How to compute density functions of $E(Y_1|Y_2)$
$begingroup$
Good afternoon. I've trying to do some exercises. Here is one that I can't continue with.
Here is the question: "Find the density functions of $E(Y_1|Y_2)$". My function that I got is $f(y_1,y_2)=exp(-y_2)$ where $0leq y_1 leq y_2$.
My own idea is that I need to find the $textit{conditional expectation}$ which I did, but then I can't continue from here.
probability-theory
asked Dec 23 '18 at 22:59
Joey AdamsJoey Adams
457
$endgroup$
add a comment |
$begingroup$
Good afternoon. I've trying to do some exercises. Here is one that I can't continue with.
Here is the question: "Find the density functions of $E(Y_1|Y_2)$". My function that I got is $f(y_1,y_2)=exp(-y_2)$ where $0leq y_1 leq y_2$.
My own idea is that I need to find the $textit{conditional expectation}$ which I did, but then I can't continue from here.
probability-theory
asked Dec 23 '18 at 22:59
Joey AdamsJoey Adams
457
$endgroup$
add a comment |
$begingroup$
Good afternoon. I've trying to do some exercises. Here is one that I can't continue with.
Here is the question: "Find the density functions of $E(Y_1|Y_2)$". My function that I got is $f(y_1,y_2)=exp(-y_2)$ where $0leq y_1 leq y_2$.
My own idea is that I need to find the $textit{conditional expectation}$ which I did, but then I can't continue from here.
probability-theory
asked Dec 23 '18 at 22:59
Joey AdamsJoey Adams
457
$endgroup$
Good afternoon. I've trying to do some exercises. Here is one that I can't continue with.
Here is the question: "Find the density functions of $E(Y_1|Y_2)$". My function that I got is $f(y_1,y_2)=exp(-y_2)$ where $0leq y_1 leq y_2$.
My own idea is that I need to find the $textit{conditional expectation}$ which I did, but then I can't continue from here.
probability-theory
probability-theory
asked Dec 23 '18 at 22:59
Joey AdamsJoey Adams
457
asked Dec 23 '18 at 22:59
Joey AdamsJoey Adams
457
asked Dec 23 '18 at 22:59
Joey AdamsJoey Adams
457
asked Dec 23 '18 at 22:59
Joey AdamsJoey Adams
457
asked Dec 23 '18 at 22:59
Joey AdamsJoey Adams
457
457
add a comment |
add a comment |
1 Answer
1
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$begingroup$
The conditional expectation $E(Y_1|Y_2)$ is a function of $Y_2$ that you have presumably figured out. Call it $g$, so $E(Y_1|Y_2)=g(Y_2)$. Since $Y_2$ is a random variable whose distribution you presumably know, you should be able to work out the distribution of the random variable $g(Y_2)$.
answered Dec 23 '18 at 23:17
kimchi loverkimchi lover
10.4k31128
$endgroup$
$begingroup$
So in this case, $g(Y_2)=frac{y}{2}$ and according to a definition in my text, I need to integrate this function with my previuous bounds?
$endgroup$
– Joey Adams
Dec 23 '18 at 23:19
1
$begingroup$
I like to do these kinds of problems step-by-step. What is the density function (the marginal density function) of $Y_2$? Once you know that you can work out the density function for $Y_2/2$.
$endgroup$
– kimchi lover
Dec 23 '18 at 23:22
$begingroup$
Sorry, typo. $g(Y_2)=frac{y_2}{2}$. Note that I computed $E(Y_1|Y_2=y_2)=frac{y_2}{2}$. I'm not sure that I understand you comment very well.
$endgroup$
– Joey Adams
Dec 23 '18 at 23:52
$begingroup$
@Joey To reiterate, $E(Y_1mid Y_2=y_2)=frac{y_2}{2}$ is a real number; $E(Y_1mid Y_2)=frac{Y_2}{2}$ is a random variable whose distribution you are asked to find.
$endgroup$
– StubbornAtom
Dec 24 '18 at 7:11
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The conditional expectation $E(Y_1|Y_2)$ is a function of $Y_2$ that you have presumably figured out. Call it $g$, so $E(Y_1|Y_2)=g(Y_2)$. Since $Y_2$ is a random variable whose distribution you presumably know, you should be able to work out the distribution of the random variable $g(Y_2)$.
answered Dec 23 '18 at 23:17
kimchi loverkimchi lover
10.4k31128
$endgroup$
$begingroup$
So in this case, $g(Y_2)=frac{y}{2}$ and according to a definition in my text, I need to integrate this function with my previuous bounds?
$endgroup$
– Joey Adams
Dec 23 '18 at 23:19
1
$begingroup$
I like to do these kinds of problems step-by-step. What is the density function (the marginal density function) of $Y_2$? Once you know that you can work out the density function for $Y_2/2$.
$endgroup$
– kimchi lover
Dec 23 '18 at 23:22
$begingroup$
Sorry, typo. $g(Y_2)=frac{y_2}{2}$. Note that I computed $E(Y_1|Y_2=y_2)=frac{y_2}{2}$. I'm not sure that I understand you comment very well.
$endgroup$
– Joey Adams
Dec 23 '18 at 23:52
$begingroup$
@Joey To reiterate, $E(Y_1mid Y_2=y_2)=frac{y_2}{2}$ is a real number; $E(Y_1mid Y_2)=frac{Y_2}{2}$ is a random variable whose distribution you are asked to find.
$endgroup$
– StubbornAtom
Dec 24 '18 at 7:11
add a comment |
$begingroup$
The conditional expectation $E(Y_1|Y_2)$ is a function of $Y_2$ that you have presumably figured out. Call it $g$, so $E(Y_1|Y_2)=g(Y_2)$. Since $Y_2$ is a random variable whose distribution you presumably know, you should be able to work out the distribution of the random variable $g(Y_2)$.
answered Dec 23 '18 at 23:17
kimchi loverkimchi lover
10.4k31128
$endgroup$
$begingroup$
So in this case, $g(Y_2)=frac{y}{2}$ and according to a definition in my text, I need to integrate this function with my previuous bounds?
$endgroup$
– Joey Adams
Dec 23 '18 at 23:19
1
$begingroup$
I like to do these kinds of problems step-by-step. What is the density function (the marginal density function) of $Y_2$? Once you know that you can work out the density function for $Y_2/2$.
$endgroup$
– kimchi lover
Dec 23 '18 at 23:22
$begingroup$
Sorry, typo. $g(Y_2)=frac{y_2}{2}$. Note that I computed $E(Y_1|Y_2=y_2)=frac{y_2}{2}$. I'm not sure that I understand you comment very well.
$endgroup$
– Joey Adams
Dec 23 '18 at 23:52
$begingroup$
@Joey To reiterate, $E(Y_1mid Y_2=y_2)=frac{y_2}{2}$ is a real number; $E(Y_1mid Y_2)=frac{Y_2}{2}$ is a random variable whose distribution you are asked to find.
$endgroup$
– StubbornAtom
Dec 24 '18 at 7:11
add a comment |
$begingroup$
The conditional expectation $E(Y_1|Y_2)$ is a function of $Y_2$ that you have presumably figured out. Call it $g$, so $E(Y_1|Y_2)=g(Y_2)$. Since $Y_2$ is a random variable whose distribution you presumably know, you should be able to work out the distribution of the random variable $g(Y_2)$.
answered Dec 23 '18 at 23:17
kimchi loverkimchi lover
10.4k31128
$endgroup$
The conditional expectation $E(Y_1|Y_2)$ is a function of $Y_2$ that you have presumably figured out. Call it $g$, so $E(Y_1|Y_2)=g(Y_2)$. Since $Y_2$ is a random variable whose distribution you presumably know, you should be able to work out the distribution of the random variable $g(Y_2)$.
answered Dec 23 '18 at 23:17
kimchi loverkimchi lover
10.4k31128
answered Dec 23 '18 at 23:17
kimchi loverkimchi lover
10.4k31128
answered Dec 23 '18 at 23:17
kimchi loverkimchi lover
10.4k31128
answered Dec 23 '18 at 23:17
kimchi loverkimchi lover
10.4k31128
10.4k31128
$begingroup$
So in this case, $g(Y_2)=frac{y}{2}$ and according to a definition in my text, I need to integrate this function with my previuous bounds?
$endgroup$
– Joey Adams
Dec 23 '18 at 23:19
1
$begingroup$
I like to do these kinds of problems step-by-step. What is the density function (the marginal density function) of $Y_2$? Once you know that you can work out the density function for $Y_2/2$.
$endgroup$
– kimchi lover
Dec 23 '18 at 23:22
$begingroup$
Sorry, typo. $g(Y_2)=frac{y_2}{2}$. Note that I computed $E(Y_1|Y_2=y_2)=frac{y_2}{2}$. I'm not sure that I understand you comment very well.
$endgroup$
– Joey Adams
Dec 23 '18 at 23:52
$begingroup$
@Joey To reiterate, $E(Y_1mid Y_2=y_2)=frac{y_2}{2}$ is a real number; $E(Y_1mid Y_2)=frac{Y_2}{2}$ is a random variable whose distribution you are asked to find.
$endgroup$
– StubbornAtom
Dec 24 '18 at 7:11
add a comment |
$begingroup$
So in this case, $g(Y_2)=frac{y}{2}$ and according to a definition in my text, I need to integrate this function with my previuous bounds?
$endgroup$
– Joey Adams
Dec 23 '18 at 23:19
1
$begingroup$
I like to do these kinds of problems step-by-step. What is the density function (the marginal density function) of $Y_2$? Once you know that you can work out the density function for $Y_2/2$.
$endgroup$
– kimchi lover
Dec 23 '18 at 23:22
$begingroup$
Sorry, typo. $g(Y_2)=frac{y_2}{2}$. Note that I computed $E(Y_1|Y_2=y_2)=frac{y_2}{2}$. I'm not sure that I understand you comment very well.
$endgroup$
– Joey Adams
Dec 23 '18 at 23:52
$begingroup$
@Joey To reiterate, $E(Y_1mid Y_2=y_2)=frac{y_2}{2}$ is a real number; $E(Y_1mid Y_2)=frac{Y_2}{2}$ is a random variable whose distribution you are asked to find.
$endgroup$
– StubbornAtom
Dec 24 '18 at 7:11
$begingroup$
So in this case, $g(Y_2)=frac{y}{2}$ and according to a definition in my text, I need to integrate this function with my previuous bounds?
$endgroup$
– Joey Adams
Dec 23 '18 at 23:19
$begingroup$
So in this case, $g(Y_2)=frac{y}{2}$ and according to a definition in my text, I need to integrate this function with my previuous bounds?
$endgroup$
– Joey Adams
Dec 23 '18 at 23:19
1
1
$begingroup$
I like to do these kinds of problems step-by-step. What is the density function (the marginal density function) of $Y_2$? Once you know that you can work out the density function for $Y_2/2$.
$endgroup$
– kimchi lover
Dec 23 '18 at 23:22
$begingroup$
I like to do these kinds of problems step-by-step. What is the density function (the marginal density function) of $Y_2$? Once you know that you can work out the density function for $Y_2/2$.
$endgroup$
– kimchi lover
Dec 23 '18 at 23:22
$begingroup$
Sorry, typo. $g(Y_2)=frac{y_2}{2}$. Note that I computed $E(Y_1|Y_2=y_2)=frac{y_2}{2}$. I'm not sure that I understand you comment very well.
$endgroup$
– Joey Adams
Dec 23 '18 at 23:52
$begingroup$
Sorry, typo. $g(Y_2)=frac{y_2}{2}$. Note that I computed $E(Y_1|Y_2=y_2)=frac{y_2}{2}$. I'm not sure that I understand you comment very well.
$endgroup$
– Joey Adams
Dec 23 '18 at 23:52
$begingroup$
@Joey To reiterate, $E(Y_1mid Y_2=y_2)=frac{y_2}{2}$ is a real number; $E(Y_1mid Y_2)=frac{Y_2}{2}$ is a random variable whose distribution you are asked to find.
$endgroup$
– StubbornAtom
Dec 24 '18 at 7:11
$begingroup$
@Joey To reiterate, $E(Y_1mid Y_2=y_2)=frac{y_2}{2}$ is a real number; $E(Y_1mid Y_2)=frac{Y_2}{2}$ is a random variable whose distribution you are asked to find.
$endgroup$
– StubbornAtom
Dec 24 '18 at 7:11
add a comment |
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