Mixing
Conditional expectation application
Clash Royale CLAN TAG#URR8PPP
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Suppose I have two non-negative r.v.s $X,Y geq 1$ and I know $mathbbE(X^2 | Y) = (Y-1) ^ 2$.
Does this mean $mathbbE(X | Y) = Y - 1$?
Cause, if i recall right the intro to probability, I have
$mathbbE(X^2 | Y = y) = (y - 1)^2$, so it obvious that $mathbbE(X | Y = y) = y - 1$ for all $y$. But then I saw that conditional expectation is actually a function - projection to some space etc. and now this doesn't seem obvious at all.
probability conditional-expectation
asked Jul 28 at 11:23
dEmigOd
1,2731512
add a comment |Â
up vote
0
down vote
favorite
Suppose I have two non-negative r.v.s $X,Y geq 1$ and I know $mathbbE(X^2 | Y) = (Y-1) ^ 2$.
Does this mean $mathbbE(X | Y) = Y - 1$?
Cause, if i recall right the intro to probability, I have
$mathbbE(X^2 | Y = y) = (y - 1)^2$, so it obvious that $mathbbE(X | Y = y) = y - 1$ for all $y$. But then I saw that conditional expectation is actually a function - projection to some space etc. and now this doesn't seem obvious at all.
probability conditional-expectation
asked Jul 28 at 11:23
dEmigOd
1,2731512
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose I have two non-negative r.v.s $X,Y geq 1$ and I know $mathbbE(X^2 | Y) = (Y-1) ^ 2$.
Does this mean $mathbbE(X | Y) = Y - 1$?
Cause, if i recall right the intro to probability, I have
$mathbbE(X^2 | Y = y) = (y - 1)^2$, so it obvious that $mathbbE(X | Y = y) = y - 1$ for all $y$. But then I saw that conditional expectation is actually a function - projection to some space etc. and now this doesn't seem obvious at all.
probability conditional-expectation
asked Jul 28 at 11:23
dEmigOd
1,2731512
Suppose I have two non-negative r.v.s $X,Y geq 1$ and I know $mathbbE(X^2 | Y) = (Y-1) ^ 2$.
Does this mean $mathbbE(X | Y) = Y - 1$?
Cause, if i recall right the intro to probability, I have
$mathbbE(X^2 | Y = y) = (y - 1)^2$, so it obvious that $mathbbE(X | Y = y) = y - 1$ for all $y$. But then I saw that conditional expectation is actually a function - projection to some space etc. and now this doesn't seem obvious at all.
probability conditional-expectation
asked Jul 28 at 11:23
dEmigOd
1,2731512
asked Jul 28 at 11:23
dEmigOd
1,2731512
asked Jul 28 at 11:23
dEmigOd
1,2731512
asked Jul 28 at 11:23
dEmigOd
1,2731512
1,2731512
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
2
down vote
accepted
No.
Let it be that $Z$ is a random variable with $Z>0.5$, $mathbb EZ^2=1$ and $0<mathbb EZ<1$. Further let it be that $Y$ and $Z$ are independent and $Y>3$.
Then for $X=(Y-1)Z>1$ and we find: $$mathbb E[X^2mid Y]=mathbb E[(Y-1)^2Z^2mid Y]=(Y-1)^2mathbb EZ^2=(Y-1)^2$$ and:$$mathbb E[Xmid Y]=mathbb E[(Y-1)Zmid Y]=(Y-1)mathbb EZ<(Y-1)$$
answered Jul 28 at 11:36
Vera
1,696313
2
You forgot the following: $X,Yge 1$
– Bob
Jul 28 at 11:39
@Bob Yes, you are right. But I am sure that it is not okay. I will delete this within some minutes, and then will try to find a suitable counterexample.
– Vera
Jul 28 at 11:41
@Bob I found another counterexample.
– Vera
Jul 28 at 12:00
add a comment |Â
up vote
3
down vote
Let $Omega=a,b, mathcalF=2^Omega$ and $mathbbP$ the uniform probability on $Omega$. Define $X(a)=sqrt3/2, X(b)=sqrt5/2$ and $Y(a)=sqrt2+1=Y(b)$. Then $X,Yge1$ and $$mathbbE(X^2|Y)=mathbbE(X^2)=frac12frac32+frac12frac52=2=(Y-1)^2$$ but
$$mathbbE(X|Y)=mathbbE(X)=frac12sqrtfrac32+frac12sqrtfrac52neqsqrt2=Y-1$$
answered Jul 28 at 11:55
Bob
1,517522
add a comment |Â
up vote
0
down vote
Generally this is not true. Even without introducing any conditioning, in all generality $mathbfE[X^2] neq (mathbfE[X])^2$.
You can still say something: from Jensen's inequality, since $x mapsto x^2$ is convex, you always have that $mathbfE[X^2] geq (mathbfE[X])^2$, which also applies to your original question for conditional probabilities:
$$mathbfE[X | Y] leq sqrt Y] = |Y-1| = Y - 1 text from Y geq 1$$
answered Jul 28 at 11:35
ippiki-ookami
303216
I suppose, if for example it is equality (after all) your claim is still true
– dEmigOd
Jul 28 at 11:46
I think Jensen's inequality for the square function can only be an equality when $X | Y$ is almost surely constant.
– ippiki-ookami
Jul 28 at 11:53
add a comment |Â
up vote
0
down vote
Let $X_0$ be $N(0,1)$, $X=X_0^2+1$ and $Y=1+sqrt 3+E(X_0)^4$. Then $E(X^2|Y)=(Y-1)^2$ and $E(X|Y)=2neq Y-1$ My idea is to give a simple example where $Y$ is a constant.
answered Jul 28 at 11:58
Kavi Rama Murthy
19.9k2829
The question requires that $X, Y geq 1$
– ippiki-ookami
Jul 28 at 11:59
@ippiki-ookami See my revised answer.
– Kavi Rama Murthy
Jul 28 at 12:08
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
No.
Let it be that $Z$ is a random variable with $Z>0.5$, $mathbb EZ^2=1$ and $0<mathbb EZ<1$. Further let it be that $Y$ and $Z$ are independent and $Y>3$.
Then for $X=(Y-1)Z>1$ and we find: $$mathbb E[X^2mid Y]=mathbb E[(Y-1)^2Z^2mid Y]=(Y-1)^2mathbb EZ^2=(Y-1)^2$$ and:$$mathbb E[Xmid Y]=mathbb E[(Y-1)Zmid Y]=(Y-1)mathbb EZ<(Y-1)$$
answered Jul 28 at 11:36
Vera
1,696313
2
You forgot the following: $X,Yge 1$
– Bob
Jul 28 at 11:39
@Bob Yes, you are right. But I am sure that it is not okay. I will delete this within some minutes, and then will try to find a suitable counterexample.
– Vera
Jul 28 at 11:41
@Bob I found another counterexample.
– Vera
Jul 28 at 12:00
add a comment |Â
up vote
2
down vote
accepted
No.
Let it be that $Z$ is a random variable with $Z>0.5$, $mathbb EZ^2=1$ and $0<mathbb EZ<1$. Further let it be that $Y$ and $Z$ are independent and $Y>3$.
Then for $X=(Y-1)Z>1$ and we find: $$mathbb E[X^2mid Y]=mathbb E[(Y-1)^2Z^2mid Y]=(Y-1)^2mathbb EZ^2=(Y-1)^2$$ and:$$mathbb E[Xmid Y]=mathbb E[(Y-1)Zmid Y]=(Y-1)mathbb EZ<(Y-1)$$
answered Jul 28 at 11:36
Vera
1,696313
2
You forgot the following: $X,Yge 1$
– Bob
Jul 28 at 11:39
@Bob Yes, you are right. But I am sure that it is not okay. I will delete this within some minutes, and then will try to find a suitable counterexample.
– Vera
Jul 28 at 11:41
@Bob I found another counterexample.
– Vera
Jul 28 at 12:00
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
No.
Let it be that $Z$ is a random variable with $Z>0.5$, $mathbb EZ^2=1$ and $0<mathbb EZ<1$. Further let it be that $Y$ and $Z$ are independent and $Y>3$.
Then for $X=(Y-1)Z>1$ and we find: $$mathbb E[X^2mid Y]=mathbb E[(Y-1)^2Z^2mid Y]=(Y-1)^2mathbb EZ^2=(Y-1)^2$$ and:$$mathbb E[Xmid Y]=mathbb E[(Y-1)Zmid Y]=(Y-1)mathbb EZ<(Y-1)$$
answered Jul 28 at 11:36
Vera
1,696313
No.
Let it be that $Z$ is a random variable with $Z>0.5$, $mathbb EZ^2=1$ and $0<mathbb EZ<1$. Further let it be that $Y$ and $Z$ are independent and $Y>3$.
Then for $X=(Y-1)Z>1$ and we find: $$mathbb E[X^2mid Y]=mathbb E[(Y-1)^2Z^2mid Y]=(Y-1)^2mathbb EZ^2=(Y-1)^2$$ and:$$mathbb E[Xmid Y]=mathbb E[(Y-1)Zmid Y]=(Y-1)mathbb EZ<(Y-1)$$
answered Jul 28 at 11:36
Vera
1,696313
edited Jul 28 at 11:59
answered Jul 28 at 11:36
Vera
1,696313
answered Jul 28 at 11:36
Vera
1,696313
answered Jul 28 at 11:36
Vera
1,696313
1,696313
2
You forgot the following: $X,Yge 1$
– Bob
Jul 28 at 11:39
@Bob Yes, you are right. But I am sure that it is not okay. I will delete this within some minutes, and then will try to find a suitable counterexample.
– Vera
Jul 28 at 11:41
@Bob I found another counterexample.
– Vera
Jul 28 at 12:00
add a comment |Â
2
You forgot the following: $X,Yge 1$
– Bob
Jul 28 at 11:39
@Bob Yes, you are right. But I am sure that it is not okay. I will delete this within some minutes, and then will try to find a suitable counterexample.
– Vera
Jul 28 at 11:41
@Bob I found another counterexample.
– Vera
Jul 28 at 12:00
2
2
You forgot the following: $X,Yge 1$
– Bob
Jul 28 at 11:39
You forgot the following: $X,Yge 1$
– Bob
Jul 28 at 11:39
@Bob Yes, you are right. But I am sure that it is not okay. I will delete this within some minutes, and then will try to find a suitable counterexample.
– Vera
Jul 28 at 11:41
@Bob Yes, you are right. But I am sure that it is not okay. I will delete this within some minutes, and then will try to find a suitable counterexample.
– Vera
Jul 28 at 11:41
@Bob I found another counterexample.
– Vera
Jul 28 at 12:00
@Bob I found another counterexample.
– Vera
Jul 28 at 12:00
add a comment |Â
up vote
3
down vote
Let $Omega=a,b, mathcalF=2^Omega$ and $mathbbP$ the uniform probability on $Omega$. Define $X(a)=sqrt3/2, X(b)=sqrt5/2$ and $Y(a)=sqrt2+1=Y(b)$. Then $X,Yge1$ and $$mathbbE(X^2|Y)=mathbbE(X^2)=frac12frac32+frac12frac52=2=(Y-1)^2$$ but
$$mathbbE(X|Y)=mathbbE(X)=frac12sqrtfrac32+frac12sqrtfrac52neqsqrt2=Y-1$$
answered Jul 28 at 11:55
Bob
1,517522
add a comment |Â
up vote
3
down vote
Let $Omega=a,b, mathcalF=2^Omega$ and $mathbbP$ the uniform probability on $Omega$. Define $X(a)=sqrt3/2, X(b)=sqrt5/2$ and $Y(a)=sqrt2+1=Y(b)$. Then $X,Yge1$ and $$mathbbE(X^2|Y)=mathbbE(X^2)=frac12frac32+frac12frac52=2=(Y-1)^2$$ but
$$mathbbE(X|Y)=mathbbE(X)=frac12sqrtfrac32+frac12sqrtfrac52neqsqrt2=Y-1$$
answered Jul 28 at 11:55
Bob
1,517522
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Let $Omega=a,b, mathcalF=2^Omega$ and $mathbbP$ the uniform probability on $Omega$. Define $X(a)=sqrt3/2, X(b)=sqrt5/2$ and $Y(a)=sqrt2+1=Y(b)$. Then $X,Yge1$ and $$mathbbE(X^2|Y)=mathbbE(X^2)=frac12frac32+frac12frac52=2=(Y-1)^2$$ but
$$mathbbE(X|Y)=mathbbE(X)=frac12sqrtfrac32+frac12sqrtfrac52neqsqrt2=Y-1$$
answered Jul 28 at 11:55
Bob
1,517522
Let $Omega=a,b, mathcalF=2^Omega$ and $mathbbP$ the uniform probability on $Omega$. Define $X(a)=sqrt3/2, X(b)=sqrt5/2$ and $Y(a)=sqrt2+1=Y(b)$. Then $X,Yge1$ and $$mathbbE(X^2|Y)=mathbbE(X^2)=frac12frac32+frac12frac52=2=(Y-1)^2$$ but
$$mathbbE(X|Y)=mathbbE(X)=frac12sqrtfrac32+frac12sqrtfrac52neqsqrt2=Y-1$$
answered Jul 28 at 11:55
Bob
1,517522
answered Jul 28 at 11:55
Bob
1,517522
answered Jul 28 at 11:55
Bob
1,517522
answered Jul 28 at 11:55
Bob
1,517522
1,517522
add a comment |Â
add a comment |Â
up vote
0
down vote
Generally this is not true. Even without introducing any conditioning, in all generality $mathbfE[X^2] neq (mathbfE[X])^2$.
You can still say something: from Jensen's inequality, since $x mapsto x^2$ is convex, you always have that $mathbfE[X^2] geq (mathbfE[X])^2$, which also applies to your original question for conditional probabilities:
$$mathbfE[X | Y] leq sqrt Y] = |Y-1| = Y - 1 text from Y geq 1$$
answered Jul 28 at 11:35
ippiki-ookami
303216
I suppose, if for example it is equality (after all) your claim is still true
– dEmigOd
Jul 28 at 11:46
I think Jensen's inequality for the square function can only be an equality when $X | Y$ is almost surely constant.
– ippiki-ookami
Jul 28 at 11:53
add a comment |Â
up vote
0
down vote
Generally this is not true. Even without introducing any conditioning, in all generality $mathbfE[X^2] neq (mathbfE[X])^2$.
You can still say something: from Jensen's inequality, since $x mapsto x^2$ is convex, you always have that $mathbfE[X^2] geq (mathbfE[X])^2$, which also applies to your original question for conditional probabilities:
$$mathbfE[X | Y] leq sqrt Y] = |Y-1| = Y - 1 text from Y geq 1$$
answered Jul 28 at 11:35
ippiki-ookami
303216
I suppose, if for example it is equality (after all) your claim is still true
– dEmigOd
Jul 28 at 11:46
I think Jensen's inequality for the square function can only be an equality when $X | Y$ is almost surely constant.
– ippiki-ookami
Jul 28 at 11:53
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Generally this is not true. Even without introducing any conditioning, in all generality $mathbfE[X^2] neq (mathbfE[X])^2$.
You can still say something: from Jensen's inequality, since $x mapsto x^2$ is convex, you always have that $mathbfE[X^2] geq (mathbfE[X])^2$, which also applies to your original question for conditional probabilities:
$$mathbfE[X | Y] leq sqrt Y] = |Y-1| = Y - 1 text from Y geq 1$$
answered Jul 28 at 11:35
ippiki-ookami
303216
Generally this is not true. Even without introducing any conditioning, in all generality $mathbfE[X^2] neq (mathbfE[X])^2$.
You can still say something: from Jensen's inequality, since $x mapsto x^2$ is convex, you always have that $mathbfE[X^2] geq (mathbfE[X])^2$, which also applies to your original question for conditional probabilities:
$$mathbfE[X | Y] leq sqrt Y] = |Y-1| = Y - 1 text from Y geq 1$$
answered Jul 28 at 11:35
ippiki-ookami
303216
edited Jul 28 at 11:46
answered Jul 28 at 11:35
ippiki-ookami
303216
answered Jul 28 at 11:35
ippiki-ookami
303216
answered Jul 28 at 11:35
ippiki-ookami
303216
303216
I suppose, if for example it is equality (after all) your claim is still true
– dEmigOd
Jul 28 at 11:46
I think Jensen's inequality for the square function can only be an equality when $X | Y$ is almost surely constant.
– ippiki-ookami
Jul 28 at 11:53
add a comment |Â
I suppose, if for example it is equality (after all) your claim is still true
– dEmigOd
Jul 28 at 11:46
I think Jensen's inequality for the square function can only be an equality when $X | Y$ is almost surely constant.
– ippiki-ookami
Jul 28 at 11:53
I suppose, if for example it is equality (after all) your claim is still true
– dEmigOd
Jul 28 at 11:46
I suppose, if for example it is equality (after all) your claim is still true
– dEmigOd
Jul 28 at 11:46
I think Jensen's inequality for the square function can only be an equality when $X | Y$ is almost surely constant.
– ippiki-ookami
Jul 28 at 11:53
I think Jensen's inequality for the square function can only be an equality when $X | Y$ is almost surely constant.
– ippiki-ookami
Jul 28 at 11:53
add a comment |Â
up vote
0
down vote
Let $X_0$ be $N(0,1)$, $X=X_0^2+1$ and $Y=1+sqrt 3+E(X_0)^4$. Then $E(X^2|Y)=(Y-1)^2$ and $E(X|Y)=2neq Y-1$ My idea is to give a simple example where $Y$ is a constant.
answered Jul 28 at 11:58
Kavi Rama Murthy
19.9k2829
The question requires that $X, Y geq 1$
– ippiki-ookami
Jul 28 at 11:59
@ippiki-ookami See my revised answer.
– Kavi Rama Murthy
Jul 28 at 12:08
add a comment |Â
up vote
0
down vote
Let $X_0$ be $N(0,1)$, $X=X_0^2+1$ and $Y=1+sqrt 3+E(X_0)^4$. Then $E(X^2|Y)=(Y-1)^2$ and $E(X|Y)=2neq Y-1$ My idea is to give a simple example where $Y$ is a constant.
answered Jul 28 at 11:58
Kavi Rama Murthy
19.9k2829
The question requires that $X, Y geq 1$
– ippiki-ookami
Jul 28 at 11:59
@ippiki-ookami See my revised answer.
– Kavi Rama Murthy
Jul 28 at 12:08
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $X_0$ be $N(0,1)$, $X=X_0^2+1$ and $Y=1+sqrt 3+E(X_0)^4$. Then $E(X^2|Y)=(Y-1)^2$ and $E(X|Y)=2neq Y-1$ My idea is to give a simple example where $Y$ is a constant.
answered Jul 28 at 11:58
Kavi Rama Murthy
19.9k2829
Let $X_0$ be $N(0,1)$, $X=X_0^2+1$ and $Y=1+sqrt 3+E(X_0)^4$. Then $E(X^2|Y)=(Y-1)^2$ and $E(X|Y)=2neq Y-1$ My idea is to give a simple example where $Y$ is a constant.
answered Jul 28 at 11:58
Kavi Rama Murthy
19.9k2829
edited Jul 28 at 12:05
answered Jul 28 at 11:58
Kavi Rama Murthy
19.9k2829
answered Jul 28 at 11:58
Kavi Rama Murthy
19.9k2829
answered Jul 28 at 11:58
Kavi Rama Murthy
19.9k2829
19.9k2829
The question requires that $X, Y geq 1$
– ippiki-ookami
Jul 28 at 11:59
@ippiki-ookami See my revised answer.
– Kavi Rama Murthy
Jul 28 at 12:08
add a comment |Â
The question requires that $X, Y geq 1$
– ippiki-ookami
Jul 28 at 11:59
@ippiki-ookami See my revised answer.
– Kavi Rama Murthy
Jul 28 at 12:08
The question requires that $X, Y geq 1$
– ippiki-ookami
Jul 28 at 11:59
The question requires that $X, Y geq 1$
– ippiki-ookami
Jul 28 at 11:59
@ippiki-ookami See my revised answer.
– Kavi Rama Murthy
Jul 28 at 12:08
@ippiki-ookami See my revised answer.
– Kavi Rama Murthy
Jul 28 at 12:08
add a comment |Â
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