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Conditional Expectation of joint pdf
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Wish to identify what I'm doing wrong when finding the E(X|Y=5) of the following:
$$f(x, y)=begincases
1/6 & textif 0<x<2, 0<y<6-3x \ 0 & textotherwise
&
endcases.$$
My working:
$$beginalignmathsf E(Xmid Y=5)~&=~displaystyleint_0^2 x fracfrac16f(y),mathsf d xendalign$$
where
$$beginalignmathsf f(y)~&=~displaystyleint_0^frac-y+63 frac16 ,mathsf d x = frac6-y18endalign$$
Therefore,
$$beginalignmathsf E(Xmid Y=5)~&=~displaystyleint_0^2 x fracfrac16frac6-y18,mathsf d x mathsf ~&=~displaystyleint_0^2 x fracfrac16frac6-518,mathsf d x~&=~displaystyleint_0^2 3x ,mathsf d x endalign = 6$$
But then when I compute Var(X|Y=5), the answer is getting negative, which is impossible. So I think that I might have done something wrong when calculating the expectation.
My working for Variance:
$$Var(X|Y=5) = E(X^2|Y=5) - (E(X|Y=5) )^2 = 8 - 6^2 = -28$$
I appreciate any help.
self-study pdf joint-distribution conditional-expectation
edited 1 hour ago
Ferdi
3,65542152
asked 1 hour ago
vic12
374
add a comment |Â
up vote
3
down vote
favorite
Wish to identify what I'm doing wrong when finding the E(X|Y=5) of the following:
$$f(x, y)=begincases
1/6 & textif 0<x<2, 0<y<6-3x \ 0 & textotherwise
&
endcases.$$
My working:
$$beginalignmathsf E(Xmid Y=5)~&=~displaystyleint_0^2 x fracfrac16f(y),mathsf d xendalign$$
where
$$beginalignmathsf f(y)~&=~displaystyleint_0^frac-y+63 frac16 ,mathsf d x = frac6-y18endalign$$
Therefore,
$$beginalignmathsf E(Xmid Y=5)~&=~displaystyleint_0^2 x fracfrac16frac6-y18,mathsf d x mathsf ~&=~displaystyleint_0^2 x fracfrac16frac6-518,mathsf d x~&=~displaystyleint_0^2 3x ,mathsf d x endalign = 6$$
But then when I compute Var(X|Y=5), the answer is getting negative, which is impossible. So I think that I might have done something wrong when calculating the expectation.
My working for Variance:
$$Var(X|Y=5) = E(X^2|Y=5) - (E(X|Y=5) )^2 = 8 - 6^2 = -28$$
I appreciate any help.
self-study pdf joint-distribution conditional-expectation
edited 1 hour ago
Ferdi
3,65542152
asked 1 hour ago
vic12
374
you forgot the constraint on the support of $X$ given $Y=y.$
– Xi'an
35 mins ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Wish to identify what I'm doing wrong when finding the E(X|Y=5) of the following:
$$f(x, y)=begincases
1/6 & textif 0<x<2, 0<y<6-3x \ 0 & textotherwise
&
endcases.$$
My working:
$$beginalignmathsf E(Xmid Y=5)~&=~displaystyleint_0^2 x fracfrac16f(y),mathsf d xendalign$$
where
$$beginalignmathsf f(y)~&=~displaystyleint_0^frac-y+63 frac16 ,mathsf d x = frac6-y18endalign$$
Therefore,
$$beginalignmathsf E(Xmid Y=5)~&=~displaystyleint_0^2 x fracfrac16frac6-y18,mathsf d x mathsf ~&=~displaystyleint_0^2 x fracfrac16frac6-518,mathsf d x~&=~displaystyleint_0^2 3x ,mathsf d x endalign = 6$$
But then when I compute Var(X|Y=5), the answer is getting negative, which is impossible. So I think that I might have done something wrong when calculating the expectation.
My working for Variance:
$$Var(X|Y=5) = E(X^2|Y=5) - (E(X|Y=5) )^2 = 8 - 6^2 = -28$$
I appreciate any help.
self-study pdf joint-distribution conditional-expectation
edited 1 hour ago
Ferdi
3,65542152
asked 1 hour ago
vic12
374
Wish to identify what I'm doing wrong when finding the E(X|Y=5) of the following:
$$f(x, y)=begincases
1/6 & textif 0<x<2, 0<y<6-3x \ 0 & textotherwise
&
endcases.$$
My working:
$$beginalignmathsf E(Xmid Y=5)~&=~displaystyleint_0^2 x fracfrac16f(y),mathsf d xendalign$$
where
$$beginalignmathsf f(y)~&=~displaystyleint_0^frac-y+63 frac16 ,mathsf d x = frac6-y18endalign$$
Therefore,
$$beginalignmathsf E(Xmid Y=5)~&=~displaystyleint_0^2 x fracfrac16frac6-y18,mathsf d x mathsf ~&=~displaystyleint_0^2 x fracfrac16frac6-518,mathsf d x~&=~displaystyleint_0^2 3x ,mathsf d x endalign = 6$$
But then when I compute Var(X|Y=5), the answer is getting negative, which is impossible. So I think that I might have done something wrong when calculating the expectation.
My working for Variance:
$$Var(X|Y=5) = E(X^2|Y=5) - (E(X|Y=5) )^2 = 8 - 6^2 = -28$$
I appreciate any help.
self-study pdf joint-distribution conditional-expectation
self-study pdf joint-distribution conditional-expectation
edited 1 hour ago
Ferdi
3,65542152
asked 1 hour ago
vic12
374
edited 1 hour ago
Ferdi
3,65542152
asked 1 hour ago
vic12
374
edited 1 hour ago
Ferdi
3,65542152
edited 1 hour ago
Ferdi
3,65542152
edited 1 hour ago
Ferdi
3,65542152
3,65542152
asked 1 hour ago
vic12
374
asked 1 hour ago
vic12
374
asked 1 hour ago
vic12
374
374
you forgot the constraint on the support of $X$ given $Y=y.$
– Xi'an
35 mins ago
add a comment |Â
you forgot the constraint on the support of $X$ given $Y=y.$
– Xi'an
35 mins ago
you forgot the constraint on the support of $X$ given $Y=y.$
– Xi'an
35 mins ago
you forgot the constraint on the support of $X$ given $Y=y.$
– Xi'an
35 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Think about this geometrically: we've got $(X,Y)$ uniformly distributed over the right triangle with vertices at $(0,0)$, $(2,0)$, and $(0,6)$.
If we imagine sampling over and over from this region uniformly, we can picture $E(X|Y=5)$ as the average $x$ coordinate of the points that end up on the horizontal line $y=5$. All of these points will necessarily have $0 < x < 1/3$, since if $x geq 1/3$ then $y = 5$ can't happen, and since the points are uniform over this line, this image suggests $E(X|Y=5) stackrel ?= 1/6$ (this is the midpoint of the range of $x$ points that can end up on this line).
Checking this with calculus, the issue with what you did is the limits of integration for $x$ also depend on $y$. If $y = 5$ is observed then the biggest that $x$ can be is $1/3$, so really
$$
E(X|Y=5) = int_0^1/3x frac36-y,text dx = 3 cdot frac 12 x^2bigg|_x=0^x=1/3 = frac 16.
$$
Another quick sanity check is that $E(X|Y)$ has to be within the range of $X$, so $E(X|Y=5)neq 6$ can be recognized as incorrect before the negative variance shows up.
answered 1 hour ago
jld
11.5k23149
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Think about this geometrically: we've got $(X,Y)$ uniformly distributed over the right triangle with vertices at $(0,0)$, $(2,0)$, and $(0,6)$.
If we imagine sampling over and over from this region uniformly, we can picture $E(X|Y=5)$ as the average $x$ coordinate of the points that end up on the horizontal line $y=5$. All of these points will necessarily have $0 < x < 1/3$, since if $x geq 1/3$ then $y = 5$ can't happen, and since the points are uniform over this line, this image suggests $E(X|Y=5) stackrel ?= 1/6$ (this is the midpoint of the range of $x$ points that can end up on this line).
Checking this with calculus, the issue with what you did is the limits of integration for $x$ also depend on $y$. If $y = 5$ is observed then the biggest that $x$ can be is $1/3$, so really
$$
E(X|Y=5) = int_0^1/3x frac36-y,text dx = 3 cdot frac 12 x^2bigg|_x=0^x=1/3 = frac 16.
$$
Another quick sanity check is that $E(X|Y)$ has to be within the range of $X$, so $E(X|Y=5)neq 6$ can be recognized as incorrect before the negative variance shows up.
answered 1 hour ago
jld
11.5k23149
add a comment |Â
up vote
3
down vote
accepted
Think about this geometrically: we've got $(X,Y)$ uniformly distributed over the right triangle with vertices at $(0,0)$, $(2,0)$, and $(0,6)$.
If we imagine sampling over and over from this region uniformly, we can picture $E(X|Y=5)$ as the average $x$ coordinate of the points that end up on the horizontal line $y=5$. All of these points will necessarily have $0 < x < 1/3$, since if $x geq 1/3$ then $y = 5$ can't happen, and since the points are uniform over this line, this image suggests $E(X|Y=5) stackrel ?= 1/6$ (this is the midpoint of the range of $x$ points that can end up on this line).
Checking this with calculus, the issue with what you did is the limits of integration for $x$ also depend on $y$. If $y = 5$ is observed then the biggest that $x$ can be is $1/3$, so really
$$
E(X|Y=5) = int_0^1/3x frac36-y,text dx = 3 cdot frac 12 x^2bigg|_x=0^x=1/3 = frac 16.
$$
Another quick sanity check is that $E(X|Y)$ has to be within the range of $X$, so $E(X|Y=5)neq 6$ can be recognized as incorrect before the negative variance shows up.
answered 1 hour ago
jld
11.5k23149
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Think about this geometrically: we've got $(X,Y)$ uniformly distributed over the right triangle with vertices at $(0,0)$, $(2,0)$, and $(0,6)$.
If we imagine sampling over and over from this region uniformly, we can picture $E(X|Y=5)$ as the average $x$ coordinate of the points that end up on the horizontal line $y=5$. All of these points will necessarily have $0 < x < 1/3$, since if $x geq 1/3$ then $y = 5$ can't happen, and since the points are uniform over this line, this image suggests $E(X|Y=5) stackrel ?= 1/6$ (this is the midpoint of the range of $x$ points that can end up on this line).
Checking this with calculus, the issue with what you did is the limits of integration for $x$ also depend on $y$. If $y = 5$ is observed then the biggest that $x$ can be is $1/3$, so really
$$
E(X|Y=5) = int_0^1/3x frac36-y,text dx = 3 cdot frac 12 x^2bigg|_x=0^x=1/3 = frac 16.
$$
Another quick sanity check is that $E(X|Y)$ has to be within the range of $X$, so $E(X|Y=5)neq 6$ can be recognized as incorrect before the negative variance shows up.
answered 1 hour ago
jld
11.5k23149
Think about this geometrically: we've got $(X,Y)$ uniformly distributed over the right triangle with vertices at $(0,0)$, $(2,0)$, and $(0,6)$.
If we imagine sampling over and over from this region uniformly, we can picture $E(X|Y=5)$ as the average $x$ coordinate of the points that end up on the horizontal line $y=5$. All of these points will necessarily have $0 < x < 1/3$, since if $x geq 1/3$ then $y = 5$ can't happen, and since the points are uniform over this line, this image suggests $E(X|Y=5) stackrel ?= 1/6$ (this is the midpoint of the range of $x$ points that can end up on this line).
Checking this with calculus, the issue with what you did is the limits of integration for $x$ also depend on $y$. If $y = 5$ is observed then the biggest that $x$ can be is $1/3$, so really
$$
E(X|Y=5) = int_0^1/3x frac36-y,text dx = 3 cdot frac 12 x^2bigg|_x=0^x=1/3 = frac 16.
$$
Another quick sanity check is that $E(X|Y)$ has to be within the range of $X$, so $E(X|Y=5)neq 6$ can be recognized as incorrect before the negative variance shows up.
answered 1 hour ago
jld
11.5k23149
edited 1 hour ago
answered 1 hour ago
jld
11.5k23149
answered 1 hour ago
jld
11.5k23149
answered 1 hour ago
jld
11.5k23149
11.5k23149
add a comment |Â
add a comment |Â
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you forgot the constraint on the support of $X$ given $Y=y.$
– Xi'an
35 mins ago