题目
已知随机变量(X,Y)的联合分布律如下.当α,β取何值时X与Y相互独立?-|||-Y-|||-x 1 2 3-|||-1 .dfrac (1)(6) .dfrac (1)(9) .dfrac (1)(18)-|||-2 .dfrac (1)(3) . α β

题目解答
答案

解析
步骤 1:计算边缘分布律
根据联合分布律,计算X和Y的边缘分布律。对于X,边缘分布律为:
$P(X=1)=\dfrac {1}{6}+\dfrac {1}{9}+\dfrac {1}{18}=\dfrac {1}{3}$
$P(X=2)=\dfrac {1}{3}+\alpha+\beta$
对于Y,边缘分布律为:
$P(Y=1)=\dfrac {1}{6}+\dfrac {1}{3}=\dfrac {1}{2}$
$P(Y=2)=\dfrac {1}{9}+\alpha$
$P(Y=3)=\dfrac {1}{18}+\beta$
步骤 2:确定α和β的值
由于X与Y相互独立,所以有:
$P(X=1,Y=2)=P(X=1)P(Y=2)$
$\dfrac {1}{9}=\dfrac {1}{3}\cdot (\dfrac {1}{9}+\alpha )$
解得:$\alpha =\dfrac {2}{9}$
$P(X=1,Y=3)=P(X=1)P(Y=3)$
$\dfrac {1}{18}=\dfrac {1}{3}\cdot (\dfrac {1}{18}+\beta )$
解得:$\beta =\dfrac {1}{9}$
步骤 3:验证α和β的值
将α和β的值代入边缘分布律中,验证X与Y的独立性。
$P(X=2)=\dfrac {1}{3}+\dfrac {2}{9}+\dfrac {1}{9}=\dfrac {2}{3}$
$P(Y=2)=\dfrac {1}{9}+\dfrac {2}{9}=\dfrac {1}{3}$
$P(Y=3)=\dfrac {1}{18}+\dfrac {1}{9}=\dfrac {1}{6}$
根据联合分布律,计算X和Y的边缘分布律。对于X,边缘分布律为:
$P(X=1)=\dfrac {1}{6}+\dfrac {1}{9}+\dfrac {1}{18}=\dfrac {1}{3}$
$P(X=2)=\dfrac {1}{3}+\alpha+\beta$
对于Y,边缘分布律为:
$P(Y=1)=\dfrac {1}{6}+\dfrac {1}{3}=\dfrac {1}{2}$
$P(Y=2)=\dfrac {1}{9}+\alpha$
$P(Y=3)=\dfrac {1}{18}+\beta$
步骤 2:确定α和β的值
由于X与Y相互独立,所以有:
$P(X=1,Y=2)=P(X=1)P(Y=2)$
$\dfrac {1}{9}=\dfrac {1}{3}\cdot (\dfrac {1}{9}+\alpha )$
解得:$\alpha =\dfrac {2}{9}$
$P(X=1,Y=3)=P(X=1)P(Y=3)$
$\dfrac {1}{18}=\dfrac {1}{3}\cdot (\dfrac {1}{18}+\beta )$
解得:$\beta =\dfrac {1}{9}$
步骤 3:验证α和β的值
将α和β的值代入边缘分布律中,验证X与Y的独立性。
$P(X=2)=\dfrac {1}{3}+\dfrac {2}{9}+\dfrac {1}{9}=\dfrac {2}{3}$
$P(Y=2)=\dfrac {1}{9}+\dfrac {2}{9}=\dfrac {1}{3}$
$P(Y=3)=\dfrac {1}{18}+\dfrac {1}{9}=\dfrac {1}{6}$