How to solve E($2X $| Y)? Where $f(x,y) = 4e^{-2y}$
$begingroup$
Help, please!
How to solve:
$E(2X| Y) = ?$
$$f(x,y) = 4e^{-2y};,0 < x<y mbox{ and }; y>0.$$
After integrating $f(x,y)$ over domain of $y$ we get marginal density of $X$:
$$f_X(x) = 2e^{-2x} ;,;x>0$$
Similarly, marginal density of $Y$
$$f_Y(y) = 4ye^{-2y};,y>0$$
Then I found out the distribution of $2X$ using Jacobian Theorem as
$$f_U(u) = e^{-u};,u>0$$
How to calculate $E(2X| Y)$?
probability probability-theory statistics probability-distributions conditional-expectation
edited Dec 3 '18 at 9:52
Davide Giraudo
126k16150261
asked Dec 3 '18 at 5:05
Cuca BeludoCuca Beludo
226
$endgroup$
add a comment |
$begingroup$
Help, please!
How to solve:
$E(2X| Y) = ?$
$$f(x,y) = 4e^{-2y};,0 < x<y mbox{ and }; y>0.$$
After integrating $f(x,y)$ over domain of $y$ we get marginal density of $X$:
$$f_X(x) = 2e^{-2x} ;,;x>0$$
Similarly, marginal density of $Y$
$$f_Y(y) = 4ye^{-2y};,y>0$$
Then I found out the distribution of $2X$ using Jacobian Theorem as
$$f_U(u) = e^{-u};,u>0$$
How to calculate $E(2X| Y)$?
probability probability-theory statistics probability-distributions conditional-expectation
edited Dec 3 '18 at 9:52
Davide Giraudo
126k16150261
asked Dec 3 '18 at 5:05
Cuca BeludoCuca Beludo
226
$endgroup$
add a comment |
$begingroup$
Help, please!
How to solve:
$E(2X| Y) = ?$
$$f(x,y) = 4e^{-2y};,0 < x<y mbox{ and }; y>0.$$
After integrating $f(x,y)$ over domain of $y$ we get marginal density of $X$:
$$f_X(x) = 2e^{-2x} ;,;x>0$$
Similarly, marginal density of $Y$
$$f_Y(y) = 4ye^{-2y};,y>0$$
Then I found out the distribution of $2X$ using Jacobian Theorem as
$$f_U(u) = e^{-u};,u>0$$
How to calculate $E(2X| Y)$?
probability probability-theory statistics probability-distributions conditional-expectation
edited Dec 3 '18 at 9:52
Davide Giraudo
126k16150261
asked Dec 3 '18 at 5:05
Cuca BeludoCuca Beludo
226
$endgroup$
Help, please!
How to solve:
$E(2X| Y) = ?$
$$f(x,y) = 4e^{-2y};,0 < x<y mbox{ and }; y>0.$$
After integrating $f(x,y)$ over domain of $y$ we get marginal density of $X$:
$$f_X(x) = 2e^{-2x} ;,;x>0$$
Similarly, marginal density of $Y$
$$f_Y(y) = 4ye^{-2y};,y>0$$
Then I found out the distribution of $2X$ using Jacobian Theorem as
$$f_U(u) = e^{-u};,u>0$$
How to calculate $E(2X| Y)$?
probability probability-theory statistics probability-distributions conditional-expectation
probability probability-theory statistics probability-distributions conditional-expectation
edited Dec 3 '18 at 9:52
Davide Giraudo
126k16150261
asked Dec 3 '18 at 5:05
Cuca BeludoCuca Beludo
226
edited Dec 3 '18 at 9:52
Davide Giraudo
126k16150261
asked Dec 3 '18 at 5:05
Cuca BeludoCuca Beludo
226
edited Dec 3 '18 at 9:52
Davide Giraudo
126k16150261
edited Dec 3 '18 at 9:52
Davide Giraudo
126k16150261
edited Dec 3 '18 at 9:52
Davide Giraudo
126k16150261
126k16150261
asked Dec 3 '18 at 5:05
Cuca BeludoCuca Beludo
226
asked Dec 3 '18 at 5:05
Cuca BeludoCuca Beludo
226
asked Dec 3 '18 at 5:05
Cuca BeludoCuca Beludo
226
226
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
Conditional density of $X$ given $Y=y >0$ is
$$
f_{X|Y}(x|y) = frac{f(x,y)}{f_Y(y)} = frac{4 e^{-2y}mathbb 1_{{0<x<y}}}{4y e^{-2y}} = frac{mathbb 1_{{0<x<y}}}{y}.
$$
This is the pdf of uniform distribution on $(0,y)$. The expectation of this distributon equals to $mathbb E(X|Y=y)=frac{y}2$, $mathbb E(2X|Y=y)=y$. You can also calulate it directly as
$$
mathbb E(2X|Y=y) = 2mathbb E(X|Y=y) = 2int x f_{X|Y}(x|y) , dx = 2int_0^y x frac1y , dx = y.
$$
Then $mathbb E(2X|Y) =Y$ a.s.
answered Dec 3 '18 at 5:30
NChNCh
6,3832723
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Conditional density of $X$ given $Y=y >0$ is
$$
f_{X|Y}(x|y) = frac{f(x,y)}{f_Y(y)} = frac{4 e^{-2y}mathbb 1_{{0<x<y}}}{4y e^{-2y}} = frac{mathbb 1_{{0<x<y}}}{y}.
$$
This is the pdf of uniform distribution on $(0,y)$. The expectation of this distributon equals to $mathbb E(X|Y=y)=frac{y}2$, $mathbb E(2X|Y=y)=y$. You can also calulate it directly as
$$
mathbb E(2X|Y=y) = 2mathbb E(X|Y=y) = 2int x f_{X|Y}(x|y) , dx = 2int_0^y x frac1y , dx = y.
$$
Then $mathbb E(2X|Y) =Y$ a.s.
answered Dec 3 '18 at 5:30
NChNCh
6,3832723
$endgroup$
add a comment |
$begingroup$
Conditional density of $X$ given $Y=y >0$ is
$$
f_{X|Y}(x|y) = frac{f(x,y)}{f_Y(y)} = frac{4 e^{-2y}mathbb 1_{{0<x<y}}}{4y e^{-2y}} = frac{mathbb 1_{{0<x<y}}}{y}.
$$
This is the pdf of uniform distribution on $(0,y)$. The expectation of this distributon equals to $mathbb E(X|Y=y)=frac{y}2$, $mathbb E(2X|Y=y)=y$. You can also calulate it directly as
$$
mathbb E(2X|Y=y) = 2mathbb E(X|Y=y) = 2int x f_{X|Y}(x|y) , dx = 2int_0^y x frac1y , dx = y.
$$
Then $mathbb E(2X|Y) =Y$ a.s.
answered Dec 3 '18 at 5:30
NChNCh
6,3832723
$endgroup$
add a comment |
$begingroup$
Conditional density of $X$ given $Y=y >0$ is
$$
f_{X|Y}(x|y) = frac{f(x,y)}{f_Y(y)} = frac{4 e^{-2y}mathbb 1_{{0<x<y}}}{4y e^{-2y}} = frac{mathbb 1_{{0<x<y}}}{y}.
$$
This is the pdf of uniform distribution on $(0,y)$. The expectation of this distributon equals to $mathbb E(X|Y=y)=frac{y}2$, $mathbb E(2X|Y=y)=y$. You can also calulate it directly as
$$
mathbb E(2X|Y=y) = 2mathbb E(X|Y=y) = 2int x f_{X|Y}(x|y) , dx = 2int_0^y x frac1y , dx = y.
$$
Then $mathbb E(2X|Y) =Y$ a.s.
answered Dec 3 '18 at 5:30
NChNCh
6,3832723
$endgroup$
Conditional density of $X$ given $Y=y >0$ is
$$
f_{X|Y}(x|y) = frac{f(x,y)}{f_Y(y)} = frac{4 e^{-2y}mathbb 1_{{0<x<y}}}{4y e^{-2y}} = frac{mathbb 1_{{0<x<y}}}{y}.
$$
This is the pdf of uniform distribution on $(0,y)$. The expectation of this distributon equals to $mathbb E(X|Y=y)=frac{y}2$, $mathbb E(2X|Y=y)=y$. You can also calulate it directly as
$$
mathbb E(2X|Y=y) = 2mathbb E(X|Y=y) = 2int x f_{X|Y}(x|y) , dx = 2int_0^y x frac1y , dx = y.
$$
Then $mathbb E(2X|Y) =Y$ a.s.
answered Dec 3 '18 at 5:30
NChNCh
6,3832723
answered Dec 3 '18 at 5:30
NChNCh
6,3832723
answered Dec 3 '18 at 5:30
NChNCh
6,3832723
answered Dec 3 '18 at 5:30
NChNCh
6,3832723
6,3832723
add a comment |
add a comment |
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