Finding $mathrm{Cov}(X,Y)$ and $E(Xmid Y=y)$ given the joint density of $(X,Y)$
$begingroup$
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
edited Dec 1 '18 at 10:21
StubbornAtom
5,54711138
asked Dec 1 '18 at 4:57
pecopeco
758
$endgroup$
add a comment |
$begingroup$
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
edited Dec 1 '18 at 10:21
StubbornAtom
5,54711138
asked Dec 1 '18 at 4:57
pecopeco
758
$endgroup$
1
$begingroup$
a) $frac12-frac13cdot frac43=color{red}{+}frac1{18}$.
$endgroup$
– farruhota
Dec 1 '18 at 6:05
add a comment |
$begingroup$
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
edited Dec 1 '18 at 10:21
StubbornAtom
5,54711138
asked Dec 1 '18 at 4:57
pecopeco
758
$endgroup$
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
edited Dec 1 '18 at 10:21
StubbornAtom
5,54711138
asked Dec 1 '18 at 4:57
pecopeco
758
edited Dec 1 '18 at 10:21
StubbornAtom
5,54711138
asked Dec 1 '18 at 4:57
pecopeco
758
edited Dec 1 '18 at 10:21
StubbornAtom
5,54711138
edited Dec 1 '18 at 10:21
StubbornAtom
5,54711138
edited Dec 1 '18 at 10:21
StubbornAtom
5,54711138
5,54711138
asked Dec 1 '18 at 4:57
pecopeco
758
asked Dec 1 '18 at 4:57
pecopeco
758
asked Dec 1 '18 at 4:57
pecopeco
758
758
1
$begingroup$
a) $frac12-frac13cdot frac43=color{red}{+}frac1{18}$.
$endgroup$
– farruhota
Dec 1 '18 at 6:05
add a comment |
1
$begingroup$
a) $frac12-frac13cdot frac43=color{red}{+}frac1{18}$.
$endgroup$
– farruhota
Dec 1 '18 at 6:05
1
1
$begingroup$
a) $frac12-frac13cdot frac43=color{red}{+}frac1{18}$.
$endgroup$
– farruhota
Dec 1 '18 at 6:05
$begingroup$
a) $frac12-frac13cdot frac43=color{red}{+}frac1{18}$.
$endgroup$
– farruhota
Dec 1 '18 at 6:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $yin(0,2)$:
$$f_{Xmid Y}(x)=frac{2,mathbf1_{0<x<y/2}}{y}$$
answered Dec 1 '18 at 10:29
StubbornAtomStubbornAtom
5,54711138
$endgroup$
$begingroup$
sorry, but can you explain why the bounds of $f_{Y}(y)$ = 0 and $frac{y}{2}$?
$endgroup$
– peco
Dec 2 '18 at 21:09
$begingroup$
@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 $x$. $$0<x<1, 2x<y<2implies 0<x<y/2,0<y<2$$
$endgroup$
– StubbornAtom
Dec 2 '18 at 21:53
$begingroup$
ok, so that means since $2x<y , x<frac{y}{2}$. thanks!
$endgroup$
– peco
Dec 2 '18 at 21:58
add a comment |
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1 Answer
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$begingroup$
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 $yin(0,2)$:
$$f_{Xmid Y}(x)=frac{2,mathbf1_{0<x<y/2}}{y}$$
answered Dec 1 '18 at 10:29
StubbornAtomStubbornAtom
5,54711138
$endgroup$
$begingroup$
sorry, but can you explain why the bounds of $f_{Y}(y)$ = 0 and $frac{y}{2}$?
$endgroup$
– peco
Dec 2 '18 at 21:09
$begingroup$
@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 $x$. $$0<x<1, 2x<y<2implies 0<x<y/2,0<y<2$$
$endgroup$
– StubbornAtom
Dec 2 '18 at 21:53
$begingroup$
ok, so that means since $2x<y , x<frac{y}{2}$. thanks!
$endgroup$
– peco
Dec 2 '18 at 21:58
add a comment |
$begingroup$
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 $yin(0,2)$:
$$f_{Xmid Y}(x)=frac{2,mathbf1_{0<x<y/2}}{y}$$
answered Dec 1 '18 at 10:29
StubbornAtomStubbornAtom
5,54711138
$endgroup$
$begingroup$
sorry, but can you explain why the bounds of $f_{Y}(y)$ = 0 and $frac{y}{2}$?
$endgroup$
– peco
Dec 2 '18 at 21:09
$begingroup$
@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 $x$. $$0<x<1, 2x<y<2implies 0<x<y/2,0<y<2$$
$endgroup$
– StubbornAtom
Dec 2 '18 at 21:53
$begingroup$
ok, so that means since $2x<y , x<frac{y}{2}$. thanks!
$endgroup$
– peco
Dec 2 '18 at 21:58
add a comment |
$begingroup$
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 $yin(0,2)$:
$$f_{Xmid Y}(x)=frac{2,mathbf1_{0<x<y/2}}{y}$$
answered Dec 1 '18 at 10:29
StubbornAtomStubbornAtom
5,54711138
$endgroup$
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 $yin(0,2)$:
$$f_{Xmid Y}(x)=frac{2,mathbf1_{0<x<y/2}}{y}$$
answered Dec 1 '18 at 10:29
StubbornAtomStubbornAtom
5,54711138
answered Dec 1 '18 at 10:29
StubbornAtomStubbornAtom
5,54711138
answered Dec 1 '18 at 10:29
StubbornAtomStubbornAtom
5,54711138
answered Dec 1 '18 at 10:29
StubbornAtomStubbornAtom
5,54711138
5,54711138
$begingroup$
sorry, but can you explain why the bounds of $f_{Y}(y)$ = 0 and $frac{y}{2}$?
$endgroup$
– peco
Dec 2 '18 at 21:09
$begingroup$
@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 $x$. $$0<x<1, 2x<y<2implies 0<x<y/2,0<y<2$$
$endgroup$
– StubbornAtom
Dec 2 '18 at 21:53
$begingroup$
ok, so that means since $2x<y , x<frac{y}{2}$. thanks!
$endgroup$
– peco
Dec 2 '18 at 21:58
add a comment |
$begingroup$
sorry, but can you explain why the bounds of $f_{Y}(y)$ = 0 and $frac{y}{2}$?
$endgroup$
– peco
Dec 2 '18 at 21:09
$begingroup$
@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 $x$. $$0<x<1, 2x<y<2implies 0<x<y/2,0<y<2$$
$endgroup$
– StubbornAtom
Dec 2 '18 at 21:53
$begingroup$
ok, so that means since $2x<y , x<frac{y}{2}$. thanks!
$endgroup$
– peco
Dec 2 '18 at 21:58
$begingroup$
sorry, but can you explain why the bounds of $f_{Y}(y)$ = 0 and $frac{y}{2}$?
$endgroup$
– peco
Dec 2 '18 at 21:09
$begingroup$
sorry, but can you explain why the bounds of $f_{Y}(y)$ = 0 and $frac{y}{2}$?
$endgroup$
– peco
Dec 2 '18 at 21:09
$begingroup$
@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 $x$. $$0<x<1, 2x<y<2implies 0<x<y/2,0<y<2$$
$endgroup$
– StubbornAtom
Dec 2 '18 at 21:53
$begingroup$
@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 $x$. $$0<x<1, 2x<y<2implies 0<x<y/2,0<y<2$$
$endgroup$
– StubbornAtom
Dec 2 '18 at 21:53
$begingroup$
ok, so that means since $2x<y , x<frac{y}{2}$. thanks!
$endgroup$
– peco
Dec 2 '18 at 21:58
$begingroup$
ok, so that means since $2x<y , x<frac{y}{2}$. thanks!
$endgroup$
– peco
Dec 2 '18 at 21:58
add a comment |
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$begingroup$
a) $frac12-frac13cdot frac43=color{red}{+}frac1{18}$.
$endgroup$
– farruhota
Dec 1 '18 at 6:05