포아송 확률과정을 고려했을 때, 특정한 사건 A A A X X X
X ∼ EXP ( λ ) X\sim \text{EXP}(\lambda) X ∼ EXP ( λ ) 1. 확률밀도함수와 확률분포함수
X ∼ EXP ( λ ) X\sim \text{EXP}(\lambda) X ∼ EXP ( λ )
1) 확률밀도함수
f ( x ) = 1 λ exp ( − x / λ ) ( x > 0 ) f(x)=\dfrac{1}{\lambda}\exp(-x/\lambda)\quad (x>0) f ( x ) = λ 1 exp ( − x / λ ) ( x > 0 ) \\[20pt]
2) 확률분포함수
F ( x ) = ∫ 0 x 1 λ exp ( − t / λ ) d t = 1 − exp ( − x / λ ) \begin{aligned} F(x) &=\int_{0}^{x}\dfrac{1}{\lambda}\exp(-t/\lambda)dt \\[10pt] &=1-\exp(-x/\lambda) \end{aligned} F ( x ) = ∫ 0 x λ 1 exp ( − t / λ ) d t = 1 − exp ( − x / λ ) \\[30pt]
2. 기댓값과 분산
X ∼ EXP ( λ ) X\sim \text{EXP}(\lambda) X ∼ EXP ( λ )
1) 기댓값
E ( X ) = ∫ 0 ∞ x f ( x ) d x = ∫ 0 ∞ x 1 λ exp ( − x / λ ) d x = − x exp ( − x λ ) ∣ 0 ∞ + ∫ 0 ∞ exp ( − x λ ) d x = λ exp ( − x λ ) ∣ 0 ∞ = λ \begin{aligned} E(X) &=\int_{0}^{\infty}xf(x)dx \\[15pt] &=\int_{0}^{\infty}x\dfrac{1}{\lambda}\exp(-x/\lambda)\ dx \\[15pt] &=\left.-x\exp\left(-\dfrac{x}{\lambda}\right)\right|_{0}^{\infty}+\int_{0}^{\infty}\exp\left(-\dfrac{x}{\lambda}\right)\ dx \\[15pt] &=\left.\lambda\exp\left(-\dfrac{x}{\lambda}\right)\right|_{0}^{\infty} \\[10pt] &=\lambda \end{aligned} E ( X ) = ∫ 0 ∞ x f ( x ) d x = ∫ 0 ∞ x λ 1 exp ( − x / λ ) d x = − x exp ( − λ x ) ∣ ∣ ∣ ∣ 0 ∞ + ∫ 0 ∞ exp ( − λ x ) d x = λ exp ( − λ x ) ∣ ∣ ∣ ∣ 0 ∞ = λ \\[20pt]
2) 분산
E ( X 2 ) = ∫ 0 ∞ x 2 f ( x ) d x = ∫ 0 ∞ x 2 1 λ exp ( − x / λ ) d x = − x 2 exp ( − x λ ) ∣ 0 ∞ + 2 ∫ 0 ∞ x exp ( − x λ ) d x = 0 + 2 λ ∫ 0 ∞ x 1 λ exp ( − x λ ) d x = 2 λ E ( X ) = 2 λ 2 ∴ Var ( X ) = 2 λ 2 − λ 2 = λ 2 \begin{aligned} E(X^2) &=\int_{0}^{\infty}x^2f(x)dx \\[15pt] &=\int_{0}^{\infty}x^2\dfrac{1}{\lambda}\exp(-x/\lambda)\ dx \\[15pt] &=\left.-x^2\exp\left(-\dfrac{x}{\lambda}\right)\right|_{0}^{\infty} + 2\int_{0}^{\infty}x\exp\left(-\dfrac{x}{\lambda}\right)\ dx \\[15pt] &= 0+ 2\lambda\int_{0}^{\infty}x\dfrac{1}{\lambda}\exp\left(-\dfrac{x}{\lambda}\right)\ dx \\[15pt] &=2\lambda E(X) \\[10pt] &= 2\lambda^2 \\[20pt] \therefore \text{Var}(X) &=2\lambda^2-\lambda^2=\lambda^2 \end{aligned} E ( X 2 ) ∴ Var ( X ) = ∫ 0 ∞ x 2 f ( x ) d x = ∫ 0 ∞ x 2 λ 1 exp ( − x / λ ) d x = − x 2 exp ( − λ x ) ∣ ∣ ∣ ∣ 0 ∞ + 2 ∫ 0 ∞ x exp ( − λ x ) d x = 0 + 2 λ ∫ 0 ∞ x λ 1 exp ( − λ x ) d x = 2 λ E ( X ) = 2 λ 2 = 2 λ 2 − λ 2 = λ 2 \\[20pt]
예 X X X θ = 1 / 2 \theta = 1/2 θ = 1 / 2 λ = 2 \lambda=2 λ = 2
f X ( x ) = ( 1 / 2 ) e − x / 2 f_X(x)=(1/2)e^{-x/2} f X ( x ) = ( 1 / 2 ) e − x / 2 따라서
P ( X ≥ 5 ) = ∫ 5 ∞ ( 1 / 2 ) e − x / 2 d x P(X\geq5)=\int_{5}^{\infty}(1/2)e^{-x/2}dx \\ P ( X ≥ 5 ) = ∫ 5 ∞ ( 1 / 2 ) e − x / 2 d x \\[30pt]
3. 적률생성함수
X ∼ EXP ( λ ) X\sim \text{EXP}(\lambda) X ∼ EXP ( λ )
M X ( t ) = E ( e t X ) = ∫ 0 ∞ e t x ⋅ 1 λ exp { ( − x / λ ) } d x = ∫ 0 ∞ 1 λ exp { ( t − 1 / λ ) x } d x = 1 λ ∣ λ λ t − 1 [ exp { ( t − 1 / λ ) x } ] ∣ 0 ∞ = 1 λ t − 1 [ lim x → ∞ exp { ( t − 1 / λ ) x } − 1 ] = 1 1 − λ t , t < 1 λ \begin{aligned} M_X(t) &=E(e^{tX}) \\[10pt] &=\int_{0}^{\infty}e^{tx}\cdot\dfrac{1}{\lambda}\exp\{(-x/\lambda)\} dx\\[15pt] &=\int_{0}^{\infty}\dfrac{1}{\lambda}\exp\{(t-1/\lambda)x\} dx\\[15pt] &=\dfrac{1}{\lambda}\left|\dfrac{\lambda}{\lambda t-1}\left[\exp\{(t-1/\lambda)x\}\right]\right|_{0}^{\infty}\\[15pt] &=\dfrac{1}{\lambda t-1}[\lim_{x\rightarrow \infty}\exp\{(t-1/\lambda)x\}-1]\\[10pt] &=\dfrac{1}{1-\lambda t},\quad t<\dfrac{1}{\lambda} \end{aligned} M X ( t ) = E ( e t X ) = ∫ 0 ∞ e t x ⋅ λ 1 exp { ( − x / λ ) } d x = ∫ 0 ∞ λ 1 exp { ( t − 1 / λ ) x } d x = λ 1 ∣ ∣ ∣ ∣ ∣ λ t − 1 λ [ exp { ( t − 1 / λ ) x } ] ∣ ∣ ∣ ∣ ∣ 0 ∞ = λ t − 1 1 [ x → ∞ lim exp { ( t − 1 / λ ) x } − 1 ] = 1 − λ t 1 , t < λ 1 따라서 적률생성함수를 활용한 기댓값과 분산은
기댓값M X ( 1 ) ( t ) = ( d d t ) ( 1 1 − λ t ) = − ( 1 − λ t ) ′ ( 1 − λ t ) 2 = λ ( 1 − λ t ) 2 ∴ E ( X ) = M X ( 1 ) ( 0 ) = λ \begin{aligned} M^{(1)}_X(t) &=\left(\dfrac{d}{dt}\right)\left(\dfrac{1}{1-\lambda t}\right)=\dfrac{-(1-\lambda t)'}{(1-\lambda t)^2}=\dfrac{\lambda}{(1-\lambda t)^2}\\[30pt] \therefore E(X) &=M^{(1)}_X(0)=\lambda \end{aligned} M X ( 1 ) ( t ) ∴ E ( X ) = ( d t d ) ( 1 − λ t 1 ) = ( 1 − λ t ) 2 − ( 1 − λ t ) ′ = ( 1 − λ t ) 2 λ = M X ( 1 ) ( 0 ) = λ
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M X ( 2 ) ( t ) = ( d d t ) ( λ ( 1 − λ t ) 2 ) = λ ⋅ − ( d / d t ) ( 1 − λ t ) 2 ( 1 − λ t ) 4 = 2 λ 2 ( 1 − λ t ) 3 M X ( 2 ) ( 0 ) = 2 λ 2 ∴ Var ( X ) = M X ( 2 ) ( 0 ) − { M X ( 1 ) ( 0 ) } 2 = λ 2 \begin{aligned} M^{(2)}_X(t) &=\left(\dfrac{d}{dt}\right)\left(\dfrac{\lambda}{(1-\lambda t)^2}\right)=\lambda\cdot\dfrac{-(d/dt)(1-\lambda t)^2}{(1-\lambda t)^4}=\dfrac{2\lambda^2}{(1-\lambda t)^3} \\[20pt] M^{(2)}_X(0)&=2\lambda^2 \\[20pt] \therefore \text{Var}(X)&=M^{(2)}_X(0)-\left\{M^{(1)}_X(0)\right\}^2=\lambda^2 \end{aligned} M X ( 2 ) ( t ) M X ( 2 ) ( 0 ) ∴ Var ( X ) = ( d t d ) ( ( 1 − λ t ) 2 λ ) = λ ⋅ ( 1 − λ t ) 4 − ( d / d t ) ( 1 − λ t ) 2 = ( 1 − λ t ) 3 2 λ 2 = 2 λ 2 = M X ( 2 ) ( 0 ) − { M X ( 1 ) ( 0 ) } 2 = λ 2 \\[20pt]
X 1 , X 2 , … , X n X_1,X_2,\ldots,X_n X 1 , X 2 , … , X n X i , ( i = 1 , 2 , … , n ) X_i,\ (i=1,2,\ldots,n) X i , ( i = 1 , 2 , … , n ) λ \lambda λ
M X i ( t ) = 1 1 − λ t , t < 1 λ M_{X_i}(t)=\dfrac{1}{1-\lambda t},\quad t<\dfrac{1}{\lambda} M X i ( t ) = 1 − λ t 1 , t < λ 1 이므로 Y = X 1 + X 2 + ⋯ + X n Y=X_1+X_2+\cdots+X_n Y = X 1 + X 2 + ⋯ + X n
M Y ( t ) = E [ exp ( t Y ) ] = E [ exp ( t ( X 1 , X 2 , … , X n ) ) ] = E [ exp ( t X 1 ) ] ⋅ E [ exp ( t X 2 ) ] … ⋅ E [ exp ( t X n ) ] = ∏ i = 1 n M X i ( t ) = ( 1 1 − λ t ) n , t < 1 λ \begin{aligned} M_Y(t)&= E[\exp(tY)]\\[10pt] &= E[\exp(t(X_1,X_2,\ldots,X_n))]\\[10pt] &= E[\exp(tX_1)]\cdot E[\exp(tX_2)]\ldots\cdot E[\exp(tX_n)] \\[5pt] &=\prod_{i=1}^n M_{X_i}(t) \\[10pt] &=\left(\dfrac{1}{1-\lambda t}\right)^n,\quad t<\dfrac{1}{\lambda} \end{aligned} M Y ( t ) = E [ exp ( t Y ) ] = E [ exp ( t ( X 1 , X 2 , … , X n ) ) ] = E [ exp ( t X 1 ) ] ⋅ E [ exp ( t X 2 ) ] … ⋅ E [ exp ( t X n ) ] = i = 1 ∏ n M X i ( t ) = ( 1 − λ t 1 ) n , t < λ 1 이며 이는 GAM ( n , λ ) \text{GAM}(n,\ \lambda) GAM ( n , λ ) Y ∼ GAM ( n , λ ) Y\sim \text{GAM}(n,\ \lambda) Y ∼ GAM ( n , λ )
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4. 표본분포의 근사
X ∼ EXP ( λ ) X\sim \text{EXP}(\lambda) X ∼ EXP ( 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}\cdot \sum_{i=1}^{n}E(X_i) \\[15pt] &=\dfrac{1}{n}\cdot n\cdot E(X) \\[10pt] &=\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 ) = 1 n 2 ⋅ n ⋅ Var ( X ) = σ 2 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{1}{n^2}\cdot n\cdot \text{Var}(X) \\[15pt] &=\dfrac{\sigma^2}{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 1 ⋅ n ⋅ Var ( X ) = n σ 2 \\[20pt]
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 Var ( X ) − n Var ( X ˉ ) ] = 1 n − 1 [ n λ 2 − n λ 2 n ] = λ 2 \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\text{Var}(X)-n\text{Var}(\bar{X})\right] \\[15pt] &=\dfrac{1}{n-1}\left[n\lambda^2-n\dfrac{\lambda^2}{n}\right] \\[15pt] &=\lambda^2 \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 Var ( X ) − n Var ( X ˉ ) ] = n − 1 1 [ n λ 2 − n n λ 2 ] = λ 2 \\[20pt]
1) 대수의 법칙
X ∼ EXP ( λ ) X\sim \text{EXP}(\lambda) X ∼ EXP ( 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 2 / 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^2/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 2 / n → 1 \\[20pt]
2) 중심극한정리
X ∼ EXP ( λ ) X\sim \text{EXP}(\lambda) X ∼ EXP ( 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 2 ) Z n = ( X ˉ n − λ ) λ 2 / n → d N ( 0 , 1 ) Z n = ∑ i = 1 n ( X i − μ ) n λ 2 → d N ( 0 , 1 ) \begin{aligned} Z_n&=\sqrt{n}(\bar{X}_n-\lambda)\xrightarrow{d}N(0,\ \lambda^2) \\[20pt] Z_n&=\dfrac{(\bar{X}_n-\lambda)}{\sqrt{\lambda^2/n}}\xrightarrow{d}N(0,\ 1) \\[20pt] Z_n&=\dfrac{\sum_{i=1}^{n}(X_i-\mu)}{\sqrt{n\lambda^2}}\xrightarrow{d}N(0,\ 1) \end{aligned} Z n Z n Z n = n ( X ˉ n − λ ) d N ( 0 , λ 2 ) = λ 2 / n ( X ˉ n − λ ) d N ( 0 , 1 ) = n λ 2 ∑ i = 1 n ( X i − μ ) d N ( 0 , 1 ) \\[20pt]
3) 델타방법
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 2 ) \sqrt{n}(X_n-\lambda)\xrightarrow{d}N(0,\lambda^2) n ( X n − λ ) d N ( 0 , λ 2 ) 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 ) 이 성립하며 이를 델타 방법이라한다. 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 2 \sigma^2=\lambda^2 σ 2 = λ 2 g ′ ( λ ) = 2 λ g'(\lambda)=2\lambda g ′ ( λ ) = 2 λ
n ( X ˉ n 2 − λ 2 ) → d N ( 0 , 4 λ 4 ) \sqrt{n}(\bar{X}_n^2-\lambda^2)\xrightarrow{d}N(0,4\lambda^4) n ( X ˉ n 2 − λ 2 ) d N ( 0 , 4 λ 4 ) \\[30pt]
5. 추정량
1) 적률 추정량
X ∼ EXP ( λ ) X\sim \text{EXP}(\lambda) X ∼ EXP ( 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 \\[20pt]
2) 최대가능도 추정량
X ∼ EXP ( λ ) X\sim \text{EXP}(\lambda) X ∼ EXP ( 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 ) = ( 1 λ ) n exp ( ∑ i = 1 n − x i / λ ) \begin{aligned} L(\lambda;x_1,x_2,\ldots,x_n) &=\prod_{i=1}^{n}f(x_i;\lambda)\\[15pt] &=\left(\dfrac{1}{\lambda}\right)^n\exp(\sum_{i=1}^{n}-x_i/\lambda) \end{aligned} L ( λ ; x 1 , x 2 , … , x n ) = i = 1 ∏ n f ( x i ; λ ) = ( λ 1 ) n exp ( i = 1 ∑ n − x i / λ ) 가 되며, 따라서 로그가능도함수는
log L ( λ ; x 1 , x 2 , … , x n ) = − n log λ − ∑ i = 1 n − x i / λ \text{log}L(\lambda;x_1,x_2,\ldots,x_n)=-n\text{log}\lambda -\sum_{i=1}^{n}-x_i/\lambda log L ( λ ; x 1 , x 2 , … , x n ) = − n log λ − i = 1 ∑ n − x i / λ 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 λ + ∑ i = 1 n x i λ 2 = 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) &=-\dfrac{n}{\lambda}+\sum_{i=1}^{n}\dfrac{x_i}{\lambda^2}=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 ) = − λ n + i = 1 ∑ n λ 2 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 = n λ 2 ∣ λ = x ˉ n − 2 ∑ i = 1 n x i λ 3 ∣ λ = x ˉ n = n x ˉ n 2 − 2 n x ˉ n 3 ⋅ 1 n ∑ i = 1 n x i = n x ˉ n 2 − 2 n x ˉ n 2 = − n x ˉ n 2 < 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} &=\left.\dfrac{n}{\lambda^2}\right|_{\lambda=\bar{x}_n}-2\sum_{i=1}^{n}\left.\dfrac{x_i}{\lambda^3}\right|_{\lambda=\bar{x}_n} \\[15pt] &=\dfrac{n}{\bar{x}_n^2}-\dfrac{2n}{\bar{x}_n^3}\cdot\dfrac{1}{n}\sum_{i=1}^{n}x_i \\[15pt] &=\dfrac{n}{\bar{x}_n^2}-\dfrac{2n}{\bar{x}_n^2} \\[15pt] &=-\dfrac{n}{\bar{x}_n^2}<0 \end{aligned} d λ 2 d 2 log L ( λ ; x 1 , x 2 , … , x n ) ∣ ∣ ∣ ∣ ∣ λ = x ˉ n = λ 2 n ∣ ∣ ∣ ∣ λ = x ˉ n − 2 i = 1 ∑ n λ 3 x i ∣ ∣ ∣ ∣ λ = x ˉ n = x ˉ n 2 n − x ˉ n 3 2 n ⋅ n 1 i = 1 ∑ n x i = x ˉ n 2 n − x ˉ n 2 2 n = − x ˉ n 2 n < 0 을 만족시키므로(모든 x ˉ n \bar{x}_n x ˉ n X ˉ n \bar{X}_n X ˉ n λ ^ \hat{\lambda} λ ^
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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 \\[30pt]
6. 좋은 추정량 조건
최대가능도 추정량인 경우만 고려하였다.
1) 비편향추정량
X ∼ EXP ( λ ) X\sim \text{EXP}(\lambda) X ∼ EXP ( 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 ∼ EXP ( λ ) X\sim \text{EXP}(\lambda) X ∼ EXP ( 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 θ \theta θ \\[10pt]
이 때 피셔의 정보량 I ( θ ) I(\theta) I ( θ )
I ( θ ) = E [ ( ∂ ∂ θ log f ( X ; θ ) ) 2 ] = − E [ ( ∂ 2 ∂ θ 2 log f ( X ; θ ) ) ] I(\theta)=E\left[\left(\dfrac{\partial}{\partial\theta}\text{log }f(X;\theta)\right)^2\right]=-E\left[\left(\dfrac{\partial^2}{\partial\theta^2}\text{log }f(X;\theta)\right)\right] I ( θ ) = E [ ( ∂ θ ∂ log f ( X ; θ ) ) 2 ] = − E [ ( ∂ θ 2 ∂ 2 log f ( X ; θ ) ) ] 계산을 하면
I ( λ ) = E [ ( ∂ ∂ λ log f ( X ; λ ) ) 2 ] = E [ ( ∂ ∂ λ { − log λ − X λ } ) 2 ] = E [ − 1 λ + X λ 2 ] 2 = E [ 1 λ 2 − 2 X λ 3 + X 2 λ 4 ] = 1 λ 2 − 2 λ 2 + 2 λ 2 ( ∵ E ( X ) = λ , E ( X 2 ) = 2 λ 2 ) = 1 λ 2 \begin{aligned} I(\lambda) &=E\left[\left(\dfrac{\partial}{\partial \lambda}\text{log }f(X;\lambda)\right)^2\right] \\[15pt] &=E\left[\left(\dfrac{\partial}{\partial \lambda}\left\{-\text{log } \lambda - \dfrac{X}{\lambda}\right\}\right)^2\right] \\[15pt] &=E\left[-\frac{1}{\lambda} + \dfrac{X}{\lambda^2}\right]^2 \\[15pt] &=E\left[\dfrac{1}{\lambda^2} - \dfrac{2X}{\lambda^3} + \dfrac{X^2}{\lambda^4}\right] \\[15pt] &=\dfrac{1}{\lambda^2} -\dfrac{2}{\lambda^2} + \dfrac{2}{\lambda^2}\quad (\because E(X)=\lambda,\ E(X^2)=2\lambda^2) \\[10pt] &=\dfrac{1}{\lambda^2} \end{aligned} I ( λ ) = E [ ( ∂ λ ∂ log f ( X ; λ ) ) 2 ] = E [ ( ∂ λ ∂ { − log λ − λ X } ) 2 ] = E [ − λ 1 + λ 2 X ] 2 = E [ λ 2 1 − λ 3 2 X + λ 4 X 2 ] = λ 2 1 − λ 2 2 + λ 2 2 ( ∵ E ( X ) = λ , E ( X 2 ) = 2 λ 2 ) = λ 2 1 따라서 Var ( X ˉ n ) = 1 n I ( μ ) = λ 2 / n \text{Var}(\bar{X}_n)=\dfrac{1}{nI(\mu)}=\lambda^2/n Var ( X ˉ n ) = n I ( μ ) 1 = λ 2 / n X ˉ n \bar{X}_n X ˉ n
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3) 일치추정량
추정량이 일치성을 가질 조건을 평균제곱오차(MSE)를 이용하여 표현할 경우, T n ( X ) T_n(X) T n ( X ) g ( θ ) g(\theta) g ( θ ) θ ∈ Ω \theta\in\Omega θ ∈ Ω
lim n → ∞ E [ T n ( X ) − g ( θ ) ] 2 = 0 \lim_{n\rightarrow\infty}E[T_n(X)-g(\theta)]^2=0 n → ∞ lim E [ T n ( X ) − g ( θ ) ] 2 = 0 이 성립하면, T n ( X ) T_n(X) T n ( X )
MSE ( T n ) = E [ T n ( X ) − g ( θ ) ] 2 = Var ( T n ) + [ E ( T n ) − g ( θ ) ] 2 \text{MSE}(T_n)=E[T_n(X)-g(\theta)]^2=\text{Var}(T_n)+[E(T_n)-g(\theta)]^2 MSE ( T n ) = E [ T n ( X ) − g ( θ ) ] 2 = Var ( T n ) + [ E ( T n ) − g ( θ ) ] 2 이므로 T n T_n T n g ( θ ) g(\theta) g ( θ ) [ E ( T n ) − g ( θ ) ] = 0 [E(T_n)-g(\theta)]=0 [ E ( T n ) − g ( θ ) ] = 0 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)
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X ∼ EXP ( λ ) X\sim \text{EXP}(\lambda) X ∼ EXP ( 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 [ E ( T n ) − g ( θ ) ] = 0 [E(T_n)-g(\theta)]=0 [ E ( T n ) − g ( θ ) ] = 0 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 2 n = 0 \lim_{n\rightarrow\infty}\text{Var}(\bar{X}_n)=\lim_{n\rightarrow\infty}\dfrac{\lambda^2}{n}=0 n → ∞ lim Var ( X ˉ n ) = n → ∞ lim n λ 2 = 0 으로
lim n → ∞ MSE ( X ˉ n ) = lim n → ∞ Var ( X ˉ n ) + [ E ( T n ) − g ( θ ) ] = 0 + 0 = 0 \begin{aligned} \lim_{n\rightarrow\infty}\text{MSE}(\bar{X}_n) &= \lim_{n\rightarrow\infty}\text{Var}(\bar{X}_n)+ [E(T_n)-g(\theta)] \\[10pt] &= 0+0 \\[10pt] &= 0 \end{aligned} n → ∞ lim MSE ( X ˉ n ) = n → ∞ lim Var ( X ˉ n ) + [ E ( T n ) − g ( θ ) ] = 0 + 0 = 0 이므로 X ˉ n \bar{X}_n X ˉ n 최대추정량 의 분산의 분모는 n n n
한편 n X ( 1 ) ∼ EXP ( λ ) nX_{(1)}\sim \text{EXP}(\lambda) n X ( 1 ) ∼ EXP ( λ ) n X ( 1 ) nX_{(1)} n X ( 1 ) λ \lambda λ ϵ < λ \epsilon<\lambda ϵ < λ ϵ > 0 \epsilon>0 ϵ > 0
P [ ∣ n X ( 1 ) − λ ∣ ≤ ϵ ] = e − 1 ( e ϵ / λ − e − ϵ / λ ) P[|nX_{(1)}-\lambda|\le\epsilon]=e^{-1}(e^{\epsilon/\lambda}-e^{-\epsilon/\lambda}) P [ ∣ n X ( 1 ) − λ ∣ ≤ ϵ ] = e − 1 ( e ϵ / λ − e − ϵ / λ ) 로 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 ∼ EXP ( λ ) X\sim \text{EXP}(\lambda) X ∼ EXP ( 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 ) = 1 λ exp ( − X / λ ) log f ( X ; λ ) = − log λ − X / λ ( ∂ ∂ λ ) log f ( X ; λ ) = − 1 λ + X λ 2 ( ∂ 2 ∂ λ 2 ) log f ( X ; λ ) = 1 λ 2 − 2 X λ 3 \begin{aligned} f(X;\lambda) &= \dfrac{1}{\lambda}\exp(-X/\lambda) \\[15pt] \log f(X;\lambda) & =-\log \lambda -X/\lambda \\[15pt] \left(\dfrac{\partial}{\partial\lambda}\right)\text{log }f(X;\lambda) &=-\dfrac{1}{\lambda}+\dfrac{X}{\lambda^2} \\[15pt] \left(\dfrac{\partial^2}{\partial\lambda^2}\right)\text{log }f(X;\lambda) &=\dfrac{1}{\lambda^2}-\dfrac{2X}{\lambda^3} \end{aligned} f ( X ; λ ) log f ( X ; λ ) ( ∂ λ ∂ ) log f ( X ; λ ) ( ∂ λ 2 ∂ 2 ) log f ( X ; λ ) = λ 1 exp ( − X / λ ) = − log λ − X / λ = − λ 1 + λ 2 X = λ 2 1 − λ 3 2 X 이므로
I ( λ ) = − E ( 1 λ 2 − 2 X λ 3 ) = 2 λ 3 E ( X ) − 1 λ 2 = 2 λ 3 ⋅ λ − 1 λ 2 = 1 λ 2 \begin{aligned} I(\lambda) &=-E\left(\dfrac{1}{\lambda^2}-\dfrac{2X}{\lambda^3}\right) \\[15pt] &=\dfrac{2}{\lambda^3}E(X)-\dfrac{1}{\lambda^2} \\[15pt] &=\dfrac{2}{\lambda^3}\cdot \lambda-\dfrac{1}{\lambda^2} \\[15pt] &=\dfrac{1}{\lambda^2} \end{aligned} I ( λ ) = − E ( λ 2 1 − λ 3 2 X ) = λ 3 2 E ( X ) − λ 2 1 = λ 3 2 ⋅ λ − λ 2 1 = λ 2 1 따라서 최대가능도 추정량 λ ^ \hat{\lambda} λ ^
n ( λ ^ − λ ) → d N ( 0 , λ 2 ) \sqrt{n}(\hat{\lambda}-\lambda)\xrightarrow{d}N\left(0,\ \lambda^2\right) n ( λ ^ − λ ) d N ( 0 , λ 2 ) 가 되고, 이것은 중심극한정리에서 얻은 결과와 동일하다.
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7. 구간추정
X ∼ EXP ( λ ) X\sim \text{EXP}(\lambda) X ∼ EXP ( 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 2 → d N ( 0 , 1 ) \dfrac{\sqrt{n}(\bar{X}_n-\lambda)}{\sqrt{\lambda^2}}\xrightarrow{d} N(0,1) λ 2 n ( X ˉ n − λ ) d N ( 0 , 1 ) 이므로 슬럿츠키 정리에 의해
Z n = n ( X ˉ n − λ ) X ˉ n 2 → d N ( 0 , 1 ) Z_n=\dfrac{\sqrt{n}(\bar{X}_n-\lambda)}{\sqrt{\bar{X}_n^2}}\xrightarrow{d} N(0,1) Z n = X ˉ n 2 n ( X ˉ n − λ ) d N ( 0 , 1 ) 로 수렴한다. 따라서 모수 λ \lambda λ
[ X ˉ n − 1.96 X ˉ n 2 n , X ˉ n + 1.96 X ˉ n 2 n ] \left[\bar{X}_n-1.96\sqrt{\dfrac{\bar{X}_n^2}{n}} , \bar{X}_n+1.96\sqrt{\dfrac{\bar{X}_n^2}{n}}\right] \\[15pt] [ X ˉ n − 1 . 9 6 n X ˉ n 2 , X ˉ n + 1 . 9 6 n X ˉ n 2 ] \\[20pt]
모분산을 알지 못하는 경우 σ 2 \sigma^2 σ 2 S n 2 S_n^2 S n 2
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8. 가설검정
1) 최강력 검정법 (네이만-피어슨 방법)
X ∼ EXP ( λ ) X\sim \text{EXP}(\lambda) X ∼ EXP ( 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 1 ( > λ 0 ) H_0:\lambda=\lambda_0\quad \text{vs}\quad H_0:\lambda=\lambda_1\ (>\lambda_0) H 0 : λ = λ 0 vs H 0 : λ = λ 1 ( > λ 0 ) 에 대한 가능도비는 다음과 같이 주어진다.
L R ( λ 0 , λ 1 ; x 1 , x 2 , … , x n ) = λ 0 − n exp ( − ∑ i = 1 n x i / λ 0 ) λ 1 − n exp ( − ∑ i = 1 n x i / λ 1 ) = ( λ 1 λ 0 ) n exp [ ( 1 λ 1 − 1 λ 0 ) ∑ i = 1 n x i ] \begin{aligned} LR(\lambda_0,\lambda_1;x_1,x_2,\ldots,x_n) &=\dfrac{\lambda_0^{-n}\exp(-\sum_{i=1}^{n}x_i/\lambda_0)} {\lambda_1^{-n}\exp(-\sum_{i=1}^{n}x_i/\lambda_1)} \\[15pt] &=\left(\dfrac{\lambda_1}{\lambda_0}\right)^n \exp\left[\left(\dfrac{1}{\lambda_1}-\dfrac{1}{\lambda_0}\right)\sum_{i=1}^{n}x_i\right] \end{aligned} L R ( λ 0 , λ 1 ; x 1 , x 2 , … , x n ) = λ 1 − n exp ( − ∑ i = 1 n x i / λ 1 ) λ 0 − n exp ( − ∑ i = 1 n x i / λ 0 ) = ( λ 0 λ 1 ) n exp [ ( λ 1 1 − λ 0 1 ) i = 1 ∑ n x i ] 그런데 λ 1 > λ 0 \lambda_1>\lambda_0 λ 1 > λ 0 1 λ 1 − 1 λ 0 < 0 \dfrac{1}{\lambda_1}-\dfrac{1}{\lambda_0}<0 λ 1 1 − λ 0 1 < 0 L R ( λ 0 , λ 1 ; x 1 , x 2 , … , x n ) ≤ k LR(\lambda_0,\lambda_1;x_1,x_2,\ldots,x_n)\le k L R ( λ 0 , λ 1 ; x 1 , x 2 , … , x n ) ≤ k ∑ i = 1 n x i ≥ k 1 \sum_{i=1}^{n}x_i\ge k_1 ∑ i = 1 n x i ≥ k 1
C = { ( x 1 , x 2 , … , x n ) : ∑ i = 1 n x i ≥ k 1 } C=\{(x_1,x_2,\ldots,x_n):\sum_{i=1}^{n}x_i\ge k_1\} C = { ( x 1 , x 2 , … , x n ) : i = 1 ∑ n x i ≥ k 1 } 의 꼴로 주어진다. 그런데 귀무가설 하에, 즉 λ = λ 0 \lambda=\lambda_0 λ = λ 0
∑ i = 1 n 2 x i λ 0 ∼ χ 2 ( 2 n ) \sum_{i=1}^{n}\dfrac{2x_i}{\lambda_0}\sim \chi^2(2n) i = 1 ∑ n λ 0 2 x i ∼ χ 2 ( 2 n ) 을 따른다. 그러므로 가설에 대한 유의수준 α \alpha α
P [ ∑ i = 1 n X i ≥ k 1 ∣ λ = λ 0 ] = α P\left[\sum_{i=1}^{n}X_i\ge k_1|\lambda=\lambda_0\right]=\alpha P [ i = 1 ∑ n X i ≥ k 1 ∣ λ = λ 0 ] = α 로부터
C = { ( x 1 , x 2 , … , x n ) : ∑ i = 1 n x i ≥ ( λ 0 / 2 ) χ 2 ( 2 n ) } C=\{(x_1,x_2,\ldots,x_n):\sum_{i=1}^{n}x_i\ge (\lambda_0/2)\chi^2(2n)\} C = { ( x 1 , x 2 , … , x n ) : i = 1 ∑ n x i ≥ ( λ 0 / 2 ) χ 2 ( 2 n ) } 으로 주어진다.
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2) 균일 최강력 검정법
X ∼ EXP ( λ ) X\sim \text{EXP}(\lambda) X ∼ EXP ( 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 ( 1 / λ 1 − 1 / λ 0 ) < 0 (1/\lambda_1-1/\lambda_0)<0 ( 1 / λ 1 − 1 / λ 0 ) < 0 L R ( λ 0 , λ 1 ; x 1 , x 2 , … , x n ) LR(\lambda_0,\lambda_1;x_1,x_2,\ldots,x_n) L R ( λ 0 , λ 1 ; x 1 , x 2 , … , x n ) 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 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 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 )
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3) 일반화 가능도비 검정법
일반화 가능도비를 적절히 변환시킴으로써 필요한 검정통계량의 분포를 구하기 어려움이 존재해 근사 가능도비 검정법을 이용하여 일반화 가능도비 통계량의 근사분포를 구할 수 있다.
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4) 근사 가능도비 검정법
X ∼ EXP ( λ ) X\sim \text{EXP}(\lambda) X ∼ EXP ( 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 1 : λ ≠ λ 0 H_0:\lambda=\lambda_0\quad\text{vs}\quad H_1:\lambda\neq \lambda_0 H 0 : λ = λ 0 vs H 1 : 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 ) = λ 0 − n exp ( − ∑ i = 1 n X i / λ 0 ) X ˉ n − n exp ( − ∑ i = 1 n X i / X ˉ n ) = ( X ˉ n λ 0 ) n exp [ ( 1 X ˉ n − 1 λ 0 ) ∑ i = 1 n X i ] \begin{aligned} \Lambda (X_1,X_2\ldots,X_n) &=\dfrac{\lambda_0^{-n}\exp\left(-\sum_{i=1}^{n}X_i/\lambda_0\right)}{\bar{X}_n^{-n}\exp\left(-\sum_{i=1}^{n}X_i/\bar{X}_n\right)} \\[15pt] &= \left(\dfrac{\bar{X}_n}{\lambda_0}\right)^n\exp\left[\left(\dfrac{1}{\bar{X}_n}-\dfrac{1}{\lambda_0}\right)\sum_{i=1}^{n}X_i\right] \end{aligned} Λ ( X 1 , X 2 … , X n ) = X ˉ n − n exp ( − ∑ i = 1 n X i / X ˉ n ) λ 0 − n exp ( − ∑ i = 1 n X i / λ 0 ) = ( λ 0 X ˉ n ) n exp [ ( X ˉ n 1 − λ 0 1 ) i = 1 ∑ n X i ] 따라서 귀무가설 하에
− 2 log Λ = − 2 { n log ( X ˉ n λ 0 ) + ( 1 X ˉ n − 1 λ 0 ) n X ˉ n } \begin{aligned} -2\text{log}\Lambda &=-2\left\{n \log\left(\dfrac{\bar{X}_n}{\lambda_0}\right) + \left(\dfrac{1}{\bar{X}_n}-\dfrac{1}{\lambda_0}\right)n\bar{X}_n\right\} \\[15pt] \end{aligned} − 2 log Λ = − 2 { n log ( λ 0 X ˉ n ) + ( X ˉ n 1 − λ 0 1 ) n X ˉ n } 이다. 테일러 정리로부터 ∣ x ∣ ≈ 0 |x|\approx 0 ∣ x ∣ ≈ 0 log ( 1 + x ) = x − x 2 / 2 \log(1+x)=x-x^2/2 log ( 1 + x ) = x − x 2 / 2
log ( X ˉ n λ 0 ) = log ( 1 + X ˉ n − λ 0 λ 0 ) ≈ X ˉ n − λ 0 λ 0 − 1 2 ( X ˉ n − λ 0 λ 0 ) 2 \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}-\dfrac{1}{2}\left(\dfrac{\bar{X}_n-\lambda_0}{\lambda_0}\right)^2 log ( λ 0 X ˉ n ) = log ( 1 + λ 0 X ˉ n − λ 0 ) ≈ λ 0 X ˉ n − λ 0 − 2 1 ( λ 0 X ˉ n − λ 0 ) 2 로 이를 대입하면
− 2 log Λ = − 2 { n log ( X ˉ n λ 0 ) + ( 1 X ˉ n − 1 λ 0 ) n ⋅ X ˉ n } = − 2 [ n { X ˉ n − λ 0 λ 0 − 1 2 ( X ˉ n − λ 0 λ 0 ) 2 } − n ⋅ ( X ˉ n − λ 0 λ 0 ) ] = n ( X ˉ n − λ 0 ) 2 λ 0 2 ≈ χ α 2 ( 1 ) \begin{aligned} -2\log\Lambda &=-2\left\{n \log\left(\dfrac{\bar{X}_n}{\lambda_0}\right) + \left(\dfrac{1}{\bar{X}_n}-\dfrac{1}{\lambda_0}\right)n\cdot\bar{X}_n\right\} \\[15pt] &=-2\left[n \left\{\dfrac{\bar{X}_n-\lambda_0}{\lambda_0}-\dfrac{1}{2}\left(\dfrac{\bar{X}_n-\lambda_0}{\lambda_0}\right)^2\right\} - n\cdot\left(\dfrac{\bar{X}_n-\lambda_0}{\lambda_0}\right)\right] \\[15pt] &=\dfrac{n\left(\bar{X}_n-\lambda_0\right)^2}{\lambda_0^2} \\[15pt] &\approx \chi_\alpha^2(1) \end{aligned} − 2 log Λ = − 2 { n log ( λ 0 X ˉ n ) + ( X ˉ n 1 − λ 0 1 ) n ⋅ X ˉ n } = − 2 [ n { λ 0 X ˉ n − λ 0 − 2 1 ( λ 0 X ˉ n − λ 0 ) 2 } − n ⋅ ( λ 0 X ˉ n − λ 0 ) ] = λ 0 2 n ( X ˉ n − λ 0 ) 2 ≈ χ α 2 ( 1 ) 이므로 − 2 log Λ -2\log\Lambda − 2 log Λ χ 2 ( α ) \chi^2(\alpha) χ 2 ( α )
