Finding $mathbb{E}(xi|eta)$
$begingroup$
Vector $(xi, eta)$ is evenly distributed in set $D={ (x,y): 0leq xleq 1, 0leq y leq 1, x+y leq 1 }$. I need to find $mathbb{E}(xi|eta)$.
So first of all I found the size of $D$. It is $int_{0}^{1}(1-x)dx= frac{1}{2}.$
Now I think I need to find the density function $p(x,y)$. How could I find it? Is it 2?
Than am I right $p_1(x) = 2(1-x)$ and $p_2(y)=2y?$
Next step maybe should be finding $p_2(y|x)$ and $p_1(x|y)$ and after than I can find $mathbb{E}(xi|eta)$. Am I right? How to find all those density functions?
probability probability-distributions random-variables conditional-expectation density-function
asked Jan 2 at 20:12
AtstovasAtstovas
1139
$endgroup$
add a comment |
$begingroup$
Vector $(xi, eta)$ is evenly distributed in set $D={ (x,y): 0leq xleq 1, 0leq y leq 1, x+y leq 1 }$. I need to find $mathbb{E}(xi|eta)$.
So first of all I found the size of $D$. It is $int_{0}^{1}(1-x)dx= frac{1}{2}.$
Now I think I need to find the density function $p(x,y)$. How could I find it? Is it 2?
Than am I right $p_1(x) = 2(1-x)$ and $p_2(y)=2y?$
Next step maybe should be finding $p_2(y|x)$ and $p_1(x|y)$ and after than I can find $mathbb{E}(xi|eta)$. Am I right? How to find all those density functions?
probability probability-distributions random-variables conditional-expectation density-function
asked Jan 2 at 20:12
AtstovasAtstovas
1139
$endgroup$
$begingroup$
Yes. The distribution possesses the value of $2$ throughout $D$
$endgroup$
– Mostafa Ayaz
Jan 2 at 20:17
$begingroup$
@MostafaAyaz yes! Is it because $1/|mathcal{D}|$?
$endgroup$
– Atstovas
Jan 2 at 20:17
2
$begingroup$
Without mentioning support, density function is meaningless.
$endgroup$
– StubbornAtom
Jan 2 at 21:33
add a comment |
$begingroup$
Vector $(xi, eta)$ is evenly distributed in set $D={ (x,y): 0leq xleq 1, 0leq y leq 1, x+y leq 1 }$. I need to find $mathbb{E}(xi|eta)$.
So first of all I found the size of $D$. It is $int_{0}^{1}(1-x)dx= frac{1}{2}.$
Now I think I need to find the density function $p(x,y)$. How could I find it? Is it 2?
Than am I right $p_1(x) = 2(1-x)$ and $p_2(y)=2y?$
Next step maybe should be finding $p_2(y|x)$ and $p_1(x|y)$ and after than I can find $mathbb{E}(xi|eta)$. Am I right? How to find all those density functions?
probability probability-distributions random-variables conditional-expectation density-function
asked Jan 2 at 20:12
AtstovasAtstovas
1139
$endgroup$
Vector $(xi, eta)$ is evenly distributed in set $D={ (x,y): 0leq xleq 1, 0leq y leq 1, x+y leq 1 }$. I need to find $mathbb{E}(xi|eta)$.
So first of all I found the size of $D$. It is $int_{0}^{1}(1-x)dx= frac{1}{2}.$
Now I think I need to find the density function $p(x,y)$. How could I find it? Is it 2?
Than am I right $p_1(x) = 2(1-x)$ and $p_2(y)=2y?$
Next step maybe should be finding $p_2(y|x)$ and $p_1(x|y)$ and after than I can find $mathbb{E}(xi|eta)$. Am I right? How to find all those density functions?
probability probability-distributions random-variables conditional-expectation density-function
probability probability-distributions random-variables conditional-expectation density-function
asked Jan 2 at 20:12
AtstovasAtstovas
1139
asked Jan 2 at 20:12
AtstovasAtstovas
1139
edited Jan 2 at 20:36
Atstovas
asked Jan 2 at 20:12
AtstovasAtstovas
1139
asked Jan 2 at 20:12
AtstovasAtstovas
1139
asked Jan 2 at 20:12
AtstovasAtstovas
1139
1139
$begingroup$
Yes. The distribution possesses the value of $2$ throughout $D$
$endgroup$
– Mostafa Ayaz
Jan 2 at 20:17
$begingroup$
@MostafaAyaz yes! Is it because $1/|mathcal{D}|$?
$endgroup$
– Atstovas
Jan 2 at 20:17
2
$begingroup$
Without mentioning support, density function is meaningless.
$endgroup$
– StubbornAtom
Jan 2 at 21:33
add a comment |
$begingroup$
Yes. The distribution possesses the value of $2$ throughout $D$
$endgroup$
– Mostafa Ayaz
Jan 2 at 20:17
$begingroup$
@MostafaAyaz yes! Is it because $1/|mathcal{D}|$?
$endgroup$
– Atstovas
Jan 2 at 20:17
2
$begingroup$
Without mentioning support, density function is meaningless.
$endgroup$
– StubbornAtom
Jan 2 at 21:33
$begingroup$
Yes. The distribution possesses the value of $2$ throughout $D$
$endgroup$
– Mostafa Ayaz
Jan 2 at 20:17
$begingroup$
Yes. The distribution possesses the value of $2$ throughout $D$
$endgroup$
– Mostafa Ayaz
Jan 2 at 20:17
$begingroup$
@MostafaAyaz yes! Is it because $1/|mathcal{D}|$?
$endgroup$
– Atstovas
Jan 2 at 20:17
$begingroup$
@MostafaAyaz yes! Is it because $1/|mathcal{D}|$?
$endgroup$
– Atstovas
Jan 2 at 20:17
2
2
$begingroup$
Without mentioning support, density function is meaningless.
$endgroup$
– StubbornAtom
Jan 2 at 21:33
$begingroup$
Without mentioning support, density function is meaningless.
$endgroup$
– StubbornAtom
Jan 2 at 21:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The area enclosed by the region $D$ is $frac{1}{2}$.
Hence the density of $(X,Y)$ uniform on $D$ is
$$f_{X,Y}(x,y)=2,mathbf1_{0<x,y<1,,,x+y<1}$$
Rewriting the above we see that the density factors as
$$f_{X,Y}(x,y)=underbrace{frac{1}{1-y}mathbf1_{0<x<1-y}}_{f_{Xmid Y}(x)},,underbrace{2(1-y)mathbf1_{0<y<1}}_{f_Y(y)}$$
So $Xmid Y$ is uniformly distributed over $(0,1-Y)$, giving $$E(Xmid Y)=frac{1-Y}{2}$$
answered Jan 2 at 21:50
StubbornAtomStubbornAtom
6,05811239
$endgroup$
$begingroup$
I don't understand how did you get the second line... I trying to find $p_1(x)$ and $p_2 (y)$.
$endgroup$
– Atstovas
Jan 3 at 9:44
$begingroup$
@Atstovas $f_{X,Y}$ is the joint density of $(X,Y)$, what you are calling $p(x,y)$. $X$ and $Y$ have the same distribution. Your $p_1,p_2$ is my $f_X,f_Y$, where $f_X(x)=int_0^{1-x} 2,dy,mathbf1_{0<x<1}$. The conditional density of $Ymid X$ is $f_{Ymid X}(y)=frac{f_{X,Y}(x,y)}{f_X(x)}$ and similarly for $Xmid Y$.
$endgroup$
– StubbornAtom
Jan 3 at 9:48
$begingroup$
Okey, so how should I do if I want also to find $mathbb{E}(eta|xi)?$
$endgroup$
– Atstovas
Jan 3 at 9:55
$begingroup$
@Atstovas That should be obvious. You tell me. I gave you pretty much everything to answer the question.
$endgroup$
– StubbornAtom
Jan 3 at 9:59
$begingroup$
is $f_Y(x)=int_{1-y}^1 2dx mathbf{1}_{0<y<1}?$
$endgroup$
– Atstovas
Jan 3 at 10:03
|
show 1 more comment
Your Answer
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
The area enclosed by the region $D$ is $frac{1}{2}$.
Hence the density of $(X,Y)$ uniform on $D$ is
$$f_{X,Y}(x,y)=2,mathbf1_{0<x,y<1,,,x+y<1}$$
Rewriting the above we see that the density factors as
$$f_{X,Y}(x,y)=underbrace{frac{1}{1-y}mathbf1_{0<x<1-y}}_{f_{Xmid Y}(x)},,underbrace{2(1-y)mathbf1_{0<y<1}}_{f_Y(y)}$$
So $Xmid Y$ is uniformly distributed over $(0,1-Y)$, giving $$E(Xmid Y)=frac{1-Y}{2}$$
answered Jan 2 at 21:50
StubbornAtomStubbornAtom
6,05811239
$endgroup$
$begingroup$
I don't understand how did you get the second line... I trying to find $p_1(x)$ and $p_2 (y)$.
$endgroup$
– Atstovas
Jan 3 at 9:44
$begingroup$
@Atstovas $f_{X,Y}$ is the joint density of $(X,Y)$, what you are calling $p(x,y)$. $X$ and $Y$ have the same distribution. Your $p_1,p_2$ is my $f_X,f_Y$, where $f_X(x)=int_0^{1-x} 2,dy,mathbf1_{0<x<1}$. The conditional density of $Ymid X$ is $f_{Ymid X}(y)=frac{f_{X,Y}(x,y)}{f_X(x)}$ and similarly for $Xmid Y$.
$endgroup$
– StubbornAtom
Jan 3 at 9:48
$begingroup$
Okey, so how should I do if I want also to find $mathbb{E}(eta|xi)?$
$endgroup$
– Atstovas
Jan 3 at 9:55
$begingroup$
@Atstovas That should be obvious. You tell me. I gave you pretty much everything to answer the question.
$endgroup$
– StubbornAtom
Jan 3 at 9:59
$begingroup$
is $f_Y(x)=int_{1-y}^1 2dx mathbf{1}_{0<y<1}?$
$endgroup$
– Atstovas
Jan 3 at 10:03
|
show 1 more comment
$begingroup$
The area enclosed by the region $D$ is $frac{1}{2}$.
Hence the density of $(X,Y)$ uniform on $D$ is
$$f_{X,Y}(x,y)=2,mathbf1_{0<x,y<1,,,x+y<1}$$
Rewriting the above we see that the density factors as
$$f_{X,Y}(x,y)=underbrace{frac{1}{1-y}mathbf1_{0<x<1-y}}_{f_{Xmid Y}(x)},,underbrace{2(1-y)mathbf1_{0<y<1}}_{f_Y(y)}$$
So $Xmid Y$ is uniformly distributed over $(0,1-Y)$, giving $$E(Xmid Y)=frac{1-Y}{2}$$
answered Jan 2 at 21:50
StubbornAtomStubbornAtom
6,05811239
$endgroup$
$begingroup$
I don't understand how did you get the second line... I trying to find $p_1(x)$ and $p_2 (y)$.
$endgroup$
– Atstovas
Jan 3 at 9:44
$begingroup$
@Atstovas $f_{X,Y}$ is the joint density of $(X,Y)$, what you are calling $p(x,y)$. $X$ and $Y$ have the same distribution. Your $p_1,p_2$ is my $f_X,f_Y$, where $f_X(x)=int_0^{1-x} 2,dy,mathbf1_{0<x<1}$. The conditional density of $Ymid X$ is $f_{Ymid X}(y)=frac{f_{X,Y}(x,y)}{f_X(x)}$ and similarly for $Xmid Y$.
$endgroup$
– StubbornAtom
Jan 3 at 9:48
$begingroup$
Okey, so how should I do if I want also to find $mathbb{E}(eta|xi)?$
$endgroup$
– Atstovas
Jan 3 at 9:55
$begingroup$
@Atstovas That should be obvious. You tell me. I gave you pretty much everything to answer the question.
$endgroup$
– StubbornAtom
Jan 3 at 9:59
$begingroup$
is $f_Y(x)=int_{1-y}^1 2dx mathbf{1}_{0<y<1}?$
$endgroup$
– Atstovas
Jan 3 at 10:03
|
show 1 more comment
$begingroup$
The area enclosed by the region $D$ is $frac{1}{2}$.
Hence the density of $(X,Y)$ uniform on $D$ is
$$f_{X,Y}(x,y)=2,mathbf1_{0<x,y<1,,,x+y<1}$$
Rewriting the above we see that the density factors as
$$f_{X,Y}(x,y)=underbrace{frac{1}{1-y}mathbf1_{0<x<1-y}}_{f_{Xmid Y}(x)},,underbrace{2(1-y)mathbf1_{0<y<1}}_{f_Y(y)}$$
So $Xmid Y$ is uniformly distributed over $(0,1-Y)$, giving $$E(Xmid Y)=frac{1-Y}{2}$$
answered Jan 2 at 21:50
StubbornAtomStubbornAtom
6,05811239
$endgroup$
The area enclosed by the region $D$ is $frac{1}{2}$.
Hence the density of $(X,Y)$ uniform on $D$ is
$$f_{X,Y}(x,y)=2,mathbf1_{0<x,y<1,,,x+y<1}$$
Rewriting the above we see that the density factors as
$$f_{X,Y}(x,y)=underbrace{frac{1}{1-y}mathbf1_{0<x<1-y}}_{f_{Xmid Y}(x)},,underbrace{2(1-y)mathbf1_{0<y<1}}_{f_Y(y)}$$
So $Xmid Y$ is uniformly distributed over $(0,1-Y)$, giving $$E(Xmid Y)=frac{1-Y}{2}$$
answered Jan 2 at 21:50
StubbornAtomStubbornAtom
6,05811239
answered Jan 2 at 21:50
StubbornAtomStubbornAtom
6,05811239
answered Jan 2 at 21:50
StubbornAtomStubbornAtom
6,05811239
answered Jan 2 at 21:50
StubbornAtomStubbornAtom
6,05811239
6,05811239
$begingroup$
I don't understand how did you get the second line... I trying to find $p_1(x)$ and $p_2 (y)$.
$endgroup$
– Atstovas
Jan 3 at 9:44
$begingroup$
@Atstovas $f_{X,Y}$ is the joint density of $(X,Y)$, what you are calling $p(x,y)$. $X$ and $Y$ have the same distribution. Your $p_1,p_2$ is my $f_X,f_Y$, where $f_X(x)=int_0^{1-x} 2,dy,mathbf1_{0<x<1}$. The conditional density of $Ymid X$ is $f_{Ymid X}(y)=frac{f_{X,Y}(x,y)}{f_X(x)}$ and similarly for $Xmid Y$.
$endgroup$
– StubbornAtom
Jan 3 at 9:48
$begingroup$
Okey, so how should I do if I want also to find $mathbb{E}(eta|xi)?$
$endgroup$
– Atstovas
Jan 3 at 9:55
$begingroup$
@Atstovas That should be obvious. You tell me. I gave you pretty much everything to answer the question.
$endgroup$
– StubbornAtom
Jan 3 at 9:59
$begingroup$
is $f_Y(x)=int_{1-y}^1 2dx mathbf{1}_{0<y<1}?$
$endgroup$
– Atstovas
Jan 3 at 10:03
|
show 1 more comment
$begingroup$
I don't understand how did you get the second line... I trying to find $p_1(x)$ and $p_2 (y)$.
$endgroup$
– Atstovas
Jan 3 at 9:44
$begingroup$
@Atstovas $f_{X,Y}$ is the joint density of $(X,Y)$, what you are calling $p(x,y)$. $X$ and $Y$ have the same distribution. Your $p_1,p_2$ is my $f_X,f_Y$, where $f_X(x)=int_0^{1-x} 2,dy,mathbf1_{0<x<1}$. The conditional density of $Ymid X$ is $f_{Ymid X}(y)=frac{f_{X,Y}(x,y)}{f_X(x)}$ and similarly for $Xmid Y$.
$endgroup$
– StubbornAtom
Jan 3 at 9:48
$begingroup$
Okey, so how should I do if I want also to find $mathbb{E}(eta|xi)?$
$endgroup$
– Atstovas
Jan 3 at 9:55
$begingroup$
@Atstovas That should be obvious. You tell me. I gave you pretty much everything to answer the question.
$endgroup$
– StubbornAtom
Jan 3 at 9:59
$begingroup$
is $f_Y(x)=int_{1-y}^1 2dx mathbf{1}_{0<y<1}?$
$endgroup$
– Atstovas
Jan 3 at 10:03
$begingroup$
I don't understand how did you get the second line... I trying to find $p_1(x)$ and $p_2 (y)$.
$endgroup$
– Atstovas
Jan 3 at 9:44
$begingroup$
I don't understand how did you get the second line... I trying to find $p_1(x)$ and $p_2 (y)$.
$endgroup$
– Atstovas
Jan 3 at 9:44
$begingroup$
@Atstovas $f_{X,Y}$ is the joint density of $(X,Y)$, what you are calling $p(x,y)$. $X$ and $Y$ have the same distribution. Your $p_1,p_2$ is my $f_X,f_Y$, where $f_X(x)=int_0^{1-x} 2,dy,mathbf1_{0<x<1}$. The conditional density of $Ymid X$ is $f_{Ymid X}(y)=frac{f_{X,Y}(x,y)}{f_X(x)}$ and similarly for $Xmid Y$.
$endgroup$
– StubbornAtom
Jan 3 at 9:48
$begingroup$
@Atstovas $f_{X,Y}$ is the joint density of $(X,Y)$, what you are calling $p(x,y)$. $X$ and $Y$ have the same distribution. Your $p_1,p_2$ is my $f_X,f_Y$, where $f_X(x)=int_0^{1-x} 2,dy,mathbf1_{0<x<1}$. The conditional density of $Ymid X$ is $f_{Ymid X}(y)=frac{f_{X,Y}(x,y)}{f_X(x)}$ and similarly for $Xmid Y$.
$endgroup$
– StubbornAtom
Jan 3 at 9:48
$begingroup$
Okey, so how should I do if I want also to find $mathbb{E}(eta|xi)?$
$endgroup$
– Atstovas
Jan 3 at 9:55
$begingroup$
Okey, so how should I do if I want also to find $mathbb{E}(eta|xi)?$
$endgroup$
– Atstovas
Jan 3 at 9:55
$begingroup$
@Atstovas That should be obvious. You tell me. I gave you pretty much everything to answer the question.
$endgroup$
– StubbornAtom
Jan 3 at 9:59
$begingroup$
@Atstovas That should be obvious. You tell me. I gave you pretty much everything to answer the question.
$endgroup$
– StubbornAtom
Jan 3 at 9:59
$begingroup$
is $f_Y(x)=int_{1-y}^1 2dx mathbf{1}_{0<y<1}?$
$endgroup$
– Atstovas
Jan 3 at 10:03
$begingroup$
is $f_Y(x)=int_{1-y}^1 2dx mathbf{1}_{0<y<1}?$
$endgroup$
– Atstovas
Jan 3 at 10:03
|
show 1 more comment
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$begingroup$
Yes. The distribution possesses the value of $2$ throughout $D$
$endgroup$
– Mostafa Ayaz
Jan 2 at 20:17
$begingroup$
@MostafaAyaz yes! Is it because $1/|mathcal{D}|$?
$endgroup$
– Atstovas
Jan 2 at 20:17
2
$begingroup$
Without mentioning support, density function is meaningless.
$endgroup$
– StubbornAtom
Jan 2 at 21:33