Definition of conditional mean of $ E[g(Y)X|Y=y]$ where Y is discrete.
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I got the following question:
$X$ and $Y$ are random variables upon the probability space $(S,F,P)$. $Y$ is discrete with range $W$. Let $g: Bbb Rto Bbb R$ be a function so that $g(Y)$ is a random variable.
Prove that for every $yin W$ for which $g(y)neq 0$, and for $X$ continuous given $Y=y$ we can obtain that:
$$E[g(Y)X|Y=y]=g(y)E[X|Y=y].$$
I managed to show that $$f_{g(Y)X|Y=y}(t)=f_{X|Y=y}left(frac{t}{g(Y)}right).$$
And now I am trying to use that in order to calculate $ E[g(Y)X|Y=y]$, but I find it difficult to understand the definition of this conditional mean:
Is it an integral (sum) of two variables $X$ and $g(Y)$ or an integral of one variable. If this is the case, what is the variable I'm integrating on? It made me confused, and it would be helpful if you can explain it to me.
probability conditional-expectation means
asked Dec 13 '18 at 12:01
Mr.OYMr.OY
16610
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add a comment |
$begingroup$
I got the following question:
$X$ and $Y$ are random variables upon the probability space $(S,F,P)$. $Y$ is discrete with range $W$. Let $g: Bbb Rto Bbb R$ be a function so that $g(Y)$ is a random variable.
Prove that for every $yin W$ for which $g(y)neq 0$, and for $X$ continuous given $Y=y$ we can obtain that:
$$E[g(Y)X|Y=y]=g(y)E[X|Y=y].$$
I managed to show that $$f_{g(Y)X|Y=y}(t)=f_{X|Y=y}left(frac{t}{g(Y)}right).$$
And now I am trying to use that in order to calculate $ E[g(Y)X|Y=y]$, but I find it difficult to understand the definition of this conditional mean:
Is it an integral (sum) of two variables $X$ and $g(Y)$ or an integral of one variable. If this is the case, what is the variable I'm integrating on? It made me confused, and it would be helpful if you can explain it to me.
probability conditional-expectation means
asked Dec 13 '18 at 12:01
Mr.OYMr.OY
16610
$endgroup$
$begingroup$
There are some confusing notations. Is $g:Wto R$, and not $Rto R$? Is $R$ the set of real numbers $mathbb{R}$?
$endgroup$
– Batominovski
Dec 13 '18 at 12:26
$begingroup$
@Batominovski it's $Bbb R to Bbb R$. fixed it.
$endgroup$
– Mr.OY
Dec 13 '18 at 12:34
add a comment |
$begingroup$
I got the following question:
$X$ and $Y$ are random variables upon the probability space $(S,F,P)$. $Y$ is discrete with range $W$. Let $g: Bbb Rto Bbb R$ be a function so that $g(Y)$ is a random variable.
Prove that for every $yin W$ for which $g(y)neq 0$, and for $X$ continuous given $Y=y$ we can obtain that:
$$E[g(Y)X|Y=y]=g(y)E[X|Y=y].$$
I managed to show that $$f_{g(Y)X|Y=y}(t)=f_{X|Y=y}left(frac{t}{g(Y)}right).$$
And now I am trying to use that in order to calculate $ E[g(Y)X|Y=y]$, but I find it difficult to understand the definition of this conditional mean:
Is it an integral (sum) of two variables $X$ and $g(Y)$ or an integral of one variable. If this is the case, what is the variable I'm integrating on? It made me confused, and it would be helpful if you can explain it to me.
probability conditional-expectation means
asked Dec 13 '18 at 12:01
Mr.OYMr.OY
16610
$endgroup$
I got the following question:
$X$ and $Y$ are random variables upon the probability space $(S,F,P)$. $Y$ is discrete with range $W$. Let $g: Bbb Rto Bbb R$ be a function so that $g(Y)$ is a random variable.
Prove that for every $yin W$ for which $g(y)neq 0$, and for $X$ continuous given $Y=y$ we can obtain that:
$$E[g(Y)X|Y=y]=g(y)E[X|Y=y].$$
I managed to show that $$f_{g(Y)X|Y=y}(t)=f_{X|Y=y}left(frac{t}{g(Y)}right).$$
And now I am trying to use that in order to calculate $ E[g(Y)X|Y=y]$, but I find it difficult to understand the definition of this conditional mean:
Is it an integral (sum) of two variables $X$ and $g(Y)$ or an integral of one variable. If this is the case, what is the variable I'm integrating on? It made me confused, and it would be helpful if you can explain it to me.
probability conditional-expectation means
probability conditional-expectation means
asked Dec 13 '18 at 12:01
Mr.OYMr.OY
16610
asked Dec 13 '18 at 12:01
Mr.OYMr.OY
16610
edited Dec 13 '18 at 12:32
Mr.OY
asked Dec 13 '18 at 12:01
Mr.OYMr.OY
16610
asked Dec 13 '18 at 12:01
Mr.OYMr.OY
16610
asked Dec 13 '18 at 12:01
Mr.OYMr.OY
16610
16610
$begingroup$
There are some confusing notations. Is $g:Wto R$, and not $Rto R$? Is $R$ the set of real numbers $mathbb{R}$?
$endgroup$
– Batominovski
Dec 13 '18 at 12:26
$begingroup$
@Batominovski it's $Bbb R to Bbb R$. fixed it.
$endgroup$
– Mr.OY
Dec 13 '18 at 12:34
add a comment |
$begingroup$
There are some confusing notations. Is $g:Wto R$, and not $Rto R$? Is $R$ the set of real numbers $mathbb{R}$?
$endgroup$
– Batominovski
Dec 13 '18 at 12:26
$begingroup$
@Batominovski it's $Bbb R to Bbb R$. fixed it.
$endgroup$
– Mr.OY
Dec 13 '18 at 12:34
$begingroup$
There are some confusing notations. Is $g:Wto R$, and not $Rto R$? Is $R$ the set of real numbers $mathbb{R}$?
$endgroup$
– Batominovski
Dec 13 '18 at 12:26
$begingroup$
There are some confusing notations. Is $g:Wto R$, and not $Rto R$? Is $R$ the set of real numbers $mathbb{R}$?
$endgroup$
– Batominovski
Dec 13 '18 at 12:26
$begingroup$
@Batominovski it's $Bbb R to Bbb R$. fixed it.
$endgroup$
– Mr.OY
Dec 13 '18 at 12:34
$begingroup$
@Batominovski it's $Bbb R to Bbb R$. fixed it.
$endgroup$
– Mr.OY
Dec 13 '18 at 12:34
add a comment |
1 Answer
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$begingroup$
It is enough to observe here that:$$mathbb E[g(Y)Xmid Y=y]=mathbb E[g(y)Xmid Y=y]=g(y)mathbb E[Xmid Y=y]$$
This also holds if $g(y)=0$ and continuity of $X$ under $Y=y$ is redundant as well.
Essential for the first equality is that under condition $Y=y$ it is immediate that $g(Y)X=g(y)X$.
The second equality is an application of $mathbb E[aZ]=amathbb E[Z]$ where $a$ denotes a constant.
edit:
Integration:
$mathbb E[g(Y)Xmid Y=y]=int g(y)x P_{X|Y=y}(dx)=g(y)int xP_{X|Y=y}(dx)=g(y)mathbb E[Xmid Y=y]$
It is enough if $g$ is a measurable function. So continuity is okay, but is not necessary.
edited Dec 13 '18 at 14:13
Did
248k23224463
answered Dec 13 '18 at 13:35
drhabdrhab
102k545136
$endgroup$
$begingroup$
it's true also for $g$ that maybe not continuous? if it's, then why? But by the way it was not the point of my question, I just tried to understand How I define $ E[g(Y)X|Y=y]$ as Integral. if you can add it, it can be wonderful. explaining also on which variable I integrate. Thank you.
$endgroup$
– Mr.OY
Dec 13 '18 at 13:48
1
$begingroup$
I have added something.
$endgroup$
– drhab
Dec 13 '18 at 14:04
add a comment |
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$begingroup$
It is enough to observe here that:$$mathbb E[g(Y)Xmid Y=y]=mathbb E[g(y)Xmid Y=y]=g(y)mathbb E[Xmid Y=y]$$
This also holds if $g(y)=0$ and continuity of $X$ under $Y=y$ is redundant as well.
Essential for the first equality is that under condition $Y=y$ it is immediate that $g(Y)X=g(y)X$.
The second equality is an application of $mathbb E[aZ]=amathbb E[Z]$ where $a$ denotes a constant.
edit:
Integration:
$mathbb E[g(Y)Xmid Y=y]=int g(y)x P_{X|Y=y}(dx)=g(y)int xP_{X|Y=y}(dx)=g(y)mathbb E[Xmid Y=y]$
It is enough if $g$ is a measurable function. So continuity is okay, but is not necessary.
edited Dec 13 '18 at 14:13
Did
248k23224463
answered Dec 13 '18 at 13:35
drhabdrhab
102k545136
$endgroup$
$begingroup$
it's true also for $g$ that maybe not continuous? if it's, then why? But by the way it was not the point of my question, I just tried to understand How I define $ E[g(Y)X|Y=y]$ as Integral. if you can add it, it can be wonderful. explaining also on which variable I integrate. Thank you.
$endgroup$
– Mr.OY
Dec 13 '18 at 13:48
1
$begingroup$
I have added something.
$endgroup$
– drhab
Dec 13 '18 at 14:04
add a comment |
$begingroup$
It is enough to observe here that:$$mathbb E[g(Y)Xmid Y=y]=mathbb E[g(y)Xmid Y=y]=g(y)mathbb E[Xmid Y=y]$$
This also holds if $g(y)=0$ and continuity of $X$ under $Y=y$ is redundant as well.
Essential for the first equality is that under condition $Y=y$ it is immediate that $g(Y)X=g(y)X$.
The second equality is an application of $mathbb E[aZ]=amathbb E[Z]$ where $a$ denotes a constant.
edit:
Integration:
$mathbb E[g(Y)Xmid Y=y]=int g(y)x P_{X|Y=y}(dx)=g(y)int xP_{X|Y=y}(dx)=g(y)mathbb E[Xmid Y=y]$
It is enough if $g$ is a measurable function. So continuity is okay, but is not necessary.
edited Dec 13 '18 at 14:13
Did
248k23224463
answered Dec 13 '18 at 13:35
drhabdrhab
102k545136
$endgroup$
$begingroup$
it's true also for $g$ that maybe not continuous? if it's, then why? But by the way it was not the point of my question, I just tried to understand How I define $ E[g(Y)X|Y=y]$ as Integral. if you can add it, it can be wonderful. explaining also on which variable I integrate. Thank you.
$endgroup$
– Mr.OY
Dec 13 '18 at 13:48
1
$begingroup$
I have added something.
$endgroup$
– drhab
Dec 13 '18 at 14:04
add a comment |
$begingroup$
It is enough to observe here that:$$mathbb E[g(Y)Xmid Y=y]=mathbb E[g(y)Xmid Y=y]=g(y)mathbb E[Xmid Y=y]$$
This also holds if $g(y)=0$ and continuity of $X$ under $Y=y$ is redundant as well.
Essential for the first equality is that under condition $Y=y$ it is immediate that $g(Y)X=g(y)X$.
The second equality is an application of $mathbb E[aZ]=amathbb E[Z]$ where $a$ denotes a constant.
edit:
Integration:
$mathbb E[g(Y)Xmid Y=y]=int g(y)x P_{X|Y=y}(dx)=g(y)int xP_{X|Y=y}(dx)=g(y)mathbb E[Xmid Y=y]$
It is enough if $g$ is a measurable function. So continuity is okay, but is not necessary.
edited Dec 13 '18 at 14:13
Did
248k23224463
answered Dec 13 '18 at 13:35
drhabdrhab
102k545136
$endgroup$
It is enough to observe here that:$$mathbb E[g(Y)Xmid Y=y]=mathbb E[g(y)Xmid Y=y]=g(y)mathbb E[Xmid Y=y]$$
This also holds if $g(y)=0$ and continuity of $X$ under $Y=y$ is redundant as well.
Essential for the first equality is that under condition $Y=y$ it is immediate that $g(Y)X=g(y)X$.
The second equality is an application of $mathbb E[aZ]=amathbb E[Z]$ where $a$ denotes a constant.
edit:
Integration:
$mathbb E[g(Y)Xmid Y=y]=int g(y)x P_{X|Y=y}(dx)=g(y)int xP_{X|Y=y}(dx)=g(y)mathbb E[Xmid Y=y]$
It is enough if $g$ is a measurable function. So continuity is okay, but is not necessary.
edited Dec 13 '18 at 14:13
Did
248k23224463
answered Dec 13 '18 at 13:35
drhabdrhab
102k545136
edited Dec 13 '18 at 14:13
Did
248k23224463
edited Dec 13 '18 at 14:13
Did
248k23224463
edited Dec 13 '18 at 14:13
Did
248k23224463
248k23224463
answered Dec 13 '18 at 13:35
drhabdrhab
102k545136
answered Dec 13 '18 at 13:35
drhabdrhab
102k545136
answered Dec 13 '18 at 13:35
drhabdrhab
102k545136
102k545136
$begingroup$
it's true also for $g$ that maybe not continuous? if it's, then why? But by the way it was not the point of my question, I just tried to understand How I define $ E[g(Y)X|Y=y]$ as Integral. if you can add it, it can be wonderful. explaining also on which variable I integrate. Thank you.
$endgroup$
– Mr.OY
Dec 13 '18 at 13:48
1
$begingroup$
I have added something.
$endgroup$
– drhab
Dec 13 '18 at 14:04
add a comment |
$begingroup$
it's true also for $g$ that maybe not continuous? if it's, then why? But by the way it was not the point of my question, I just tried to understand How I define $ E[g(Y)X|Y=y]$ as Integral. if you can add it, it can be wonderful. explaining also on which variable I integrate. Thank you.
$endgroup$
– Mr.OY
Dec 13 '18 at 13:48
1
$begingroup$
I have added something.
$endgroup$
– drhab
Dec 13 '18 at 14:04
$begingroup$
it's true also for $g$ that maybe not continuous? if it's, then why? But by the way it was not the point of my question, I just tried to understand How I define $ E[g(Y)X|Y=y]$ as Integral. if you can add it, it can be wonderful. explaining also on which variable I integrate. Thank you.
$endgroup$
– Mr.OY
Dec 13 '18 at 13:48
$begingroup$
it's true also for $g$ that maybe not continuous? if it's, then why? But by the way it was not the point of my question, I just tried to understand How I define $ E[g(Y)X|Y=y]$ as Integral. if you can add it, it can be wonderful. explaining also on which variable I integrate. Thank you.
$endgroup$
– Mr.OY
Dec 13 '18 at 13:48
1
1
$begingroup$
I have added something.
$endgroup$
– drhab
Dec 13 '18 at 14:04
$begingroup$
I have added something.
$endgroup$
– drhab
Dec 13 '18 at 14:04
add a comment |
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$begingroup$
There are some confusing notations. Is $g:Wto R$, and not $Rto R$? Is $R$ the set of real numbers $mathbb{R}$?
$endgroup$
– Batominovski
Dec 13 '18 at 12:26
$begingroup$
@Batominovski it's $Bbb R to Bbb R$. fixed it.
$endgroup$
– Mr.OY
Dec 13 '18 at 12:34