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東京工業大学 環境・社会理工学院 土木・環境工学系 2022年8月実施 基礎科目4

标签：

Author​

Miyake

Description​

Kai​

(1)​

確率を PP で表す。 x>0x \gt 0 について、

P(X≤x)=P(Y≤ln⁡x)=12πζ∫−∞ln⁡xexp⁡[−(y−λ)22ζ2]dy∴  fX(x)=ddxP(X≤x)=12πζxexp⁡[−(ln⁡x−λ)22ζ2] \begin{aligned} P \left( X \leq x \right) &= P \left( Y \leq \ln x \right) \\ &= \frac{1}{\sqrt{2 \pi} \zeta} \int_{-\infty}^{\ln x} \exp \left[ - \frac{(y-\lambda)^2}{2 \zeta^2} \right] dy \\ \therefore \ \ f_X(x) &= \frac{d}{dx} P \left( X \leq x \right) \\ &= \frac{1}{\sqrt{2 \pi} \zeta x} \exp \left[ - \frac{(\ln x - \lambda)^2}{2 \zeta^2} \right] \end{aligned}

である。

(2)​

μX=E[X]=E[eY]=12πζ∫−∞∞exp⁡[y−(y−λ)22ζ2]dy=12πζ∫−∞∞exp⁡[−y2−2(λ+ζ2)y+λ22ζ2]dy=12πζ∫−∞∞exp⁡[−(y−(λ+ζ2))2−2λζ2−ζ42ζ2]dy=exp⁡[λ+ζ22]E[X2]=E[e2Y]=12πζ∫−∞∞exp⁡[2y−(y−λ)22ζ2]dy=12πζ∫−∞∞exp⁡[−y2−2(λ+2ζ2)y+λ22ζ2]dy=12πζ∫−∞∞exp⁡[−(y−(λ+2ζ2))2−4λζ2−4ζ42ζ2]dy=exp⁡[2λ+2ζ2]σX2=E[X2]−E[X]2=exp⁡[2λ+2ζ2]−exp⁡[2λ+ζ2]=(eζ2−1)exp⁡[2λ+ζ2]\begin{aligned} \mu_X &= E \left[ X \right] \\ &= E \left[ e^Y \right] \\ &= \frac{1}{\sqrt{2 \pi} \zeta} \int_{-\infty}^\infty \exp \left[ y - \frac{(y-\lambda)^2}{2 \zeta^2} \right] dy \\ &= \frac{1}{\sqrt{2 \pi} \zeta} \int_{-\infty}^\infty \exp \left[ - \frac{y^2-2(\lambda + \zeta^2) y + \lambda^2} {2 \zeta^2} \right] dy \\ &= \frac{1}{\sqrt{2 \pi} \zeta} \int_{-\infty}^\infty \exp \left[ - \frac{(y-(\lambda+\zeta^2))^2 - 2 \lambda \zeta^2 - \zeta^4} {2 \zeta^2} \right] dy \\ &= \exp \left[ \lambda + \frac{\zeta^2}{2} \right] \\ E \left[ X^2 \right] &= E \left[ e^{2Y} \right] \\ &= \frac{1}{\sqrt{2 \pi} \zeta} \int_{-\infty}^\infty \exp \left[ 2y - \frac{(y-\lambda)^2}{2 \zeta^2} \right] dy \\ &= \frac{1}{\sqrt{2 \pi} \zeta} \int_{-\infty}^\infty \exp \left[ - \frac{y^2-2(\lambda + 2 \zeta^2) y + \lambda^2} {2 \zeta^2} \right] dy \\ &= \frac{1}{\sqrt{2 \pi} \zeta} \int_{-\infty}^\infty \exp \left[ - \frac{(y-(\lambda+2 \zeta^2))^2 - 4 \lambda \zeta^2 - 4 \zeta^4} {2 \zeta^2} \right] dy \\ &= \exp \left[ 2 \lambda + 2 \zeta^2 \right] \\ \sigma_X^2 &= E \left[ X^2 \right] - E \left[ X \right]^2 \\ &= \exp \left[ 2 \lambda + 2 \zeta^2 \right] - \exp \left[ 2 \lambda + \zeta^2 \right] \\ &= \left( e^{\zeta^2} - 1 \right) \exp \left[ 2 \lambda + \zeta^2 \right] \end{aligned}

(3)​

(a)​

L(x1,x2,⋯ ,xn;λ,ζ)=∏i=1n12πζxiexp⁡[−(ln⁡xi−λ)22ζ2]=(2πζ)−n(∏i=1nxi)−1exp⁡[−12ζ2∑j=1n(ln⁡xj−λ)2]\begin{aligned} L (x_1, x_2, \cdots, x_n ; \lambda, \zeta) &= \prod_{i=1}^n \frac{1}{\sqrt{2 \pi} \zeta x_i} \exp \left[ - \frac{(\ln x_i - \lambda)^2}{2 \zeta^2} \right] \\ &= \left( \sqrt{2 \pi} \zeta \right)^{-n} \left( \prod_{i=1}^n x_i \right)^{-1} \exp \left[ - \frac{1}{2 \zeta^2} \sum_{j=1}^n (\ln x_j - \lambda)^2 \right] \end{aligned}

(b)​

ln⁡L(x1,x2,⋯ ,xn;λ,ζ)=−n2ln⁡(2π)−nln⁡ζ−∑i=1nln⁡xi−12ζ2∑j=1n(ln⁡xj−λ)2∂∂λln⁡L(x1,x2,⋯ ,xn;λ,ζ)=1ζ2∑j=1n(ln⁡xj−λ)=nζ2(1n∑j=1nln⁡xj−λ)∂∂ζln⁡L(x1,x2,⋯ ,xn;λ,ζ)=−nζ+1ζ3∑j=1n(ln⁡xj−λ)2=−nζ3(ζ2−1n∑j=1n(ln⁡xj−λ)2)\begin{aligned} \ln L (x_1, x_2, \cdots, x_n ; \lambda, \zeta) &= - \frac{n}{2} \ln (2 \pi) - n \ln \zeta - \sum_{i=1}^n \ln x_i - \frac{1}{2 \zeta^2} \sum_{j=1}^n (\ln x_j - \lambda)^2 \\ \frac{\partial}{\partial \lambda} \ln L (x_1, x_2, \cdots, x_n ; \lambda, \zeta) &= \frac{1}{\zeta^2} \sum_{j=1}^n (\ln x_j - \lambda) \\ &= \frac{n}{\zeta^2} \left( \frac{1}{n} \sum_{j=1}^n \ln x_j - \lambda \right) \\ \frac{\partial}{\partial \zeta} \ln L (x_1, x_2, \cdots, x_n ; \lambda, \zeta) &= - \frac{n}{\zeta} + \frac{1}{\zeta^3} \sum_{j=1}^n (\ln x_j - \lambda)^2 \\ &= - \frac{n}{\zeta^3} \left( \zeta^2 - \frac{1}{n} \sum_{j=1}^n (\ln x_j - \lambda)^2 \right) \end{aligned}

なので、

λ^=1n∑j=1nln⁡xjζ^=1n∑j=1n(ln⁡xj−λ^)2\begin{aligned} \hat{\lambda} &= \frac{1}{n} \sum_{j=1}^n \ln x_j \\ \hat{\zeta} &= \sqrt{ \frac{1}{n} \sum_{j=1}^n \left( \ln x_j - \hat{\lambda} \right)^2 } \end{aligned}

がわかる。

((c))​

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