Let Ч and Y be absolutely continuous and discrete random variables, respectively. Does r

lobht98

lobht98

Answered question

2022-06-21

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Let Ч and Y be absolutely continuous and discrete random variables, respectively. Does random vector ( X , Y ) have to be absolutely continuous with respect to the product measure on the respective supports of Ч and Y?
I think this is true because the following equality holds
f X , Y ( x , y ) = f X | Y ( x ) P ( Y = y ) ,
so f X , Y ( x , y ) is the probability density function with respect to the product measure. However, I am not able to prove it formally. I would appreciate some help on this. Thanks.

Answer & Explanation

plodno8n

plodno8n

Beginner2022-06-22Added 17 answers

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document assuming
f X , Y = f X f Y
product f X , Y ( x , y ) can't be absolutely continuous on Y axis - since Y is not absolutely continuous then
x 0 , y 0 y 1 , ϵ 0 : | f ( y 0 ) f ( y 1 ) | > ϵ 0   =>   | f X , Y ( x 0 , y 0 ) f X , Y ( x 0 , y 1 ) | > ϵ 0
example: X is normal and Y B e r n o u l l i ( 0.5 ) then f X , Y ( 0 , 0 ) = f X ( 0 ) 0.5 and f X , Y ( ϵ , 0 ) = 0

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