设随机变量ᵢ X ᵢ , i = 1 , 2 的分布列是： Xᵢ -1 0 1 P 0.25 0.5 0.25 且满足₁₂ P X ₁ X ₂ = 0 = 1. 求₁₂ ( X ₁ , X ₂ ) 的联合分布列.

根据题目中给出的条件，可以列出以下方程组：
P(X₁=-1,X₂=-1) + P(X₁=-1,X₂=0) + P(X₁=-1,X₂=1) = 0
P(X₁=0,X₂=-1) + P(X₁=0,X₂=0) + P(X₁=0,X₂=1) = 0
P(X₁=1,X₂=-1) + P(X₁=1,X₂=0) + P(X₁=1,X₂=1) = 0
P(X₁=-1,X₂=-1) + P(X₁=0,X₂=-1) + P(X₁=1,X₂=-1) = 0
P(X₁=-1,X₂=0) + P(X₁=0,X₂=0) + P(X₁=1,X₂=0) = 1
P(X₁=-1,X₂=1) + P(X₁=0,X₂=1) + P(X₁=1,X₂=1) = 0
P(X₁=-1,X₂=-1) + P(X₁=-1,X₂=0) + P(X₁=-1,X₂=1) = 0
P(X₁=0,X₂=-1) + P(X₁=0,X₂=0) + P(X₁=0,X₂=1) = 1
P(X₁=1,X₂=-1) + P(X₁=1,X₂=0) + P(X₁=1,X₂=1) = 0

解方程组得到：
P(X₁=-1,X₂=-1) = 0.25
P(X₁=-1,X₂=0) = 0
P(X₁=-1,X₂=1) = 0.25
P(X₁=0,X₂=-1) = 0
P(X₁=0,X₂=0) = 0
P(X₁=0,X₂=1) = 0
P(X₁=1,X₂=-1) = 0.25
P(X₁=1,X₂=0) = 0
P(X₁=1,X₂=1) = 0.25

因此，₁₂(X₁,X₂)的联合分布列为：
X₂ -1 0 1
X₁
-1 0.25 0 0.25
0 0 0 0
1 0.25 0 0.25