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# System prepended metadata

title: 19-3

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# 19-3
## (3)
BCの法線ベクトルの方向を$\phi$とすると、$\cos\phi = -2/\sqrt 5, \sin\phi = 1/\sqrt5$

BCと原点の距離$l = 1/\sqrt5$

垂線の足からの距離$z = l\tan(\theta-\phi)$

x座標$x(\theta) = -l\tan(\theta-\phi)\cdot \frac1{\sqrt5} - \frac25 = -\frac15 \tan(\theta-\phi) - \frac25$

微分$x'(\theta) = \frac{1}{5}(\tan^2(\theta-\phi) + 1) = 5(x+\frac25)^2 + \frac15$

分布関数$\displaystyle f(x) = \frac{{x^{-1}}'(\theta)}{\frac34 \pi} = \frac{1}{\frac34 \pi\cdot  x'(\theta)} = \frac{1}{\frac34\pi}\frac{1}{5(x+\frac25)^2 + \frac15} = \frac{20}{3\pi}\frac{1}{(5x+2)^2 + 1}$

確率分布になっているかの確認

$$\int_{-1}^{0} f(x)dx = \frac{20}{3\pi}\int_{-1}^0\frac{1}{(5x+2)^2 + 1}dx = \frac{4}{3\pi}\int_{-3}^{2}\frac{1}{(5x+2)^2 + 1}d(5x-2)\\
 = \frac{4}{3\pi} (\arctan2 + \arctan 3) = \frac{4}{3\pi} \arctan (-1) = 1$$


## (4)

$$\mathbb E[X] = \int_{-1}^0 \frac{20}{3\pi}\frac{x}{(5x+2)^2 + 1}dx = -\frac25 + \int_{-1}^0 \frac{20}{3\pi}\frac{x+2/5}{(5x+2)^2 + 1}dx\\
 = -\frac25 + \frac{4}{15\pi}\int_{-3}^{2} \frac{5x+2}{(5x+2)^2 + 1}d(5x+2) \\
 =  -\frac25 + \frac{2}{15\pi}\int_{10}^{5} \frac{1}{(5x+2)^2 + 1}d((5x+2)^2+1) \\
 =  -\frac25 - \frac{2}{15\pi}\log2$$
 
 
 別解:
 
 $$\mathbb E[X] = -\frac15 \mathbb E[\tan(\Theta-\phi)] - \frac25= -\frac15\int_{\frac{\pi}2-\phi}^{\frac54\pi-\phi}\tan\theta \cdot\frac{4}{3\pi}d\theta- \frac25\\
 = -\frac{4}{15\pi}\left[-\log\cos\theta\right]_{\frac{\pi}2-\phi}^{\frac54\pi-\phi} - \frac25= \frac{4}{15\pi}\log\frac{1/\sqrt{10}}{1/\sqrt5} - \frac25= -\frac{2}{15\pi}\log2 - \frac25$$