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 Nov 19 at 14:51
Kass
2008
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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 Nov 19 at 14:51
Kass
2008
It is not an easy question, though might be slightly easier if $Y,Z_1,Z_2$ had independent normal distributions
– Henry
Nov 19 at 15:38
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
Nov 19 at 15:41
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 Nov 19 at 14:51
Kass
2008
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 Nov 19 at 14:51
Kass
2008
asked Nov 19 at 14:51
Kass
2008
edited Nov 19 at 15:22
asked Nov 19 at 14:51
Kass
2008
asked Nov 19 at 14:51
Kass
2008
asked Nov 19 at 14:51
Kass
2008
2008
It is not an easy question, though might be slightly easier if $Y,Z_1,Z_2$ had independent normal distributions
– Henry
Nov 19 at 15:38
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
Nov 19 at 15:41
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
Nov 19 at 15:38
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
Nov 19 at 15:41
It is not an easy question, though might be slightly easier if $Y,Z_1,Z_2$ had independent normal distributions
– Henry
Nov 19 at 15:38
It is not an easy question, though might be slightly easier if $Y,Z_1,Z_2$ had independent normal distributions
– Henry
Nov 19 at 15:38
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
Nov 19 at 15:41
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
Nov 19 at 15:41
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
Nov 19 at 15:38
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
Nov 19 at 15:41