ ) 2 ] = 1 n − 1 [ ∑ i = 1 n E ( X i − λ ) 2 − n E ( X ˉ n − λ ) 2 ] = 1 n − 1 [ n λ − n λ n ] = λ \begin{aligned} E(S_n^2) &=E\left[\dfrac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar{X}_n)^2\right] \\[15pt] &=E\left[\dfrac{1}{n-1}\sum_{i=1}^{n}(X_i-\lambda+\lambda-\bar{X}_n)^2\right] \\[15pt] &=E\left[\dfrac{1}{n-1}\sum_{i=1}^{n}[(X_i-\lambda)^2+(\lambda-\bar{X}_n)^2+2(X_i-\lambda)(\lambda-\bar{X}_n)]\right] \\[15pt] &=\dfrac{1}{n-1}E\left[\sum_{i=1}^{n}(X_i-\lambda)^2-n(\bar{X}_n-\lambda)^2\right] \\[15pt] &=\dfrac{1}{n-1}\left[\sum_{i=1}^{n}E(X_i-\lambda)^2-nE(\bar{X}_n-\lambda)^2\right] \\[15pt] &=\dfrac{1}{n-1}\left[n\lambda-n\dfrac{\lambda}{n}\right] \\[15pt] &=\lambda \end{aligned} E ( S n 2 ) = E [ n − 1 1 i = 1 ∑ n ( X i − X ˉ n ) 2 ] = E [ n − 1 1 i = 1 ∑ n ( X i − λ + λ − X ˉ n ) 2 ] = E [ n − 1 1 i = 1 ∑ n [ ( X i − λ ) 2 + ( λ − X ˉ n ) 2 + 2 ( X i − λ ) ( λ − X ˉ n ) ] ] = n − 1 1 E [ i = 1 ∑ n ( X i − λ ) 2 − n ( X ˉ n − λ ) 2 ] = n − 1 1 [ i = 1 ∑ n E ( X i − λ ) 2 − n E ( X ˉ n − λ ) 2 ] = n − 1 1 [ n λ − n n λ ] = λ \\[40pt]
1) 대수의 법칙
X ∼ POI ( λ ) X\sim \text{POI}(\lambda) X ∼ POI ( λ ) X 1 , X 2 , … , X n X_1,X_2,\ldots,X_n X 1 , X 2 , … , X n
X ˉ n → p λ \bar{X}_n \xrightarrow{p}\lambda X ˉ n p λ \\[20pt]
∵ P [ ∣ X n ˉ − λ ∣ < ϵ ] = P [ ∣ X n ˉ − λ ∣ 2 < ϵ 2 ] ≥ 1 − E ( X n ˉ − λ ) 2 ϵ 2 = 1 − λ / n ϵ 2 → 1 \begin{aligned} \because \quad P[|\bar{X_n}-\lambda|<\epsilon] &=P[|\bar{X_n}-\lambda|^2<\epsilon^2] \\[10pt] &\ge 1-\dfrac{E(\bar{X_n}-\lambda)^2}{\epsilon^2} \\[10pt] &= 1-\dfrac{\lambda/n}{\epsilon^2}\rightarrow1 \end{aligned} ∵ P [ ∣ X n ˉ − λ ∣ < ϵ ] = P [ ∣ X n ˉ − λ ∣ 2 < ϵ 2 ] ≥ 1 − ϵ 2 E ( X n ˉ − λ ) 2 = 1 − ϵ 2 λ / n → 1 \\[40pt]
2) 중심극한정리
X ∼ POI ( λ ) X\sim \text{POI}(\lambda) X ∼ POI ( λ ) X 1 , X 2 , … , X n X_1,X_2,\ldots,X_n X 1 , X 2 , … , X n n n n Z n Z_n Z n
Z n = n ( X ˉ n − λ ) → d N ( 0 , λ ) Z n = ( X ˉ n − λ ) λ / n → d N ( 0 , 1 ) Z n = ∑ i = 1 n ( X i − λ ) n λ → d N ( 0 , 1 ) \begin{aligned} Z_n&=\sqrt{n}(\bar{X}_n-\lambda)\xrightarrow{d}N(0,\ \lambda) \\[20pt] Z_n&=\dfrac{(\bar{X}_n-\lambda)}{\sqrt{\lambda/n}}\xrightarrow{d}N(0,\ 1) \\[20pt] Z_n&=\dfrac{\sum_{i=1}^{n}(X_i-\lambda)}{\sqrt{n\lambda}}\xrightarrow{d}N(0,\ 1) \end{aligned} Z n Z n Z n = n ( X ˉ n − λ ) d N ( 0 , λ ) = λ / n ( X ˉ n − λ ) d N ( 0 , 1 ) = n λ ∑ i = 1 n ( X i − λ ) d N ( 0 , 1 ) \\[40pt]
3) 델타 방법
확률변수 열 X 1 , X 2 , … , X n , … X_1,X_2,\ldots,X_n,\ldots X 1 , X 2 , … , X n , …
n ( X n − λ ) → d N ( 0 , λ ) \sqrt{n}(X_n-\lambda)\xrightarrow{d}N(0,\lambda) n ( X n − λ ) d N ( 0 , λ ) 이때 함수 g ( λ ) g(\lambda) g ( λ ) g ′ ( λ ) g'(\lambda) g ′ ( λ )
n ( g ( X n ) − g ( λ ) ) → d N ( 0 , σ 2 [ g ′ ( λ ) ] 2 ) \sqrt{n}(g(X_n)-g(\lambda))\xrightarrow{d}N(0,\sigma^2[g'(\lambda)]^2) n ( g ( X n ) − g ( λ ) ) d N ( 0 , σ 2 [ g ′ ( λ ) ] 2 ) 이제 함수 g ( x ) = x 2 g(x)=x^2 g ( x ) = x 2 g ( λ ) = λ 2 g(\lambda)=\lambda^2 g ( λ ) = λ 2 σ 2 = λ \sigma^2=\lambda σ 2 = λ g ′ ( λ ) = 2 λ g'(\lambda)=2\lambda g ′ ( λ ) = 2 λ
n ( X ˉ n 2 − λ 2 ) → d N ( 0 , 4 λ 3 ) \sqrt{n}(\bar{X}_n^2-\lambda^2)\xrightarrow{d}N(0,4\lambda^3) n ( X ˉ n 2 − λ 2 ) d N ( 0 , 4 λ 3 ) \\[60pt]
5. 추정량
1) 적률 추정량
X ∼ POI ( λ ) X\sim \text{POI}(\lambda) X ∼ POI ( λ ) X 1 , X 2 , … , X n X_1,X_2,\ldots,X_n X 1 , X 2 , … , X n λ \lambda λ
m 1 ′ = μ 1 ′ ( λ ) = λ → ∑ i = 1 n X i 1 / n = λ ^ ∴ λ ^ = X ˉ n \begin{aligned} m_1'&=\mu_1'(\lambda)=\lambda \\[10pt] \rightarrow \sum_{i=1}^{n}X_i^1/n &= \hat{\lambda} \\[15pt] \therefore \hat{\lambda} &= \bar{X}_n \end{aligned} m 1 ′ → i = 1 ∑ n X i 1 / n ∴ λ ^ = μ 1 ′ ( λ ) = λ = λ ^ = X ˉ n \\[40pt]
2) 최대가능도 추정량
X ∼ POI ( λ ) X\sim \text{POI}(\lambda) X ∼ POI ( λ ) X 1 , X 2 , … , X n X_1,X_2,\ldots,X_n X 1 , X 2 , … , X n λ \lambda λ
λ \lambda λ
L ( λ ; x 1 , x 2 , … , x n ) = ∏ i = 1 n f ( x i ; λ ) = ( e − n λ ⋅ λ ∑ x i ) / ∏ i = 1 n x i ! \begin{aligned} L(\lambda;x_1,x_2,\ldots,x_n) &=\prod_{i=1}^{n}f(x_i;\lambda)\\[15pt] &=\left.\left(e^{-n\lambda}\cdot \lambda^{\sum x_i}\right)\right/\prod_{i=1}^{n}x_i! \end{aligned} L ( λ ; x 1 , x 2 , … , x n ) = i = 1 ∏ n f ( x i ; λ ) = ( e − n λ ⋅ λ ∑ x i ) / i = 1 ∏ n x i ! 가 되며, 따라서 로그가능도함수는
log L ( λ ; x 1 , x 2 , … , x n ) = log ( e − n λ ) + log ( λ ∑ x i ) − log ( ∏ i = 1 n x i ! ) = − n λ + ∑ i = 1 n x i ⋅ log λ − log ( ∏ i = 1 n x i ! ) \begin{aligned} \log L(\lambda;x_1,x_2,\ldots,x_n) &=\text{log}(e^{-n\lambda})+ \log(\lambda^{\sum x_i})-\log\left(\prod_{i=1}^{n}x_i!\right)\\[15pt] &=-n\lambda + \sum_{i=1}^{n}x_i\cdot\log \lambda-\log\left(\prod_{i=1}^{n}x_i!\right) \end{aligned} log L ( λ ; x 1 , x 2 , … , x n ) = log ( e − n λ ) + log ( λ ∑ x i ) − log ( i = 1 ∏ n x i ! ) = − n λ + i = 1 ∑ n x i ⋅ log λ − log ( i = 1 ∏ n x i ! ) 가 된다. 이제 로그가능도 함수를 최대화하는 값을 찾기 위해 다음 방적식의 해를 찾는다.
d d λ log L ( λ ; x 1 , x 2 , … , x n ) = − n + ∑ i = 1 n x i λ = 0 = ∑ i = 1 n x i = n λ = λ ^ = x ˉ n \begin{aligned} \dfrac{d}{d\lambda}\text{log}L(\lambda;x_1,x_2,\ldots,x_n) &=-n+\sum_{i=1}^{n}\dfrac{x_i}{\lambda}=0 \\[15pt] &=\sum_{i=1}^{n}x_i=n\lambda \\[15pt] &=\hat{\lambda}=\bar{x}_n \end{aligned} d λ d log L ( λ ; x 1 , x 2 , … , x n ) = − n + i = 1 ∑ n λ x i = 0 = i = 1 ∑ n x i = n λ = λ ^ = x ˉ n 이 방정식이 방정식을 만족하는 λ ^ = x ˉ n \hat{\lambda}=\bar{x}_n λ ^ = x ˉ n
d 2 d λ 2 log L ( λ ; x 1 , x 2 , … , x n ) ∣ λ = x ˉ n = − ∑ i = 1 n x i λ 2 ∣ λ = x ˉ n = − n x ˉ n 2 ⋅ 1 n ∑ i = 1 n x i = − n x ˉ n < 0 \begin{aligned} \left.\dfrac{d^2}{d\lambda^2}\text{log}L(\lambda;x_1,x_2,\ldots,x_n)\right|_{\lambda=\bar{x}_n} &=-\sum_{i=1}^{n}\left.\dfrac{x_i}{\lambda^2}\right|_{\lambda=\bar{x}_n} \\[15pt] &=-\dfrac{n}{\bar{x}_n^2}\cdot\dfrac{1}{n}\sum_{i=1}^{n}x_i \\[15pt] &=-\dfrac{n}{\bar{x}_n}<0 \end{aligned} d λ 2 d 2 log L ( λ ; x 1 , x 2 , … , x n ) ∣ ∣ ∣ ∣ ∣ λ = x ˉ n = − i = 1 ∑ n λ 2 x i ∣ ∣ ∣ ∣ λ = x ˉ n = − x ˉ n 2 n ⋅ n 1 i = 1 ∑ n x i = − x ˉ n n < 0 을 만족시키므로(단 x ˉ n > 0 \bar{x}_n>0 x ˉ n > 0 X ˉ n \bar{X}_n X ˉ n x ˉ n > 0 \bar{x}_n>0 x ˉ n > 0
최대가능도 추정량의 불변성
λ ^ \hat{\lambda} λ ^ λ \lambda λ λ \lambda λ g ( λ ) g(\lambda) g ( λ ) g ( λ ^ ) g(\hat{\lambda}) g ( λ ^ ) g ( λ ) g(\lambda) g ( λ ) g ( λ ) = λ 2 g(\lambda)=\lambda^2 g ( λ ) = λ 2
g ( λ ^ ) = λ 2 ^ = X ˉ n 2 g(\hat{\lambda})=\hat{\lambda^2}=\bar{X}_n^2 g ( λ ^ ) = λ 2 ^ = X ˉ n 2 \\[60pt]
6. 좋은 추정량 조건
최대가능도 추정량인 경우만 고려하였다.
1) 비편향추정량
X ∼ POI ( λ ) X\sim \text{POI}(\lambda) X ∼ POI ( λ ) X 1 , X 2 , … , X n X_1,X_2,\ldots,X_n X 1 , X 2 , … , X n E [ X ˉ n ] E[\bar{X}_n] E [ X ˉ n ] λ \lambda λ
E [ X ˉ n ] − λ = 0 E[\bar{X}_n]-\lambda=0 E [ X ˉ n ] − λ = 0 비편향추정량을 만족한다.
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2) 최소분산 비편향추정량
2-1. 크래머-라오 방법
X ∼ POI ( λ ) X\sim \text{POI}(\lambda) X ∼ POI ( λ ) X 1 , X 2 , … , X n X_1,X_2,\ldots,X_n X 1 , X 2 , … , X n
Var ( X ˉ n ) ≥ [ g ′ ( λ ) ] 2 n I ( λ ) \text{Var}(\bar{X}_n)\ge\dfrac{[g'(\lambda)]^2}{nI(\lambda)} Var ( X ˉ n ) ≥ n I ( λ ) [ g ′ ( λ ) ] 2 T ( X ) T(X) T ( X ) g ( θ ) g(\theta) g ( θ ) θ \theta θ
Var ( X ˉ n ) ≥ 1 n I ( λ ) \text{Var}(\bar{X}_n)\ge\dfrac{1}{nI(\lambda)} Var ( X ˉ n ) ≥ n I ( λ ) 1 을 얻을 수 있다. 그러므로 어떤 비편향추정량 X ˉ n \bar{X}_n X ˉ n 1 n I ( λ ) \dfrac{1}{nI(\lambda)} n I ( λ ) 1 λ \lambda λ \\[10pt]
이 때 피셔의 정보량 I ( λ ) I(\lambda) I ( λ )
I ( λ ) = E [ ( ∂ ∂ λ log f ( X ; λ ) ) 2 ] = − E [ ( ∂ 2 ∂ λ 2 log f ( X ; λ ) ) ] I(\lambda)=E\left[\left(\dfrac{\partial}{\partial\lambda}\text{log }f(X;\lambda)\right)^2\right]=-E\left[\left(\dfrac{\partial^2}{\partial\lambda^2}\text{log }f(X;\lambda)\right)\right] I ( λ ) = E [ ( ∂ λ ∂ log f ( X ; λ ) ) 2 ] = − E [ ( ∂ λ 2 ∂ 2 log f ( X ; λ ) ) ] 계산을 하면
I ( λ ) = − E [ ( ∂ 2 ∂ λ 2 log f ( X ; λ ) ) ] = − E [ ∂ ∂ λ ( ∂ ∂ λ { − λ + x log λ − x ! } ) ] = − E [ ∂ ∂ λ ( − 1 + X λ ) ] = − E [ − X λ 2 ] = 1 λ 2 E [ X ] = 1 λ \begin{aligned} I(\lambda) &=-E\left[\left(\dfrac{\partial^2}{\partial\lambda^2}\text{log }f(X;\lambda)\right)\right] \\[15pt] &=-E\left[\dfrac{\partial}{\partial \lambda}\left(\dfrac{\partial}{\partial \lambda}\left\{-\lambda+x\text{log } \lambda - x!\right\}\right)\right] \\[15pt] &=-E\left[\dfrac{\partial}{\partial \lambda}\left(-1 + \dfrac{X}{\lambda}\right)\right] \\[15pt] &=-E\left[-\dfrac{X}{\lambda^2}\right] \\[15pt] &=\dfrac{1}{\lambda^2}E\left[X\right] \\[15pt] &=\dfrac{1}{\lambda} \end{aligned} I ( λ ) = − E [ ( ∂ λ 2 ∂ 2 log f ( X ; λ ) ) ] = − E [ ∂ λ ∂ ( ∂ λ ∂ { − λ + x log λ − x ! } ) ] = − E [ ∂ λ ∂ ( − 1 + λ X ) ] = − E [ − λ 2 X ] = λ 2 1 E [ X ] = λ 1 따라서 Var ( X ˉ n ) = 1 n I ( μ ) = λ / n \text{Var}(\bar{X}_n)=\dfrac{1}{nI(\mu)}=\lambda/n Var ( X ˉ n ) = n I ( μ ) 1 = λ / n X ˉ n \bar{X}_n X ˉ n
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3) 일치추정량
T n ( X ) T_n(X) T n ( X ) g ( λ ) g(\lambda) g ( λ ) λ ∈ Ω \lambda\in\Omega λ ∈ Ω
MSE ( T n ) = Var ( T n ) + [ E ( T n ) − g ( λ ) ] 2 \text{MSE}(T_n)=\text{Var}(T_n)+[E(T_n)-g(\lambda)]^2 MSE ( T n ) = Var ( T n ) + [ E ( T n ) − g ( λ ) ] 2 이므로 T n T_n T n g ( λ ) g(\lambda) g ( λ ) lim n → ∞ Var ( T n ) = 0 \lim_{n\rightarrow\infty}\text{Var}(T_n)=0 lim n → ∞ Var ( T n ) = 0 T n T_n T n 정리 4.9 참조)
X ∼ POI ( λ ) X\sim \text{POI}(\lambda) X ∼ POI ( λ ) X 1 , X 2 , … , X n X_1,X_2,\ldots,X_n X 1 , X 2 , … , X n X ˉ n \bar{X}_n X ˉ n MSE \text{MSE} MSE n n n Var ( X ˉ n ) \text{Var}(\bar{X}_n) Var ( X ˉ n )
lim n → ∞ Var ( X ˉ n ) = lim n → ∞ λ n = 0 \lim_{n\rightarrow\infty}\text{Var}(\bar{X}_n)=\lim_{n\rightarrow\infty}\dfrac{\lambda}{n}=0 n → ∞ lim Var ( X ˉ n ) = n → ∞ lim n λ = 0 으로
MSE ( X ˉ n ) = Var ( X ˉ n ) = 0 \text{MSE}(\bar{X}_n)=\text{Var}(\bar{X}_n)=0 MSE ( X ˉ n ) = Var ( X ˉ n ) = 0 이므로 X ˉ n \bar{X}_n X ˉ n n n n
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직접적으로 일치성을 입증하기 어려운 경우 적절한 조건하에서 점근적 성질을 이용한다면 최대가능도 추정량은 일치성을 가질 뿐 아니라 점근적으로 정규분포를 따르기 때문에 쉽게 근사적 구간추정과 검정이 가능하다.
n ( λ ^ n − λ ) → d N ( 0 , 1 I ( λ ) ) \sqrt{n}(\hat{\lambda}_n-\lambda)\xrightarrow{d}N\left(0,\ \dfrac{1}{I(\lambda)}\right) n ( λ ^ n − λ ) d N ( 0 , I ( λ ) 1 ) X ∼ POI ( λ ) X\sim \text{POI}(\lambda) X ∼ POI ( λ ) X 1 , X 2 , … , X n X_1,X_2,\ldots,X_n X 1 , X 2 , … , X n λ ^ n = X ˉ n \hat{\lambda}_n=\bar{X}_n λ ^ n = X ˉ n
이 때 I ( λ ) I(\lambda) I ( λ )
I ( λ ) = − E [ ( ∂ 2 ∂ λ 2 log f ( X ; λ ) ) ] I(\lambda)=-E\left[\left(\dfrac{\partial^2}{\partial\lambda^2}\text{log }f(X;\lambda)\right)\right] I ( λ ) = − E [ ( ∂ λ 2 ∂ 2 log f ( X ; λ ) ) ] 으로 log f ( X ; λ ) \log f(X;\lambda) log f ( X ; λ )
f ( X ; λ ) = λ X e − λ X ! log f ( X ; λ ) = X log λ + λ − log X ! ( ∂ ∂ λ ) log f ( X ; λ ) = X λ − 1 ( ∂ 2 ∂ λ 2 ) log f ( X ; λ ) = − X λ 2 \begin{aligned} f(X;\lambda) &= \dfrac{\lambda^X e^{-\lambda}}{X!} \\[15pt] \log f(X;\lambda) & =X\log \lambda +\lambda - \log X! \\[15pt] \left(\dfrac{\partial}{\partial\lambda}\right)\text{log }f(X;\lambda) &=\dfrac{X}{\lambda}-1 \\[15pt] \left(\dfrac{\partial^2}{\partial\lambda^2}\right)\text{log }f(X;\lambda) &=-\dfrac{X}{\lambda^2} \end{aligned} f ( X ; λ ) log f ( X ; λ ) ( ∂ λ ∂ ) log f ( X ; λ ) ( ∂ λ 2 ∂ 2 ) log f ( X ; λ ) = X ! λ X e − λ = X log λ + λ − log X ! = λ X − 1 = − λ 2 X 이므로
I ( λ ) = − E ( − X λ 2 ) = 1 λ 2 E ( X ) = 1 λ \begin{aligned} I(\lambda) &=-E\left(-\dfrac{X}{\lambda^2}\right) \\[15pt] &=\dfrac{1}{\lambda^2}E(X) \\[15pt] &=\dfrac{1}{\lambda} \end{aligned} I ( λ ) = − E ( − λ 2 X ) = λ 2 1 E ( X ) = λ 1 따라서 최대가능도 추정량 λ ^ \hat{\lambda} λ ^
n ( λ ^ − λ ) → d N ( 0 , λ ) \sqrt{n}(\hat{\lambda}-\lambda)\xrightarrow{d}N\left(0,\ \lambda\right) n ( λ ^ − λ ) d N ( 0 , λ ) 가 되고, 이것은 중심극한정리에서 얻은 결과와 동일하다.
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7. 구간추정
X ∼ POI ( λ ) X\sim \text{POI}(\lambda) X ∼ POI ( λ ) X 1 , X 2 , … , X n X_1,X_2,\ldots,X_n X 1 , X 2 , … , X n
모분산을 아는경우
표본의 크기가 커질 때 대수의 법칙에 따라 표본평균 X ˉ n → p λ \bar{X}_n\xrightarrow{p}\lambda X ˉ n p λ
n ( X ˉ n − λ ) λ → d N ( 0 , 1 ) \dfrac{\sqrt{n}(\bar{X}_n-\lambda)}{\sqrt{\lambda}}\xrightarrow{d} N(0,1) λ n ( X ˉ n − λ ) d N ( 0 , 1 ) 이므로 슬럿츠키 정리에 의해
Z n = n ( X ˉ n − λ ) X ˉ n → d N ( 0 , 1 ) Z_n=\dfrac{\sqrt{n}(\bar{X}_n-\lambda)}{\sqrt{\bar{X}_n}}\xrightarrow{d} N(0,1) Z n = X ˉ n n ( X ˉ n − λ ) d N ( 0 , 1 ) 로 수렴한다. 따라서 모수 λ \lambda λ
[ X ˉ n − 1.96 X ˉ n n , X ˉ n + 1.96 X ˉ n n ] \left[\bar{X}_n-1.96\sqrt{\dfrac{\bar{X}_n}{n}} , \bar{X}_n+1.96\sqrt{\dfrac{\bar{X}_n}{n}}\right] \\[15pt] [ X ˉ n − 1 . 9 6 n X ˉ n , X ˉ n + 1 . 9 6 n X ˉ n ] \\[40pt]
모분산을 알지 못하는 경우 σ 2 \sigma^2 σ 2 S n 2 S_n^2 S n 2
8. 가설검정
(1) 균일 최강력 검정법
X ∼ POI ( λ ) X\sim \text{POI}(\lambda) X ∼ POI ( λ ) X 1 , X 2 , … , X n X_1,X_2,\ldots,X_n X 1 , X 2 , … , X n
H 0 : λ ≤ λ 0 vs H 0 : λ > λ 0 H_0:\lambda\le \lambda_0\quad\text{vs}\quad H_0:\lambda > \lambda_0 H 0 : λ ≤ λ 0 vs H 0 : λ > λ 0 그러면 λ 1 < λ 2 \lambda_1<\lambda_2 λ 1 < λ 2
L R ( λ 1 , λ 2 ; x 1 , x 2 , … , x n ) = exp ( − n λ 1 ) λ 1 ∑ i = 1 n x i / ∏ i = 1 n x i ! exp ( − n λ 2 ) λ 2 ∑ i = 1 n x i / ∏ i = 1 n x i ! = exp ( − n λ 1 ) λ 1 ∑ i = 1 n x i exp ( − n λ 2 ) λ 2 ∑ i = 1 n x i = exp ( − n ( λ 1 − λ 2 ) ) ( λ 1 λ 2 ) ∑ i = 1 n x i \begin{aligned} LR(\lambda_1,\lambda_2;x_1,x_2,\ldots,x_n) &=\dfrac{\exp(-n\lambda_1)\lambda_1^{\sum_{i=1}^{n}x_i}/\prod_{i=1}^{n}x_i!}{\exp(-n\lambda_2)\lambda_2^{\sum_{i=1}^{n}x_i}/\prod_{i=1}^{n}x_i!} \\[20pt] &= \frac{\exp(-n\lambda_1) \lambda_1^{\sum_{i=1}^n x_i}}{\exp(-n\lambda_2) \lambda_2^{\sum_{i=1}^n x_i}} \\[15pt] &= \exp(-n(\lambda_1-\lambda_2)) \left(\frac{\lambda_1}{\lambda_2}\right)^{\sum_{i=1}^n x_i} \end{aligned} L R ( λ 1 , λ 2 ; x 1 , x 2 , … , x n ) = exp ( − n λ 2 ) λ 2 ∑ i = 1 n x i / ∏ i = 1 n x i ! exp ( − n λ 1 ) λ 1 ∑ i = 1 n x i / ∏ i = 1 n x i ! = exp ( − n λ 2 ) λ 2 ∑ i = 1 n x i exp ( − n λ 1 ) λ 1 ∑ i = 1 n x i = exp ( − n ( λ 1 − λ 2 ) ) ( λ 2 λ 1 ) ∑ i = 1 n x i 따라서 이 경우 가능도함수는 T ( x 1 , x 2 , … , x n ) = ∑ i = 1 n x i T(x_1,x_2,\ldots,x_n)=\sum_{i=1}^{n}x_i T ( x 1 , x 2 , … , x n ) = ∑ i = 1 n x i ( λ 1 < λ 2 ) (\lambda_1<\lambda_2) ( λ 1 < λ 2 ) α = P ( ∑ i = 1 n x i ≥ k ∣ λ 0 ) \alpha=P(\sum_{i=1}^{n}x_i\ge k|\lambda_0) α = P ( ∑ i = 1 n x i ≥ k ∣ λ 0 )
C = { ( x 1 , x 2 , … , x n ) : ∑ i = 1 n x i ≥ k } C=\{(x_1,x_2,\ldots,x_n):\sum_{i=1}^{n}x_i\ge k\} C = { ( x 1 , x 2 , … , x n ) : i = 1 ∑ n x i ≥ k } 로 주어지며, 상수 k k k α = P ( ∑ i = 1 n x i ≥ k ∣ λ 0 ) \alpha=P(\sum_{i=1}^{n}x_i\ge k|\lambda_0) α = P ( ∑ i = 1 n x i ≥ k ∣ λ 0 )
P ( ∑ i = 1 n x i ≥ k ∣ λ 0 ) = P [ ∑ i = 1 n x i − E [ ∑ i = 1 n x i ] Var ( ∑ i = 1 n x i ) ≥ k − E [ ∑ i = 1 n x i ] Var ( ∑ i = 1 n x i ) ∣ λ 0 ] = P [ ∑ i = 1 n x i − n λ 0 n λ 0 ≥ k − n λ 0 n λ 0 ∣ λ 0 ] ≈ P ( Z ≥ k − n λ 0 n λ 0 ∣ λ 0 ) = α \begin{aligned} P(\sum_{i=1}^{n}x_i\ge k|\lambda_0) &=P\left[\left.\dfrac{\sum_{i=1}^{n}x_i - E[\sum_{i=1}^{n}x_i]}{\sqrt{\text{Var}(\sum_{i=1}^{n}x_i)}}\ge\dfrac{k - E[\sum_{i=1}^{n}x_i]}{\sqrt{\text{Var}(\sum_{i=1}^{n}x_i)}}\right|\lambda_0\right] \\[15pt] &=P\left[\left.\dfrac{\sum_{i=1}^{n}x_i-n\lambda_0}{\sqrt{n\lambda_0}}\ge\dfrac{k-n\lambda_0}{\sqrt{n\lambda_0}}\right|\lambda_0\right] \\[15pt] &\approx P\left(\left.Z\ge\dfrac{k-n\lambda_0}{\sqrt{n\lambda_0}}\right|\lambda_0\right) \\[15pt] &=\alpha \end{aligned} P ( i = 1 ∑ n x i ≥ k ∣ λ 0 ) = P [ Var ( ∑ i = 1 n x i ) ∑ i = 1 n x i − E [ ∑ i = 1 n x i ] ≥ Var ( ∑ i = 1 n x i ) k − E [ ∑ i = 1 n x i ] ∣ ∣ ∣ ∣ ∣ ∣ λ 0 ] = P [ n λ 0 ∑ i = 1 n x i − n λ 0 ≥ n λ 0 k − n λ 0 ∣ ∣ ∣ ∣ ∣ λ 0 ] ≈ P ( Z ≥ n λ 0 k − n λ 0 ∣ ∣ ∣ ∣ ∣ λ 0 ) = α 가 성립하므로, k = n λ 0 + z α n λ 0 k=n\lambda_0+z_{\alpha}\sqrt{n\lambda_0} k = n λ 0 + z α n λ 0
C = { ( x 1 , x 2 , … , x n ) : ∑ i = 1 n x i ≥ n λ 0 + z α n λ 0 } C=\{(x_1,x_2,\ldots,x_n):\sum_{i=1}^{n}x_i\ge n\lambda_0 + z_{\alpha}\sqrt{n\lambda_0}\} C = { ( x 1 , x 2 , … , x n ) : i = 1 ∑ n x i ≥ n λ 0 + z α n λ 0 } \\[40pt]
(2) 근사 가능도비 검정법
일반화 가능도비를 적절히 변환시킴으로써 필요한 검정통계량의 분포를 구하기 어려움이 존재해 근사 가능도비 검정법을 이용하여 일반화 가능도비 통계량의 근사분포를 구할 수 있다.
X ∼ POI ( λ ) X\sim \text{POI}(\lambda) X ∼ POI ( λ ) X 1 , X 2 , … , X n X_1,X_2,\ldots,X_n X 1 , X 2 , … , X n
H 0 : λ ≤ λ 0 vs H 0 : λ > λ 0 H_0:\lambda\le \lambda_0\quad\text{vs}\quad H_0:\lambda > \lambda_0 H 0 : λ ≤ λ 0 vs H 0 : λ > λ 0 점근적 정규성 (정리4.13) 조건을 만족할 때, − 2 log Λ -2\text{log}\Lambda − 2 log Λ k k k − 2 log Λ -2\text{log}\Lambda − 2 log Λ Λ \Lambda Λ Λ \Lambda Λ − 2 log Λ -2\text{log}\Lambda − 2 log Λ
− 2 log Λ ≥ χ α ( k ) -2\text{log}\Lambda \ge \chi_{\alpha}(k) − 2 log Λ ≥ χ α ( k ) 따라서 유의수준 α \alpha α { ( x 1 , x 2 , … , x n ) : − 2 log Λ ≥ χ α ( 1 ) } \left\{(x_1,x_2,\ldots,x_n):-2\text{log}\Lambda\ge \chi_{\alpha}(1)\right\} { ( x 1 , x 2 , … , x n ) : − 2 log Λ ≥ χ α ( 1 ) }
최대가능도 추정량이 X ˉ n \bar{X}_n X ˉ n
Λ ( X 1 , X 2 … , X n ) = exp ( − n λ 0 ) λ 0 ∑ i = 1 n x i / ∏ i = 1 n x i ! exp ( − n X ˉ n ) X ˉ n ∑ i = 1 n x i / ∏ i = 1 n x i ! = exp ( − n λ 0 ) λ 0 ∑ i = 1 n x i exp ( − n X ˉ n ) X ˉ n ∑ i = 1 n x i = exp ( − n ( λ 0 − X ˉ n ) ) ( λ 0 X ˉ n ) ∑ i = 1 n x i \begin{aligned} \Lambda (X_1,X_2\ldots,X_n) &=\dfrac{\exp(-n\lambda_0)\lambda_0^{\sum_{i=1}^{n}x_i}/\prod_{i=1}^{n}x_i!}{\exp(-n\bar{X}_n)\bar{X}_n^{\sum_{i=1}^{n}x_i}/\prod_{i=1}^{n}x_i!} \\[20pt] &= \frac{\exp(-n\lambda_0) \lambda_0^{\sum_{i=1}^n x_i}}{\exp(-n\bar{X}_n) \bar{X}_n^{\sum_{i=1}^n x_i}} \\[15pt] &= \exp(-n(\lambda_0-\bar{X}_n)) \left(\dfrac{\lambda_0}{\bar{X}_n}\right)^{\sum_{i=1}^n x_i} \end{aligned} Λ ( X 1 , X 2 … , X n ) = exp ( − n X ˉ n ) X ˉ n ∑ i = 1 n x i / ∏ i = 1 n x i ! exp ( − n λ 0 ) λ 0 ∑ i = 1 n x i / ∏ i = 1 n x i ! = exp ( − n X ˉ n ) X ˉ n ∑ i = 1 n x i exp ( − n λ 0 ) λ 0 ∑ i = 1 n x i = exp ( − n ( λ 0 − X ˉ n ) ) ( X ˉ n λ 0 ) ∑ i = 1 n x i 따라서 귀무가설 하에
− 2 log Λ = − 2 { − n ( λ 0 − X ˉ n ) − n X ˉ n ⋅ log ( X ˉ n λ 0 ) } \begin{aligned} -2\log\Lambda &=-2\left\{-n(\lambda_0-\bar{X}_n) - n\bar{X}_n\cdot \log \left(\dfrac{\bar{X}_n}{\lambda_0}\right)\right\} \\[15pt] \end{aligned} − 2 log Λ = − 2 { − n ( λ 0 − X ˉ n ) − n X ˉ n ⋅ log ( λ 0 X ˉ n ) } 이다. 테일러 정리로부터 ∣ x ∣ ≈ 0 |x|\approx 0 ∣ x ∣ ≈ 0 log ( 1 + x ) = x \log(1+x)=x log ( 1 + x ) = x
log ( X ˉ n λ 0 ) = log ( 1 + X ˉ n − λ 0 λ 0 ) ≈ X ˉ n − λ 0 λ 0 \log\left(\dfrac{\bar{X}_n}{\lambda_0}\right) = \log\left(1+\dfrac{\bar{X}_n-\lambda_0}{\lambda_0}\right) \approx\dfrac{\bar{X}_n-\lambda_0}{\lambda_0} log ( λ 0 X ˉ n ) = log ( 1 + λ 0 X ˉ n − λ 0 ) ≈ λ 0 X ˉ n − λ 0 로 이를 대입하면
− 2 log Λ = − 2 { − n ( λ 0 − X ˉ n ) − n X ˉ n ⋅ log ( X ˉ n λ 0 ) } = − 2 { − n ( λ 0 − X ˉ n ) − n X ˉ n ⋅ ( X ˉ n − λ 0 λ 0 ) } = − 2 { − n λ 0 ( λ 0 − X ˉ n ) λ 0 − n X ˉ n ( X ˉ n − λ 0 ) λ 0 } = − 2 { − n ( X ˉ n 2 − 2 λ 0 X ˉ n + λ 0 2 ) λ 0 } = 2 n ( X ˉ n − λ 0 λ 0 ) 2 ≈ χ α 2 ( 1 ) \begin{aligned} -2\log\Lambda &=-2\left\{-n(\lambda_0-\bar{X}_n) - n\bar{X}_n \cdot \log \left(\dfrac{\bar{X}_n}{\lambda_0}\right)\right\} \\[15pt] &=-2\left\{-n(\lambda_0-\bar{X}_n) - n\bar{X}_n \cdot \left(\dfrac{\bar{X}_n-\lambda_0}{\lambda_0}\right)\right\} \\[15pt] &=-2\left\{\dfrac{-n\lambda_0(\lambda_0-\bar{X}_n)}{\lambda_0} - \dfrac{n\bar{X}_n\left(\bar{X}_n-\lambda_0\right)}{\lambda_0}\right\} \\[15pt] &=-2\left\{\dfrac{-n(\bar{X}_n^2-2\lambda_0\bar{X}_n+\lambda_0^2)}{\lambda_0}\right\} \\[15pt] &=2n\left(\dfrac{\bar{X}_n-\lambda_0}{\sqrt{\lambda_0}}\right)^2 \\[15pt] &\approx\chi_\alpha^2(1) \end{aligned} − 2 log Λ = − 2 { − n ( λ 0 − X ˉ n ) − n X ˉ n ⋅ log ( λ 0 X ˉ n ) } = − 2 { − n ( λ 0 − X ˉ n ) − n X ˉ n ⋅ ( λ 0 X ˉ n − λ 0 ) } = − 2 { λ 0 − n λ 0 ( λ 0 − X ˉ n ) − λ 0 n X ˉ n ( X ˉ n − λ 0 ) } = − 2 { λ 0 − n ( X ˉ n 2 − 2 λ 0 X ˉ n + λ 0 2 ) } = 2 n ( λ 0 X ˉ n − λ 0 ) 2 ≈ χ α 2 ( 1 ) 이므로 − 2 log Λ -2\log\Lambda − 2 log Λ 2 ⋅ χ 2 ( α ) 2\cdot\chi^2(\alpha) 2 ⋅ χ 2 ( α )
