ここで勉強すれば数学検定１級の壁は超えられるか。

Xの期待値は

$E\left[ X\right] =\int ^{1}_{0}dx\int ^{x}_{0}dy\cdot xf\left( x,y\right)=\int ^{1}_{0}dx\int ^{x}_{0}8x^{2}ydy$

$=\int ^{1}_{0}\left[ 4x^{2}y^{2}\right] ^{x}_{0}=\int ^{1}_{0}4x^{4}dx=\left[ \dfrac {4}{5}x^{4}\right] ^{1}_{0}=\dfrac {4}{5}$

Yの期待値は

$E\left[ Y\right] =\int ^{1}_{0}dx\int ^{x}_{0}8x^{2}y^{2}dy$

$\begin{aligned}=\int ^{1}_{0}dx\left[ \dfrac {8x}{3}y^{3}\right] ^{x}_{0}=\int ^{1}_{0}\dfrac {8}{3}x^{4}dx\\ =\left[ \dfrac {8}{15}x^{4}\right] ^{1}_{0}=\dfrac {8}{15}\end{aligned}$

XYの期待値は

$\begin{aligned}E\left[XY\right] =\int ^{1}_{0}dx\int ^{x}_{0}dy\cdot xyf\left( x,y\right) \\ =\int ^{1}_{0}dx\int ^{x}_{0}8x^{2}y^{2}dy\end{aligned}$

$\begin{aligned}=\int ^{1}_{0}dx\cdot \left[ \dfrac {8x^{2}}{3}y^{3}\right] ^{x}_{0}=\int ^{1}_{0}\dfrac {8x^{5}}{3}dx\\ =\left[ \dfrac {8}{18}\right] ^{1}_{0}=\dfrac {4}{9}\end{aligned}$

XとYの共分散は

$\begin{aligned}Cov\left( X,Y\right) =E\left( XY\right) -E\left[ X\right] E\left[ Y\right] \\ =\dfrac {4}{9}-\dfrac {4}{5}\cdot \dfrac {8}{15}\ =\dfrac {4}{225}\end{aligned}$