Mixing
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
asked yesterday
Kass
1958
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up vote
3
down vote
favorite
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
asked yesterday
Kass
1958
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
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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
asked yesterday
Kass
1958
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
asked yesterday
Kass
1958
asked yesterday
Kass
1958
edited yesterday
asked yesterday
Kass
1958
asked yesterday
Kass
1958
asked yesterday
Kass
1958
1958
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
add a comment |
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
add a comment |
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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