4.1
4.2
4.3
4.4
4.5
(1) $f$を全領域で積分すると \begin{align*} 1&=\int dxdy\ f\left( x.y\right)\\ &=\int_0^\infty dx\int_0^\infty dy\ cx^{3}\exp[-x(1+y)]\\ &=\int_0^\infty dx\ \left[ cx^{3}\cdot \dfrac {1} {-x}\exp
\left[ -x\left( 1+y\right) \right] \right]_0^\infty\\ &=\int_0^\infty dx\ cx^{2}\exp \left( -x\right)\\ &=\Bigl[\left(-cx^2-2cx-2c\right)\exp(-x)\Bigr]_{0}^{\infty}=2c \end{align*} となるので,
$c=\dfrac{1}{2}$
(2) $f_1(x)$は \begin{align*} f_1(x)&=\int dy\ f\left(x.y\right)\\ &=\int_0^\infty dy\ \frac{1}{2}x^{3}\exp[-x(1+y)]\\ &=\left[\frac{1}{2}x^{3}\cdot \dfrac{1}{-x}\exp\left[-x(1+y)\right] \right]_0^\infty=\frac{1}{2}x^{2}e^{-x} \end{align*} となる. また$f_2(y)$は, \begin{align*} f_2(y)&=\int dx\ f\left(x.y\right)\\ &=\int_0^\infty dx\ \frac{1}{2}x^{3}\exp[-x(1+y)]\\ &=\left[\frac{1}{2}x^{3}\cdot\frac{1}{-(1+y)}\exp[-x(1+y)]\right]_{0}^{\infty}+\int_{0}^{\infty} dx\ \frac{3}{2(1+y)}x^2\exp[-x(1+y)]\\ &=\left[\frac{3}{2(1+y)}x^2\cdot\frac{1}{-(1+y)}\exp[-x(1+y)]\right]_{0}^{\infty}+\int_{0}^{\infty} dx\ \frac{3}{(1+y)^2}x\exp[-x(1+y)]\\ &=\left[\frac{3}{(1+y)^2}x\cdot\frac{1}{-(1+y)}\exp[-x(1+y)]\right]_{0}^{\infty}+\int_{0}^{\infty} dx\ \frac{3}{(1+y)^3}\exp[-x(1+y)]\\ &=\left[\frac{3}{(1+y)^3}\cdot\frac{1}{-(1+y)}\exp[-x(1+y)]\right]_{0}^{\infty}=\frac{3}{(1+y)^4} \end{align*} となる.
(3) $X$の平均は \begin{align*} E(X)&=\int_0^\infty dx\ xf_1(x)\\ &=\int_0^\infty dx\ \frac{1}{2}x^{3}e^{-x}\\ &=\left[\left(-\frac{1}{2}x^{3}-\frac{3}{2}x^2-3x-3\right)e^{-x}\right]_0^\infty=3 \end{align*} となり, \begin{align*} E\left(X^2\right)&=\int_0^\infty dx\ x^2f_1(x)\\ &=\int_0^\infty dx\ \frac{1}{2}x^{4}e^{-x}\\ &=\left[\left(-\frac{1}{2}x^{4}-2x^3-6x^2-12x-12\right)e^{-x}\right]_0^\infty=12 \end{align*} であるので, $X$の分散は \begin{align*} V(X)&=E\left(X^2\right)-\left[E(X)\right]^2=12-3^2=3 \end{align*} となる. 次に, $Y$の平均は \begin{align*} E(Y)&=\int_0^\infty dy\ yf_2(y)\\ &=\int_0^\infty dy\ \frac{3y}{(1+y)^4}\\ &=\left[\frac{3y}{-3(1+y)^3}\right]_{0}^{\infty}+\int_0^\infty dy\ \frac{1}{(1+y)^3}\\ &=\left[\frac{1}{-2(1+y)^2}\right]_{0}^{\infty}=\frac{1}{2} \end{align*}となり, \begin{align*} E\left(Y^2\right)&=\int_0^\infty dy\ y^2f_2(y)\\ &=\int_0^\infty dy\ \frac{3y^2}{(1+y)^4}\\ &=\left[\frac{3y^2}{-3(1+y)^3}\right]_{0}^{\infty}+\int_0^\infty dy\ \frac{2y}{(1+y)^3}\\ &=\left[\frac{2y}{-2(1+y)^2}\right]_{0}^{\infty}+\int_0^\infty dy\ \frac{y}{(1+y)^2}\\ &=\left[\frac{1}{-(1+y)}\right]_{0}^{\infty}=1 \end{align*} であるので, $Y$の分散は \begin{align*} V(Y)&=E\left(Y^2\right)-\left[E(Y)\right]^2=1-{\left(\frac{1}{2}\right)}^2=\frac{3}{4} \end{align*}となる. さらに$XY$の平均は \begin{align*} E(XY)&=\int dxdy\ xyf\left( x.y\right)\\ &=\int_0^\infty dx\int_0^\infty dy\ \frac{1}{2}x^{4}y\exp[-x(1+y)]\\ &=\int_0^\infty dx\left(\left[\frac{1}{2}x^{4}y\cdot\dfrac{1}{-x}\exp\left[-x\left(1+y\right)\right]\right]_0^\infty+\int_0^\infty dy\ \frac{1}{2}x^{3}\exp\left[-x\left(1+y\right)\right]\right)\\ &=\int_0^\infty dx\left[\frac{1}{2}x^{3}\cdot \dfrac{1}{-x}\exp\left[-x(1+y)\right] \right]_0^\infty\\ &=\int_0^\infty dx\ \frac{1}{2}x^{2}e^{-x}\\ &=\left[\left(-\frac{1}{2}x^{2}-x-1\right)e^{-x}\right]_0^\infty=1 \end{align*}となるので$XとY$の共分散は \begin{align*} Cov(X,Y)&=E(XY)-E(X)E(Y)=1-3\cdot\frac{1}{2}=-\frac{1}{2} \end{align*} である. よって$XとY$の相関係数は \begin{align*} Corr(X,Y)&=\frac{Cov(X,Y)}{\sqrt{V(X)V(Y)}}=\frac{-\dfrac{1}{2}}{\sqrt{3\cdot\dfrac{3}{4}}}=-\frac{1}{3} \end{align*} となる.
(2) $f_1(x)$は \begin{align*} f_1(x)&=\int dy\ f\left(x.y\right)\\ &=\int_0^\infty dy\ \frac{1}{2}x^{3}\exp[-x(1+y)]\\ &=\left[\frac{1}{2}x^{3}\cdot \dfrac{1}{-x}\exp\left[-x(1+y)\right] \right]_0^\infty=\frac{1}{2}x^{2}e^{-x} \end{align*} となる. また$f_2(y)$は, \begin{align*} f_2(y)&=\int dx\ f\left(x.y\right)\\ &=\int_0^\infty dx\ \frac{1}{2}x^{3}\exp[-x(1+y)]\\ &=\left[\frac{1}{2}x^{3}\cdot\frac{1}{-(1+y)}\exp[-x(1+y)]\right]_{0}^{\infty}+\int_{0}^{\infty} dx\ \frac{3}{2(1+y)}x^2\exp[-x(1+y)]\\ &=\left[\frac{3}{2(1+y)}x^2\cdot\frac{1}{-(1+y)}\exp[-x(1+y)]\right]_{0}^{\infty}+\int_{0}^{\infty} dx\ \frac{3}{(1+y)^2}x\exp[-x(1+y)]\\ &=\left[\frac{3}{(1+y)^2}x\cdot\frac{1}{-(1+y)}\exp[-x(1+y)]\right]_{0}^{\infty}+\int_{0}^{\infty} dx\ \frac{3}{(1+y)^3}\exp[-x(1+y)]\\ &=\left[\frac{3}{(1+y)^3}\cdot\frac{1}{-(1+y)}\exp[-x(1+y)]\right]_{0}^{\infty}=\frac{3}{(1+y)^4} \end{align*} となる.
(3) $X$の平均は \begin{align*} E(X)&=\int_0^\infty dx\ xf_1(x)\\ &=\int_0^\infty dx\ \frac{1}{2}x^{3}e^{-x}\\ &=\left[\left(-\frac{1}{2}x^{3}-\frac{3}{2}x^2-3x-3\right)e^{-x}\right]_0^\infty=3 \end{align*} となり, \begin{align*} E\left(X^2\right)&=\int_0^\infty dx\ x^2f_1(x)\\ &=\int_0^\infty dx\ \frac{1}{2}x^{4}e^{-x}\\ &=\left[\left(-\frac{1}{2}x^{4}-2x^3-6x^2-12x-12\right)e^{-x}\right]_0^\infty=12 \end{align*} であるので, $X$の分散は \begin{align*} V(X)&=E\left(X^2\right)-\left[E(X)\right]^2=12-3^2=3 \end{align*} となる. 次に, $Y$の平均は \begin{align*} E(Y)&=\int_0^\infty dy\ yf_2(y)\\ &=\int_0^\infty dy\ \frac{3y}{(1+y)^4}\\ &=\left[\frac{3y}{-3(1+y)^3}\right]_{0}^{\infty}+\int_0^\infty dy\ \frac{1}{(1+y)^3}\\ &=\left[\frac{1}{-2(1+y)^2}\right]_{0}^{\infty}=\frac{1}{2} \end{align*}となり, \begin{align*} E\left(Y^2\right)&=\int_0^\infty dy\ y^2f_2(y)\\ &=\int_0^\infty dy\ \frac{3y^2}{(1+y)^4}\\ &=\left[\frac{3y^2}{-3(1+y)^3}\right]_{0}^{\infty}+\int_0^\infty dy\ \frac{2y}{(1+y)^3}\\ &=\left[\frac{2y}{-2(1+y)^2}\right]_{0}^{\infty}+\int_0^\infty dy\ \frac{y}{(1+y)^2}\\ &=\left[\frac{1}{-(1+y)}\right]_{0}^{\infty}=1 \end{align*} であるので, $Y$の分散は \begin{align*} V(Y)&=E\left(Y^2\right)-\left[E(Y)\right]^2=1-{\left(\frac{1}{2}\right)}^2=\frac{3}{4} \end{align*}となる. さらに$XY$の平均は \begin{align*} E(XY)&=\int dxdy\ xyf\left( x.y\right)\\ &=\int_0^\infty dx\int_0^\infty dy\ \frac{1}{2}x^{4}y\exp[-x(1+y)]\\ &=\int_0^\infty dx\left(\left[\frac{1}{2}x^{4}y\cdot\dfrac{1}{-x}\exp\left[-x\left(1+y\right)\right]\right]_0^\infty+\int_0^\infty dy\ \frac{1}{2}x^{3}\exp\left[-x\left(1+y\right)\right]\right)\\ &=\int_0^\infty dx\left[\frac{1}{2}x^{3}\cdot \dfrac{1}{-x}\exp\left[-x(1+y)\right] \right]_0^\infty\\ &=\int_0^\infty dx\ \frac{1}{2}x^{2}e^{-x}\\ &=\left[\left(-\frac{1}{2}x^{2}-x-1\right)e^{-x}\right]_0^\infty=1 \end{align*}となるので$XとY$の共分散は \begin{align*} Cov(X,Y)&=E(XY)-E(X)E(Y)=1-3\cdot\frac{1}{2}=-\frac{1}{2} \end{align*} である. よって$XとY$の相関係数は \begin{align*} Corr(X,Y)&=\frac{Cov(X,Y)}{\sqrt{V(X)V(Y)}}=\frac{-\dfrac{1}{2}}{\sqrt{3\cdot\dfrac{3}{4}}}=-\frac{1}{3} \end{align*} となる.
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