Question

一道大学的概率论问题

Answer 1

设随机变量U的分布函数为F(u)，F'(u)=g(u)
F(u)=P{U<u}=P{x+y<u}
=P{x=1,y<u-1}+P{x=2,y<u-2}
因为随机变量X与Y独立
所以上式=P{x=1}P{y<u-1}+P{x=2}P{y<u-2}
=0.3*P{y<u-1}+0.7*P{y<u-2}
=0.3*∫[-∝,u-1]f(y)dy+0.7*∫[-∝,u-2]f(y)dy
所以g(u)=F'(u)=[ 0.3*∫[-∝,u-1]f(y)dy+0.7*∫[-∝,u-2]f(y)dy ]'
= 0.3 * [ ∫ [-∝,u-1] f(y) dy ]' + 0.7 * [ ∫ [-∝,u-2] f(y) dy ]'
= 0.3 * [ f(u-1)*(u-1)'-lim[A→-∝] f(A)*A' ] + 0.7 * [ f(u-2)*(u-2)'-lim[B→-∝] f(B)*B' ]
因为 ∫ [-∝,+∝] f(y) dy = 1，又因为f(y)>=0
所以f(y)有界，所以lim[A→-∝] f(A)*A'=lim[B→-∝] f(B)*B'=0
所以g(u) = 0.3 * f(u-1) + 0.7 * f(u-2)

Answer 2

y的取值是整个实轴，即负无穷到正无穷。对随见变量U，取u对应。u的分布如下：
F（t）=p（u<t）=p(x+y<t)=p(x=1,y<t-1)+p(x=2,y<t-2)=0.3积分f（y）从负无穷到t-1+0.7积分f（y）从负无穷到t-2；再求导即得所需概率密度