题目
5.已知随机变量(X,Y)的联合概率分布列为-|||-X-|||-Y 1 2 3-|||--1 .dfrac (1)(6) . .dfrac (1)(9) .dfrac (1)(18)-|||-1 .dfrac (1)(3) . α β-|||-且X,Y相互独立,求参数α,β的值。

题目解答
答案

解析
步骤 1:计算边缘概率分布
根据联合概率分布表,我们可以计算出X和Y的边缘概率分布。对于X,我们有:
- P(X=1) = $\dfrac {1}{6} + \dfrac {1}{3} = \dfrac {1}{2}$
- P(X=2) = $\dfrac {1}{9} + \alpha$
- P(X=3) = $\dfrac {1}{18} + \beta$
对于Y,我们有:
- P(Y=-1) = $\dfrac {1}{6} + \dfrac {1}{9} + \dfrac {1}{18} = \dfrac {1}{2}$
- P(Y=1) = $\dfrac {1}{3} + \alpha + \beta$
步骤 2:利用独立性条件
由于X和Y相互独立,我们有P(X=x,Y=y) = P(X=x)P(Y=y)。因此,我们可以根据这个条件来求解α和β。
- 对于P(X=2,Y=-1) = $\dfrac {1}{9}$,我们有$\dfrac {1}{9} = P(X=2)P(Y=-1) = (\dfrac {1}{9} + \alpha) \times \dfrac {1}{2}$
- 对于P(X=3,Y=-1) = $\dfrac {1}{18}$,我们有$\dfrac {1}{18} = P(X=3)P(Y=-1) = (\dfrac {1}{18} + \beta) \times \dfrac {1}{2}$
步骤 3:求解α和β
从步骤2中的方程,我们可以解出α和β。
- $\dfrac {1}{9} = (\dfrac {1}{9} + \alpha) \times \dfrac {1}{2}$,解得$\alpha = \dfrac {2}{9}$
- $\dfrac {1}{18} = (\dfrac {1}{18} + \beta) \times \dfrac {1}{2}$,解得$\beta = \dfrac {1}{9}$
根据联合概率分布表,我们可以计算出X和Y的边缘概率分布。对于X,我们有:
- P(X=1) = $\dfrac {1}{6} + \dfrac {1}{3} = \dfrac {1}{2}$
- P(X=2) = $\dfrac {1}{9} + \alpha$
- P(X=3) = $\dfrac {1}{18} + \beta$
对于Y,我们有:
- P(Y=-1) = $\dfrac {1}{6} + \dfrac {1}{9} + \dfrac {1}{18} = \dfrac {1}{2}$
- P(Y=1) = $\dfrac {1}{3} + \alpha + \beta$
步骤 2:利用独立性条件
由于X和Y相互独立,我们有P(X=x,Y=y) = P(X=x)P(Y=y)。因此,我们可以根据这个条件来求解α和β。
- 对于P(X=2,Y=-1) = $\dfrac {1}{9}$,我们有$\dfrac {1}{9} = P(X=2)P(Y=-1) = (\dfrac {1}{9} + \alpha) \times \dfrac {1}{2}$
- 对于P(X=3,Y=-1) = $\dfrac {1}{18}$,我们有$\dfrac {1}{18} = P(X=3)P(Y=-1) = (\dfrac {1}{18} + \beta) \times \dfrac {1}{2}$
步骤 3:求解α和β
从步骤2中的方程,我们可以解出α和β。
- $\dfrac {1}{9} = (\dfrac {1}{9} + \alpha) \times \dfrac {1}{2}$,解得$\alpha = \dfrac {2}{9}$
- $\dfrac {1}{18} = (\dfrac {1}{18} + \beta) \times \dfrac {1}{2}$,解得$\beta = \dfrac {1}{9}$