Mixing
The random variables $X$ and $Y$ have the joint density $f_{X,Y}(x,y)=…$ [duplicate]
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This question already has an answer here:
Given a joint density function, what is the conditional expectation $E[Y|x]$?
2 answers
The random variables $X$ and $Y$ have the joint density $$f_{X,Y}(x,y)=
left{begin{matrix}e^{-y}, mbox{ } 0leq x leq y le infty
\ 0, mbox{ otherwise}end{matrix}right.$$
Evaluate the conditional expectation $E[X|y]$.
I'm not sure if I'm doing this correctly. Please let me know if something is off in my solution. Thanks for reading.
My solution:
I'm not sure if my approach is correct since $f_{Y}(y)$ turns out to be an infinite number. Also in line 4, is the domain correct?
$$E[X|y] = int_{-infty}^{infty}xf_{X|Y}(x|y)dx.$$
$$f_{X|Y}(x|y) = frac{f_{X,Y}(x,y)}{f_{Y}(y)}.$$
$$f_{Y}(y) = int_{-infty}^{infty}f_{X,Y}(x,y)dx$$
$$= int_{0}^{infty}e^{-y}dx = e^{-y}int_{0}^{infty}1dx $$
$$=e^{-y}left [ x right]Big|_0^infty $$
$$ = ??? $$
Thanks in advance for reading this and responding!
probability random-variables conditional-expectation
asked Jan 23 at 6:54
beepboopbeepboopbeepboopbeepboop
10810
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Jan 23 at 9:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Given a joint density function, what is the conditional expectation $E[Y|x]$?
2 answers
The random variables $X$ and $Y$ have the joint density $$f_{X,Y}(x,y)=
left{begin{matrix}e^{-y}, mbox{ } 0leq x leq y le infty
\ 0, mbox{ otherwise}end{matrix}right.$$
Evaluate the conditional expectation $E[X|y]$.
I'm not sure if I'm doing this correctly. Please let me know if something is off in my solution. Thanks for reading.
My solution:
I'm not sure if my approach is correct since $f_{Y}(y)$ turns out to be an infinite number. Also in line 4, is the domain correct?
$$E[X|y] = int_{-infty}^{infty}xf_{X|Y}(x|y)dx.$$
$$f_{X|Y}(x|y) = frac{f_{X,Y}(x,y)}{f_{Y}(y)}.$$
$$f_{Y}(y) = int_{-infty}^{infty}f_{X,Y}(x,y)dx$$
$$= int_{0}^{infty}e^{-y}dx = e^{-y}int_{0}^{infty}1dx $$
$$=e^{-y}left [ x right]Big|_0^infty $$
$$ = ??? $$
Thanks in advance for reading this and responding!
probability random-variables conditional-expectation
asked Jan 23 at 6:54
beepboopbeepboopbeepboopbeepboop
10810
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marked as duplicate by Did probability
Users with the probability badge can single-handedly close probability questions as duplicates and reopen them as needed.
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Jan 23 at 9:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
To avoid mistakes like integrating from $0$ to $infty$ instead of $0$ to $y$ for finding $f_Y(y)$, you should always write the support of the joint density where the indicator functions come in handy.
$endgroup$
– StubbornAtom
Jan 23 at 7:56
1
$begingroup$
Please avoid multiplying structural duplicates. If you can answer the other question, surely you can answer the present one, right?
$endgroup$
– Did
Jan 23 at 9:25
add a comment |
$begingroup$
This question already has an answer here:
Given a joint density function, what is the conditional expectation $E[Y|x]$?
2 answers
The random variables $X$ and $Y$ have the joint density $$f_{X,Y}(x,y)=
left{begin{matrix}e^{-y}, mbox{ } 0leq x leq y le infty
\ 0, mbox{ otherwise}end{matrix}right.$$
Evaluate the conditional expectation $E[X|y]$.
I'm not sure if I'm doing this correctly. Please let me know if something is off in my solution. Thanks for reading.
My solution:
I'm not sure if my approach is correct since $f_{Y}(y)$ turns out to be an infinite number. Also in line 4, is the domain correct?
$$E[X|y] = int_{-infty}^{infty}xf_{X|Y}(x|y)dx.$$
$$f_{X|Y}(x|y) = frac{f_{X,Y}(x,y)}{f_{Y}(y)}.$$
$$f_{Y}(y) = int_{-infty}^{infty}f_{X,Y}(x,y)dx$$
$$= int_{0}^{infty}e^{-y}dx = e^{-y}int_{0}^{infty}1dx $$
$$=e^{-y}left [ x right]Big|_0^infty $$
$$ = ??? $$
Thanks in advance for reading this and responding!
probability random-variables conditional-expectation
asked Jan 23 at 6:54
beepboopbeepboopbeepboopbeepboop
10810
$endgroup$
This question already has an answer here:
Given a joint density function, what is the conditional expectation $E[Y|x]$?
2 answers
The random variables $X$ and $Y$ have the joint density $$f_{X,Y}(x,y)=
left{begin{matrix}e^{-y}, mbox{ } 0leq x leq y le infty
\ 0, mbox{ otherwise}end{matrix}right.$$
Evaluate the conditional expectation $E[X|y]$.
I'm not sure if I'm doing this correctly. Please let me know if something is off in my solution. Thanks for reading.
My solution:
I'm not sure if my approach is correct since $f_{Y}(y)$ turns out to be an infinite number. Also in line 4, is the domain correct?
$$E[X|y] = int_{-infty}^{infty}xf_{X|Y}(x|y)dx.$$
$$f_{X|Y}(x|y) = frac{f_{X,Y}(x,y)}{f_{Y}(y)}.$$
$$f_{Y}(y) = int_{-infty}^{infty}f_{X,Y}(x,y)dx$$
$$= int_{0}^{infty}e^{-y}dx = e^{-y}int_{0}^{infty}1dx $$
$$=e^{-y}left [ x right]Big|_0^infty $$
$$ = ??? $$
Thanks in advance for reading this and responding!
This question already has an answer here:
Given a joint density function, what is the conditional expectation $E[Y|x]$?
2 answers
probability random-variables conditional-expectation
probability random-variables conditional-expectation
asked Jan 23 at 6:54
beepboopbeepboopbeepboopbeepboop
10810
asked Jan 23 at 6:54
beepboopbeepboopbeepboopbeepboop
10810
edited Jan 23 at 7:38
beepboopbeepboop
asked Jan 23 at 6:54
beepboopbeepboopbeepboopbeepboop
10810
asked Jan 23 at 6:54
beepboopbeepboopbeepboopbeepboop
10810
asked Jan 23 at 6:54
beepboopbeepboopbeepboopbeepboop
10810
10810
marked as duplicate by Did probability
Users with the probability badge can single-handedly close probability questions as duplicates and reopen them as needed.
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Jan 23 at 9:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Did probability
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Jan 23 at 9:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
To avoid mistakes like integrating from $0$ to $infty$ instead of $0$ to $y$ for finding $f_Y(y)$, you should always write the support of the joint density where the indicator functions come in handy.
$endgroup$
– StubbornAtom
Jan 23 at 7:56
1
$begingroup$
Please avoid multiplying structural duplicates. If you can answer the other question, surely you can answer the present one, right?
$endgroup$
– Did
Jan 23 at 9:25
add a comment |
$begingroup$
To avoid mistakes like integrating from $0$ to $infty$ instead of $0$ to $y$ for finding $f_Y(y)$, you should always write the support of the joint density where the indicator functions come in handy.
$endgroup$
– StubbornAtom
Jan 23 at 7:56
1
$begingroup$
Please avoid multiplying structural duplicates. If you can answer the other question, surely you can answer the present one, right?
$endgroup$
– Did
Jan 23 at 9:25
$begingroup$
To avoid mistakes like integrating from $0$ to $infty$ instead of $0$ to $y$ for finding $f_Y(y)$, you should always write the support of the joint density where the indicator functions come in handy.
$endgroup$
– StubbornAtom
Jan 23 at 7:56
$begingroup$
To avoid mistakes like integrating from $0$ to $infty$ instead of $0$ to $y$ for finding $f_Y(y)$, you should always write the support of the joint density where the indicator functions come in handy.
$endgroup$
– StubbornAtom
Jan 23 at 7:56
1
1
$begingroup$
Please avoid multiplying structural duplicates. If you can answer the other question, surely you can answer the present one, right?
$endgroup$
– Did
Jan 23 at 9:25
$begingroup$
Please avoid multiplying structural duplicates. If you can answer the other question, surely you can answer the present one, right?
$endgroup$
– Did
Jan 23 at 9:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I suggest rewriting the joint density using the indicator function $mathbf1_{ain A}$, which equals $1$ if $ain A$ and equals $0$ otherwise.
So you have
begin{align}
f_{X,Y}(x,y)=e^{-y}mathbf1_{0<x<y}&=begin{cases}frac{1}{y}mathbf1_{0<x<y},ye^{-y}mathbf1_{y>0}
\\e^{-(y-x)}mathbf1_{y>x},e^{-x}mathbf1_{x>0}end{cases}
end{align}
We have expressed the joint density in two different ways, in each case $f_{X,Y}$ factors as the product of a conditional density and a marginal density.
From the first case, $$f_{Xmid Y=y}(x)=frac{1}{y}mathbf1_{0<x<y}$$
And from the second case, $$f_{Ymid X=x}(y)=e^{-(y-x)}mathbf1_{y>x}$$
At this point you can calculate the conditional means simply from definition. Or you might identify $Xmid Y$ as having a uniform distribution over $(0,Y)$ and $Ymid X$ as having a shifted exponential distribution (i.e. $[Ymid X]-X$ has an exponential distribution with mean $1$), from which the expectations follow easily.
answered Jan 23 at 7:49
StubbornAtomStubbornAtom
6,16811339
$endgroup$
add a comment |
$begingroup$
For $mathbb{E}[X|Y=y]$:
$$
f_Y(y) = int_{[0, y]}e^{-y}dx=e^{-y}(y-0)=ye^{-y},
$$
$Y sim mathcal{G}amma(2,1)$. Hence,
$$
f_{X|Y}(x|y)= frac{e^{-y}}{ye^{-y}} = 1/y, quad xin(0,y),
$$
namely, $X|Y=y$ is Uniform on $[0,y]$.
answered Jan 23 at 7:04
V. VancakV. Vancak
11.3k3926
$endgroup$
$begingroup$
I'm failing to see how you brought in the gamma. Do you mind explaining that bit? thanks.
$endgroup$
– beepboopbeepboop
Jan 23 at 7:15
$begingroup$
$f_{Gamma(2,1)}(y) = 1^2/1! e^{-1y}y^{2-1}$.
$endgroup$
– V. Vancak
Jan 23 at 7:16
$begingroup$
Do you mind explaining it a different way.
$endgroup$
– beepboopbeepboop
Jan 23 at 7:19
$begingroup$
I have not learned it this way. Is it possible to show this using:
$endgroup$
– beepboopbeepboop
Jan 23 at 7:21
1
$begingroup$
@beepboopbeepboop Exactly. If $X|Y=y sim U[0,y]$, then $mathbb{E}[X|Y=y]=y/2$.
$endgroup$
– V. Vancak
Jan 23 at 13:20
|
show 5 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I suggest rewriting the joint density using the indicator function $mathbf1_{ain A}$, which equals $1$ if $ain A$ and equals $0$ otherwise.
So you have
begin{align}
f_{X,Y}(x,y)=e^{-y}mathbf1_{0<x<y}&=begin{cases}frac{1}{y}mathbf1_{0<x<y},ye^{-y}mathbf1_{y>0}
\\e^{-(y-x)}mathbf1_{y>x},e^{-x}mathbf1_{x>0}end{cases}
end{align}
We have expressed the joint density in two different ways, in each case $f_{X,Y}$ factors as the product of a conditional density and a marginal density.
From the first case, $$f_{Xmid Y=y}(x)=frac{1}{y}mathbf1_{0<x<y}$$
And from the second case, $$f_{Ymid X=x}(y)=e^{-(y-x)}mathbf1_{y>x}$$
At this point you can calculate the conditional means simply from definition. Or you might identify $Xmid Y$ as having a uniform distribution over $(0,Y)$ and $Ymid X$ as having a shifted exponential distribution (i.e. $[Ymid X]-X$ has an exponential distribution with mean $1$), from which the expectations follow easily.
answered Jan 23 at 7:49
StubbornAtomStubbornAtom
6,16811339
$endgroup$
add a comment |
$begingroup$
I suggest rewriting the joint density using the indicator function $mathbf1_{ain A}$, which equals $1$ if $ain A$ and equals $0$ otherwise.
So you have
begin{align}
f_{X,Y}(x,y)=e^{-y}mathbf1_{0<x<y}&=begin{cases}frac{1}{y}mathbf1_{0<x<y},ye^{-y}mathbf1_{y>0}
\\e^{-(y-x)}mathbf1_{y>x},e^{-x}mathbf1_{x>0}end{cases}
end{align}
We have expressed the joint density in two different ways, in each case $f_{X,Y}$ factors as the product of a conditional density and a marginal density.
From the first case, $$f_{Xmid Y=y}(x)=frac{1}{y}mathbf1_{0<x<y}$$
And from the second case, $$f_{Ymid X=x}(y)=e^{-(y-x)}mathbf1_{y>x}$$
At this point you can calculate the conditional means simply from definition. Or you might identify $Xmid Y$ as having a uniform distribution over $(0,Y)$ and $Ymid X$ as having a shifted exponential distribution (i.e. $[Ymid X]-X$ has an exponential distribution with mean $1$), from which the expectations follow easily.
answered Jan 23 at 7:49
StubbornAtomStubbornAtom
6,16811339
$endgroup$
add a comment |
$begingroup$
I suggest rewriting the joint density using the indicator function $mathbf1_{ain A}$, which equals $1$ if $ain A$ and equals $0$ otherwise.
So you have
begin{align}
f_{X,Y}(x,y)=e^{-y}mathbf1_{0<x<y}&=begin{cases}frac{1}{y}mathbf1_{0<x<y},ye^{-y}mathbf1_{y>0}
\\e^{-(y-x)}mathbf1_{y>x},e^{-x}mathbf1_{x>0}end{cases}
end{align}
We have expressed the joint density in two different ways, in each case $f_{X,Y}$ factors as the product of a conditional density and a marginal density.
From the first case, $$f_{Xmid Y=y}(x)=frac{1}{y}mathbf1_{0<x<y}$$
And from the second case, $$f_{Ymid X=x}(y)=e^{-(y-x)}mathbf1_{y>x}$$
At this point you can calculate the conditional means simply from definition. Or you might identify $Xmid Y$ as having a uniform distribution over $(0,Y)$ and $Ymid X$ as having a shifted exponential distribution (i.e. $[Ymid X]-X$ has an exponential distribution with mean $1$), from which the expectations follow easily.
answered Jan 23 at 7:49
StubbornAtomStubbornAtom
6,16811339
$endgroup$
I suggest rewriting the joint density using the indicator function $mathbf1_{ain A}$, which equals $1$ if $ain A$ and equals $0$ otherwise.
So you have
begin{align}
f_{X,Y}(x,y)=e^{-y}mathbf1_{0<x<y}&=begin{cases}frac{1}{y}mathbf1_{0<x<y},ye^{-y}mathbf1_{y>0}
\\e^{-(y-x)}mathbf1_{y>x},e^{-x}mathbf1_{x>0}end{cases}
end{align}
We have expressed the joint density in two different ways, in each case $f_{X,Y}$ factors as the product of a conditional density and a marginal density.
From the first case, $$f_{Xmid Y=y}(x)=frac{1}{y}mathbf1_{0<x<y}$$
And from the second case, $$f_{Ymid X=x}(y)=e^{-(y-x)}mathbf1_{y>x}$$
At this point you can calculate the conditional means simply from definition. Or you might identify $Xmid Y$ as having a uniform distribution over $(0,Y)$ and $Ymid X$ as having a shifted exponential distribution (i.e. $[Ymid X]-X$ has an exponential distribution with mean $1$), from which the expectations follow easily.
answered Jan 23 at 7:49
StubbornAtomStubbornAtom
6,16811339
answered Jan 23 at 7:49
StubbornAtomStubbornAtom
6,16811339
answered Jan 23 at 7:49
StubbornAtomStubbornAtom
6,16811339
answered Jan 23 at 7:49
StubbornAtomStubbornAtom
6,16811339
6,16811339
add a comment |
add a comment |
$begingroup$
For $mathbb{E}[X|Y=y]$:
$$
f_Y(y) = int_{[0, y]}e^{-y}dx=e^{-y}(y-0)=ye^{-y},
$$
$Y sim mathcal{G}amma(2,1)$. Hence,
$$
f_{X|Y}(x|y)= frac{e^{-y}}{ye^{-y}} = 1/y, quad xin(0,y),
$$
namely, $X|Y=y$ is Uniform on $[0,y]$.
answered Jan 23 at 7:04
V. VancakV. Vancak
11.3k3926
$endgroup$
$begingroup$
I'm failing to see how you brought in the gamma. Do you mind explaining that bit? thanks.
$endgroup$
– beepboopbeepboop
Jan 23 at 7:15
$begingroup$
$f_{Gamma(2,1)}(y) = 1^2/1! e^{-1y}y^{2-1}$.
$endgroup$
– V. Vancak
Jan 23 at 7:16
$begingroup$
Do you mind explaining it a different way.
$endgroup$
– beepboopbeepboop
Jan 23 at 7:19
$begingroup$
I have not learned it this way. Is it possible to show this using:
$endgroup$
– beepboopbeepboop
Jan 23 at 7:21
1
$begingroup$
@beepboopbeepboop Exactly. If $X|Y=y sim U[0,y]$, then $mathbb{E}[X|Y=y]=y/2$.
$endgroup$
– V. Vancak
Jan 23 at 13:20
|
show 5 more comments
$begingroup$
For $mathbb{E}[X|Y=y]$:
$$
f_Y(y) = int_{[0, y]}e^{-y}dx=e^{-y}(y-0)=ye^{-y},
$$
$Y sim mathcal{G}amma(2,1)$. Hence,
$$
f_{X|Y}(x|y)= frac{e^{-y}}{ye^{-y}} = 1/y, quad xin(0,y),
$$
namely, $X|Y=y$ is Uniform on $[0,y]$.
answered Jan 23 at 7:04
V. VancakV. Vancak
11.3k3926
$endgroup$
$begingroup$
I'm failing to see how you brought in the gamma. Do you mind explaining that bit? thanks.
$endgroup$
– beepboopbeepboop
Jan 23 at 7:15
$begingroup$
$f_{Gamma(2,1)}(y) = 1^2/1! e^{-1y}y^{2-1}$.
$endgroup$
– V. Vancak
Jan 23 at 7:16
$begingroup$
Do you mind explaining it a different way.
$endgroup$
– beepboopbeepboop
Jan 23 at 7:19
$begingroup$
I have not learned it this way. Is it possible to show this using:
$endgroup$
– beepboopbeepboop
Jan 23 at 7:21
1
$begingroup$
@beepboopbeepboop Exactly. If $X|Y=y sim U[0,y]$, then $mathbb{E}[X|Y=y]=y/2$.
$endgroup$
– V. Vancak
Jan 23 at 13:20
|
show 5 more comments
$begingroup$
For $mathbb{E}[X|Y=y]$:
$$
f_Y(y) = int_{[0, y]}e^{-y}dx=e^{-y}(y-0)=ye^{-y},
$$
$Y sim mathcal{G}amma(2,1)$. Hence,
$$
f_{X|Y}(x|y)= frac{e^{-y}}{ye^{-y}} = 1/y, quad xin(0,y),
$$
namely, $X|Y=y$ is Uniform on $[0,y]$.
answered Jan 23 at 7:04
V. VancakV. Vancak
11.3k3926
$endgroup$
For $mathbb{E}[X|Y=y]$:
$$
f_Y(y) = int_{[0, y]}e^{-y}dx=e^{-y}(y-0)=ye^{-y},
$$
$Y sim mathcal{G}amma(2,1)$. Hence,
$$
f_{X|Y}(x|y)= frac{e^{-y}}{ye^{-y}} = 1/y, quad xin(0,y),
$$
namely, $X|Y=y$ is Uniform on $[0,y]$.
answered Jan 23 at 7:04
V. VancakV. Vancak
11.3k3926
edited Jan 23 at 7:26
answered Jan 23 at 7:04
V. VancakV. Vancak
11.3k3926
answered Jan 23 at 7:04
V. VancakV. Vancak
11.3k3926
answered Jan 23 at 7:04
V. VancakV. Vancak
11.3k3926
11.3k3926
$begingroup$
I'm failing to see how you brought in the gamma. Do you mind explaining that bit? thanks.
$endgroup$
– beepboopbeepboop
Jan 23 at 7:15
$begingroup$
$f_{Gamma(2,1)}(y) = 1^2/1! e^{-1y}y^{2-1}$.
$endgroup$
– V. Vancak
Jan 23 at 7:16
$begingroup$
Do you mind explaining it a different way.
$endgroup$
– beepboopbeepboop
Jan 23 at 7:19
$begingroup$
I have not learned it this way. Is it possible to show this using:
$endgroup$
– beepboopbeepboop
Jan 23 at 7:21
1
$begingroup$
@beepboopbeepboop Exactly. If $X|Y=y sim U[0,y]$, then $mathbb{E}[X|Y=y]=y/2$.
$endgroup$
– V. Vancak
Jan 23 at 13:20
|
show 5 more comments
$begingroup$
I'm failing to see how you brought in the gamma. Do you mind explaining that bit? thanks.
$endgroup$
– beepboopbeepboop
Jan 23 at 7:15
$begingroup$
$f_{Gamma(2,1)}(y) = 1^2/1! e^{-1y}y^{2-1}$.
$endgroup$
– V. Vancak
Jan 23 at 7:16
$begingroup$
Do you mind explaining it a different way.
$endgroup$
– beepboopbeepboop
Jan 23 at 7:19
$begingroup$
I have not learned it this way. Is it possible to show this using:
$endgroup$
– beepboopbeepboop
Jan 23 at 7:21
1
$begingroup$
@beepboopbeepboop Exactly. If $X|Y=y sim U[0,y]$, then $mathbb{E}[X|Y=y]=y/2$.
$endgroup$
– V. Vancak
Jan 23 at 13:20
$begingroup$
I'm failing to see how you brought in the gamma. Do you mind explaining that bit? thanks.
$endgroup$
– beepboopbeepboop
Jan 23 at 7:15
$begingroup$
I'm failing to see how you brought in the gamma. Do you mind explaining that bit? thanks.
$endgroup$
– beepboopbeepboop
Jan 23 at 7:15
$begingroup$
$f_{Gamma(2,1)}(y) = 1^2/1! e^{-1y}y^{2-1}$.
$endgroup$
– V. Vancak
Jan 23 at 7:16
$begingroup$
$f_{Gamma(2,1)}(y) = 1^2/1! e^{-1y}y^{2-1}$.
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– V. Vancak
Jan 23 at 7:16
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Do you mind explaining it a different way.
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– beepboopbeepboop
Jan 23 at 7:19
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Do you mind explaining it a different way.
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– beepboopbeepboop
Jan 23 at 7:19
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I have not learned it this way. Is it possible to show this using:
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– beepboopbeepboop
Jan 23 at 7:21
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I have not learned it this way. Is it possible to show this using:
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– beepboopbeepboop
Jan 23 at 7:21
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@beepboopbeepboop Exactly. If $X|Y=y sim U[0,y]$, then $mathbb{E}[X|Y=y]=y/2$.
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– V. Vancak
Jan 23 at 13:20
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@beepboopbeepboop Exactly. If $X|Y=y sim U[0,y]$, then $mathbb{E}[X|Y=y]=y/2$.
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– V. Vancak
Jan 23 at 13:20
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To avoid mistakes like integrating from $0$ to $infty$ instead of $0$ to $y$ for finding $f_Y(y)$, you should always write the support of the joint density where the indicator functions come in handy.
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– StubbornAtom
Jan 23 at 7:56
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Please avoid multiplying structural duplicates. If you can answer the other question, surely you can answer the present one, right?
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– Did
Jan 23 at 9:25