Conditional distribution from the sum of uniform distributions











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I am trying to find the conditional probability distribution function of $Y$ $$F(Ymid X_1,X_2)$$
given that $Y$ is distributed uniformly on $[0,1]$,
$$X_1=Y+Z_1$$ and $$X_2=Y+Z_2$$
where $Z_1$ and $Z_2$ are also drawn independently from each other and from $Y$ according to the uniform distribution on $[0,1]$.



How does the result change if we have $X_1=m Y+nZ_1$ for some real $m,n$?



It seems to be an easy question, but I cannot come to any meaningful answer. Thank you for any hints.






probability probability-distributions convolution conditional-probability







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  • It is not an easy question, though might be slightly easier if $Y,Z_1,Z_2$ had independent normal distributions
    – Henry
    yesterday










  • Thank you. For the normal distributions it is easy, I know the answer... It is also easy for $F(Ymid X_1)$. But either I am too dizzy or it is indeed difficult, I really cannot see for $F(Ymid X_1,X_2)$..
    – Kass
    yesterday















up vote
3
down vote

favorite
1












I am trying to find the conditional probability distribution function of $Y$ $$F(Ymid X_1,X_2)$$
given that $Y$ is distributed uniformly on $[0,1]$,
$$X_1=Y+Z_1$$ and $$X_2=Y+Z_2$$
where $Z_1$ and $Z_2$ are also drawn independently from each other and from $Y$ according to the uniform distribution on $[0,1]$.



How does the result change if we have $X_1=m Y+nZ_1$ for some real $m,n$?



It seems to be an easy question, but I cannot come to any meaningful answer. Thank you for any hints.






probability probability-distributions convolution conditional-probability







share|cite|improve this question
























  • It is not an easy question, though might be slightly easier if $Y,Z_1,Z_2$ had independent normal distributions
    – Henry
    yesterday










  • Thank you. For the normal distributions it is easy, I know the answer... It is also easy for $F(Ymid X_1)$. But either I am too dizzy or it is indeed difficult, I really cannot see for $F(Ymid X_1,X_2)$..
    – Kass
    yesterday













up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





I am trying to find the conditional probability distribution function of $Y$ $$F(Ymid X_1,X_2)$$
given that $Y$ is distributed uniformly on $[0,1]$,
$$X_1=Y+Z_1$$ and $$X_2=Y+Z_2$$
where $Z_1$ and $Z_2$ are also drawn independently from each other and from $Y$ according to the uniform distribution on $[0,1]$.



How does the result change if we have $X_1=m Y+nZ_1$ for some real $m,n$?



It seems to be an easy question, but I cannot come to any meaningful answer. Thank you for any hints.






probability probability-distributions convolution conditional-probability







share|cite|improve this question















I am trying to find the conditional probability distribution function of $Y$ $$F(Ymid X_1,X_2)$$
given that $Y$ is distributed uniformly on $[0,1]$,
$$X_1=Y+Z_1$$ and $$X_2=Y+Z_2$$
where $Z_1$ and $Z_2$ are also drawn independently from each other and from $Y$ according to the uniform distribution on $[0,1]$.



How does the result change if we have $X_1=m Y+nZ_1$ for some real $m,n$?



It seems to be an easy question, but I cannot come to any meaningful answer. Thank you for any hints.






probability probability-distributions convolution conditional-probability




probability probability-distributions convolution conditional-probability






share|cite|improve this question















share|cite|improve this question













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Kass

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  • It is not an easy question, though might be slightly easier if $Y,Z_1,Z_2$ had independent normal distributions
    – Henry
    yesterday










  • Thank you. For the normal distributions it is easy, I know the answer... It is also easy for $F(Ymid X_1)$. But either I am too dizzy or it is indeed difficult, I really cannot see for $F(Ymid X_1,X_2)$..
    – Kass
    yesterday


















  • It is not an easy question, though might be slightly easier if $Y,Z_1,Z_2$ had independent normal distributions
    – Henry
    yesterday










  • Thank you. For the normal distributions it is easy, I know the answer... It is also easy for $F(Ymid X_1)$. But either I am too dizzy or it is indeed difficult, I really cannot see for $F(Ymid X_1,X_2)$..
    – Kass
    yesterday
















It is not an easy question, though might be slightly easier if $Y,Z_1,Z_2$ had independent normal distributions
– Henry
yesterday




It is not an easy question, though might be slightly easier if $Y,Z_1,Z_2$ had independent normal distributions
– Henry
yesterday












Thank you. For the normal distributions it is easy, I know the answer... It is also easy for $F(Ymid X_1)$. But either I am too dizzy or it is indeed difficult, I really cannot see for $F(Ymid X_1,X_2)$..
– Kass
yesterday




Thank you. For the normal distributions it is easy, I know the answer... It is also easy for $F(Ymid X_1)$. But either I am too dizzy or it is indeed difficult, I really cannot see for $F(Ymid X_1,X_2)$..
– Kass
yesterday















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