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$f(x) = \lambda e^{-\lambda x} I(x>0)$

(1)

$p_1 = P(X>1)$

—> $\int_1^\infty \lambda e ^{-\lambda x}I(x>0)=e^{-\lambda}$

$\hat p_1$의 최대가능도추정량

$e^{-\lambda}$

f(x)의 최대가능도추정량

$L(x) = \lambda^n e^{-\lambda \sum x}$

$l(x) = n log \lambda - \lambda \sum x$

$l'(x) = \dfrac{n}{\lambda} - \sum x = 0$

$\rightarrow n - \hat \lambda \sum x = 0$

$\rightarrow \hat \lambda = \dfrac{n}{\sum x} = \dfrac{1}{\bar X}$

$e^{-\hat \lambda} = e^{\bar X}$

$\hat p_1$의 기댓값

$E(\hat p_1) = E(e^{\bar X}) = e^{E(ln^{e^{\bar X}})} = e^{E(\bar X)}$

$E(\bar X) = E(X)$

$E(X) = \int_{-\infty}^{\infty} x f(x)dx$

$\hat p_1$의 점근분포

$ H_0: = 1 $ 가능도비 검정

$X_i \sim Ber(p)$이고 $\sum Y = \bar X$ <- y이게맞나 기억이 잘.

$\sqrt{n}(\bar Y(1-\bar Y) - p(1-p)) \sim N(0, p(1-p)(1-2p)^2)$임을 증명

$\sqrt{n}(g(\hat \theta) - g(\theta_0)) \sim N(0, \dfrac{(g'(\theta))^2}{I(\theta)})$ 델타분포이용

$g(\theta) = p(1-p)$

$g'(\theta) = 1-2p$

$I(p) = p(1-p)$

$f(x) = p^x(1-p)^{1-x}$

$I(p) = E(\dfrac{df(x)}{dp} log f(x))^2$

$log f(x) = x log p + (1-x) log (1-p)$

미분하면 $\dfrac{x}{p} - \dfrac{1-x}{1-p} = \dfrac{x-p}{p(1-p)}$

$I(p) = E(\dfrac{(x-p)^2}{(p(1-p))^2}= \dfrac{Var(X)}{(p(1-p))^2}=\dfrac{1}{p(1-p)}$

$\because Var(X) = p(1-p)$