Finding $mathbb{E}(xi|eta)$












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 20:36







Atstovas

















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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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 $$E(Xmid Y)=frac{1-Y}{2}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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













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1 Answer
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1 Answer
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active

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active

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active

oldest

votes









1












$begingroup$

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 $$E(Xmid Y)=frac{1-Y}{2}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















1












$begingroup$

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 $$E(Xmid Y)=frac{1-Y}{2}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












1








1





$begingroup$

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 $$E(Xmid Y)=frac{1-Y}{2}$$






share|cite|improve this answer









$endgroup$



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 $$E(Xmid Y)=frac{1-Y}{2}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • $begingroup$
    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




















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.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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