Finding $mathrm{Cov}(X,Y)$ and $E(Xmid Y=y)$ given the joint density of $(X,Y)$












1














The continuous random variables $X$ and $Y$ have joint density function $f_{xy} = 1$ for $0 < x < 1$ and $2x < y < 2$, and zero otherwise.



I am stuck on the above question, with parts a and b below:



a) Find the $cov(x,y)$.



Work:



$cov(x,y) = E(XY) - E(X)E(Y)$



$= int_0^1int_{2x}^2 xydydx - int_0^1int_{2x}^2 xdydxint_0^1int_{2x}^2 ydydx$
= $frac{-1}{18}$



However I am given the answer $frac{1}{18}$. Did I make a mistake in my computation or is the answer incorrect?



b) Find $E(X|Y = y)$.



Work:



$E(X|Y=y) = int_{-infty}^{infty}xf_{x|y}(x|y)dx$



This is where I am puzzled. I know that $f_{x|y}(x|y) = frac{f_{xy}(xy)}{f_yy}$, and $f_yy= int_{-infty}^{infty}f_{x,y}(x,y)dx$. That means $f_yy = 1$ and therefore $f_{x|y}(x|y) = 1$. However I am given the answer $E(X|Y=y) = frac{y}4$. Again, did I make a computational mistake or am I not getting something?



Thanks.






probability-distributions







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  • 1




    a) $frac12-frac13cdot frac43=color{red}{+}frac1{18}$.
    – farruhota
    Dec 1 at 6:05
















1














The continuous random variables $X$ and $Y$ have joint density function $f_{xy} = 1$ for $0 < x < 1$ and $2x < y < 2$, and zero otherwise.



I am stuck on the above question, with parts a and b below:



a) Find the $cov(x,y)$.



Work:



$cov(x,y) = E(XY) - E(X)E(Y)$



$= int_0^1int_{2x}^2 xydydx - int_0^1int_{2x}^2 xdydxint_0^1int_{2x}^2 ydydx$
= $frac{-1}{18}$



However I am given the answer $frac{1}{18}$. Did I make a mistake in my computation or is the answer incorrect?



b) Find $E(X|Y = y)$.



Work:



$E(X|Y=y) = int_{-infty}^{infty}xf_{x|y}(x|y)dx$



This is where I am puzzled. I know that $f_{x|y}(x|y) = frac{f_{xy}(xy)}{f_yy}$, and $f_yy= int_{-infty}^{infty}f_{x,y}(x,y)dx$. That means $f_yy = 1$ and therefore $f_{x|y}(x|y) = 1$. However I am given the answer $E(X|Y=y) = frac{y}4$. Again, did I make a computational mistake or am I not getting something?



Thanks.






probability-distributions







share|cite|improve this question




















  • 1




    a) $frac12-frac13cdot frac43=color{red}{+}frac1{18}$.
    – farruhota
    Dec 1 at 6:05














1












1








1


0





The continuous random variables $X$ and $Y$ have joint density function $f_{xy} = 1$ for $0 < x < 1$ and $2x < y < 2$, and zero otherwise.



I am stuck on the above question, with parts a and b below:



a) Find the $cov(x,y)$.



Work:



$cov(x,y) = E(XY) - E(X)E(Y)$



$= int_0^1int_{2x}^2 xydydx - int_0^1int_{2x}^2 xdydxint_0^1int_{2x}^2 ydydx$
= $frac{-1}{18}$



However I am given the answer $frac{1}{18}$. Did I make a mistake in my computation or is the answer incorrect?



b) Find $E(X|Y = y)$.



Work:



$E(X|Y=y) = int_{-infty}^{infty}xf_{x|y}(x|y)dx$



This is where I am puzzled. I know that $f_{x|y}(x|y) = frac{f_{xy}(xy)}{f_yy}$, and $f_yy= int_{-infty}^{infty}f_{x,y}(x,y)dx$. That means $f_yy = 1$ and therefore $f_{x|y}(x|y) = 1$. However I am given the answer $E(X|Y=y) = frac{y}4$. Again, did I make a computational mistake or am I not getting something?



Thanks.






probability-distributions







share|cite|improve this question















The continuous random variables $X$ and $Y$ have joint density function $f_{xy} = 1$ for $0 < x < 1$ and $2x < y < 2$, and zero otherwise.



I am stuck on the above question, with parts a and b below:



a) Find the $cov(x,y)$.



Work:



$cov(x,y) = E(XY) - E(X)E(Y)$



$= int_0^1int_{2x}^2 xydydx - int_0^1int_{2x}^2 xdydxint_0^1int_{2x}^2 ydydx$
= $frac{-1}{18}$



However I am given the answer $frac{1}{18}$. Did I make a mistake in my computation or is the answer incorrect?



b) Find $E(X|Y = y)$.



Work:



$E(X|Y=y) = int_{-infty}^{infty}xf_{x|y}(x|y)dx$



This is where I am puzzled. I know that $f_{x|y}(x|y) = frac{f_{xy}(xy)}{f_yy}$, and $f_yy= int_{-infty}^{infty}f_{x,y}(x,y)dx$. That means $f_yy = 1$ and therefore $f_{x|y}(x|y) = 1$. However I am given the answer $E(X|Y=y) = frac{y}4$. Again, did I make a computational mistake or am I not getting something?



Thanks.






probability-distributions




probability-distributions






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edited Dec 1 at 10:21









StubbornAtom

5,22211138




5,22211138










asked Dec 1 at 4:57









peco

758




758








  • 1




    a) $frac12-frac13cdot frac43=color{red}{+}frac1{18}$.
    – farruhota
    Dec 1 at 6:05














  • 1




    a) $frac12-frac13cdot frac43=color{red}{+}frac1{18}$.
    – farruhota
    Dec 1 at 6:05








1




1




a) $frac12-frac13cdot frac43=color{red}{+}frac1{18}$.
– farruhota
Dec 1 at 6:05




a) $frac12-frac13cdot frac43=color{red}{+}frac1{18}$.
– farruhota
Dec 1 at 6:05










1 Answer
1






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oldest

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0














The densities are usually denoted with subscripts in capital letters for the random variables.



For part (a), it is an algebraic error on your part. For part (b), you are not paying attention to the supports of the random variables in your calculations. Hence the absurd answers.



Define the indicator $$mathbf1_{xin A}=begin{cases}1&,text{ if }xin A\0&,text{ otherwise }end{cases}$$



Then joint density of $(X,Y)$ is $$f_{X,Y}(x,y)=mathbf1_{0<x<1,,,2x<y<2}$$



So density of $Y$ is



begin{align}
f_{Y}(y)&=int f_{X,Y}(x,y),dx,mathbf1_{0<y<2}
\&=int_0^{y/2},dx,mathbf1_{0<y<2}
\&=frac{y,mathbf1_{0<y<2}}{2}
end{align}



Thus giving the density of $Xmid Y$ for each $yin(0,2)$:



$$f_{Xmid Y}(x)=frac{2,mathbf1_{0<x<y/2}}{y}$$






share|cite|improve this answer





















  • sorry, but can you explain why the bounds of $f_{Y}(y)$ = 0 and $frac{y}{2}$?
    – peco
    Dec 2 at 21:09










  • @peco Look at the support of $(X,Y)$. There is dependency between $x$ and $y$, so you have to account for that when you integrate over $x$. $$0<x<1, 2x<y<2implies 0<x<y/2,0<y<2$$
    – StubbornAtom
    Dec 2 at 21:53










  • ok, so that means since $2x<y , x<frac{y}{2}$. thanks!
    – peco
    Dec 2 at 21:58













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0














The densities are usually denoted with subscripts in capital letters for the random variables.



For part (a), it is an algebraic error on your part. For part (b), you are not paying attention to the supports of the random variables in your calculations. Hence the absurd answers.



Define the indicator $$mathbf1_{xin A}=begin{cases}1&,text{ if }xin A\0&,text{ otherwise }end{cases}$$



Then joint density of $(X,Y)$ is $$f_{X,Y}(x,y)=mathbf1_{0<x<1,,,2x<y<2}$$



So density of $Y$ is



begin{align}
f_{Y}(y)&=int f_{X,Y}(x,y),dx,mathbf1_{0<y<2}
\&=int_0^{y/2},dx,mathbf1_{0<y<2}
\&=frac{y,mathbf1_{0<y<2}}{2}
end{align}



Thus giving the density of $Xmid Y$ for each $yin(0,2)$:



$$f_{Xmid Y}(x)=frac{2,mathbf1_{0<x<y/2}}{y}$$






share|cite|improve this answer





















  • sorry, but can you explain why the bounds of $f_{Y}(y)$ = 0 and $frac{y}{2}$?
    – peco
    Dec 2 at 21:09










  • @peco Look at the support of $(X,Y)$. There is dependency between $x$ and $y$, so you have to account for that when you integrate over $x$. $$0<x<1, 2x<y<2implies 0<x<y/2,0<y<2$$
    – StubbornAtom
    Dec 2 at 21:53










  • ok, so that means since $2x<y , x<frac{y}{2}$. thanks!
    – peco
    Dec 2 at 21:58


















0














The densities are usually denoted with subscripts in capital letters for the random variables.



For part (a), it is an algebraic error on your part. For part (b), you are not paying attention to the supports of the random variables in your calculations. Hence the absurd answers.



Define the indicator $$mathbf1_{xin A}=begin{cases}1&,text{ if }xin A\0&,text{ otherwise }end{cases}$$



Then joint density of $(X,Y)$ is $$f_{X,Y}(x,y)=mathbf1_{0<x<1,,,2x<y<2}$$



So density of $Y$ is



begin{align}
f_{Y}(y)&=int f_{X,Y}(x,y),dx,mathbf1_{0<y<2}
\&=int_0^{y/2},dx,mathbf1_{0<y<2}
\&=frac{y,mathbf1_{0<y<2}}{2}
end{align}



Thus giving the density of $Xmid Y$ for each $yin(0,2)$:



$$f_{Xmid Y}(x)=frac{2,mathbf1_{0<x<y/2}}{y}$$






share|cite|improve this answer





















  • sorry, but can you explain why the bounds of $f_{Y}(y)$ = 0 and $frac{y}{2}$?
    – peco
    Dec 2 at 21:09










  • @peco Look at the support of $(X,Y)$. There is dependency between $x$ and $y$, so you have to account for that when you integrate over $x$. $$0<x<1, 2x<y<2implies 0<x<y/2,0<y<2$$
    – StubbornAtom
    Dec 2 at 21:53










  • ok, so that means since $2x<y , x<frac{y}{2}$. thanks!
    – peco
    Dec 2 at 21:58
















0












0








0






The densities are usually denoted with subscripts in capital letters for the random variables.



For part (a), it is an algebraic error on your part. For part (b), you are not paying attention to the supports of the random variables in your calculations. Hence the absurd answers.



Define the indicator $$mathbf1_{xin A}=begin{cases}1&,text{ if }xin A\0&,text{ otherwise }end{cases}$$



Then joint density of $(X,Y)$ is $$f_{X,Y}(x,y)=mathbf1_{0<x<1,,,2x<y<2}$$



So density of $Y$ is



begin{align}
f_{Y}(y)&=int f_{X,Y}(x,y),dx,mathbf1_{0<y<2}
\&=int_0^{y/2},dx,mathbf1_{0<y<2}
\&=frac{y,mathbf1_{0<y<2}}{2}
end{align}



Thus giving the density of $Xmid Y$ for each $yin(0,2)$:



$$f_{Xmid Y}(x)=frac{2,mathbf1_{0<x<y/2}}{y}$$






share|cite|improve this answer












The densities are usually denoted with subscripts in capital letters for the random variables.



For part (a), it is an algebraic error on your part. For part (b), you are not paying attention to the supports of the random variables in your calculations. Hence the absurd answers.



Define the indicator $$mathbf1_{xin A}=begin{cases}1&,text{ if }xin A\0&,text{ otherwise }end{cases}$$



Then joint density of $(X,Y)$ is $$f_{X,Y}(x,y)=mathbf1_{0<x<1,,,2x<y<2}$$



So density of $Y$ is



begin{align}
f_{Y}(y)&=int f_{X,Y}(x,y),dx,mathbf1_{0<y<2}
\&=int_0^{y/2},dx,mathbf1_{0<y<2}
\&=frac{y,mathbf1_{0<y<2}}{2}
end{align}



Thus giving the density of $Xmid Y$ for each $yin(0,2)$:



$$f_{Xmid Y}(x)=frac{2,mathbf1_{0<x<y/2}}{y}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 1 at 10:29









StubbornAtom

5,22211138




5,22211138












  • sorry, but can you explain why the bounds of $f_{Y}(y)$ = 0 and $frac{y}{2}$?
    – peco
    Dec 2 at 21:09










  • @peco Look at the support of $(X,Y)$. There is dependency between $x$ and $y$, so you have to account for that when you integrate over $x$. $$0<x<1, 2x<y<2implies 0<x<y/2,0<y<2$$
    – StubbornAtom
    Dec 2 at 21:53










  • ok, so that means since $2x<y , x<frac{y}{2}$. thanks!
    – peco
    Dec 2 at 21:58




















  • sorry, but can you explain why the bounds of $f_{Y}(y)$ = 0 and $frac{y}{2}$?
    – peco
    Dec 2 at 21:09










  • @peco Look at the support of $(X,Y)$. There is dependency between $x$ and $y$, so you have to account for that when you integrate over $x$. $$0<x<1, 2x<y<2implies 0<x<y/2,0<y<2$$
    – StubbornAtom
    Dec 2 at 21:53










  • ok, so that means since $2x<y , x<frac{y}{2}$. thanks!
    – peco
    Dec 2 at 21:58


















sorry, but can you explain why the bounds of $f_{Y}(y)$ = 0 and $frac{y}{2}$?
– peco
Dec 2 at 21:09




sorry, but can you explain why the bounds of $f_{Y}(y)$ = 0 and $frac{y}{2}$?
– peco
Dec 2 at 21:09












@peco Look at the support of $(X,Y)$. There is dependency between $x$ and $y$, so you have to account for that when you integrate over $x$. $$0<x<1, 2x<y<2implies 0<x<y/2,0<y<2$$
– StubbornAtom
Dec 2 at 21:53




@peco Look at the support of $(X,Y)$. There is dependency between $x$ and $y$, so you have to account for that when you integrate over $x$. $$0<x<1, 2x<y<2implies 0<x<y/2,0<y<2$$
– StubbornAtom
Dec 2 at 21:53












ok, so that means since $2x<y , x<frac{y}{2}$. thanks!
– peco
Dec 2 at 21:58






ok, so that means since $2x<y , x<frac{y}{2}$. thanks!
– peco
Dec 2 at 21:58




















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