$mathbb{E}X^2 = mathbb{E}Y^2$ and $mathbb{E}(Y|mathcal{A}) = X$ P-a.s. $Rightarrow X=Y$ P.-a.s.
$begingroup$
I have a question and hope you can help me. The problem is about stochastic variables $X,Y$, which are square integrable, independent and identically distributed on $(Omega, mathcal{S}, P)$. Furthermore $mathcal{A} subsetmathcal{S}$ is a sub-sigma-algebra.
Now I have to show three statements:
a) $mathbb{V}(X) = mathbb{E}(X|mathcal{A}) + mathbb{E}[X-mathbb{E}(X|mathcal{A})]^2 Rightarrow mathbb{V}(mathbb{E}(X|mathcal{A})) leq mathbb{V}(X)$
b) $mathbb{E}X^2 = mathbb{E}Y^2$ and $mathbb{E}(Y|mathcal{A}) = X$ P-a.s. $Rightarrow X=Y$ P.-a.s.
c) $mathbb{E}(X|Y) = Y$ P-a.s. and $mathbb{E}(Y|X) = X$ P-a.s. $Rightarrow X=Y$ P-a.s.
I already managed to proof a) by using variance-decomposition-theorem and a little bit of calculation.
Furthermore I think I have an idea how to solve c):
$mathbb{E}[XY|X] = X mathbb{E}[Y|X] = X X = X^2$ P.-almost sure, where the second equal-sign uses the assumption. Similarly there one can conclude $mathbb{E}[XY|Y] = Y^2$ P-a.s.
Applying $mathbb{E}$ on both sides gives me:
$mathbb{E}[X^2]=mathbb{E}[mathbb{E}[XY|X]] = mathbb{E}[XY] = mathbb{E}[mathbb{E}[XY|Y]] = mathbb{E}[Y^2]$ P-a.s. Now, by using this result, I can conclude $mathbb{E}[(X-Y)^2] = 0$ to get the required result.
Sadly I don't have a clue how to prove b) to finish this task.
I would be very glad if anyone could help me solving b) too!
Thanks in advance for your help!
measure-theory conditional-expectation conditional-probability variance
asked Dec 17 '18 at 20:57
TNickyTNicky
85
$endgroup$
|
show 1 more comment
$begingroup$
I have a question and hope you can help me. The problem is about stochastic variables $X,Y$, which are square integrable, independent and identically distributed on $(Omega, mathcal{S}, P)$. Furthermore $mathcal{A} subsetmathcal{S}$ is a sub-sigma-algebra.
Now I have to show three statements:
a) $mathbb{V}(X) = mathbb{E}(X|mathcal{A}) + mathbb{E}[X-mathbb{E}(X|mathcal{A})]^2 Rightarrow mathbb{V}(mathbb{E}(X|mathcal{A})) leq mathbb{V}(X)$
b) $mathbb{E}X^2 = mathbb{E}Y^2$ and $mathbb{E}(Y|mathcal{A}) = X$ P-a.s. $Rightarrow X=Y$ P.-a.s.
c) $mathbb{E}(X|Y) = Y$ P-a.s. and $mathbb{E}(Y|X) = X$ P-a.s. $Rightarrow X=Y$ P-a.s.
I already managed to proof a) by using variance-decomposition-theorem and a little bit of calculation.
Furthermore I think I have an idea how to solve c):
$mathbb{E}[XY|X] = X mathbb{E}[Y|X] = X X = X^2$ P.-almost sure, where the second equal-sign uses the assumption. Similarly there one can conclude $mathbb{E}[XY|Y] = Y^2$ P-a.s.
Applying $mathbb{E}$ on both sides gives me:
$mathbb{E}[X^2]=mathbb{E}[mathbb{E}[XY|X]] = mathbb{E}[XY] = mathbb{E}[mathbb{E}[XY|Y]] = mathbb{E}[Y^2]$ P-a.s. Now, by using this result, I can conclude $mathbb{E}[(X-Y)^2] = 0$ to get the required result.
Sadly I don't have a clue how to prove b) to finish this task.
I would be very glad if anyone could help me solving b) too!
Thanks in advance for your help!
measure-theory conditional-expectation conditional-probability variance
asked Dec 17 '18 at 20:57
TNickyTNicky
85
$endgroup$
$begingroup$
Implication c) holds for integrable random variables X and Y not necessarily square integrable, as shown in answers to several questions on this site.
$endgroup$
– Did
Dec 17 '18 at 22:19
$begingroup$
@Did I believe c) it to be true in general, but I failed to provide a proof. Do you have any of this links or a proof-sketch?
$endgroup$
– Will M.
Dec 17 '18 at 22:37
$begingroup$
I thought, that my proof for c) (given above) is valid. Is there anything wrong with it? The only step I need square integrability is (at least I think so) $mathbb{E}[XY|X] = X mathbb{E}[Y|X]$. With Hölder's inequality I can conclude, that if X,Y are square-integrable then XY is integrable: $int XY dP leq int |XY| dp stackrel{Hölder}{=} left( int |X|^2 dP right)^{frac{1}{2}} left( int |Y|^2 dP right)^{frac{1}{2}} < infty$. Or am I wrong?
$endgroup$
– TNicky
Dec 17 '18 at 22:42
$begingroup$
It is true that if $X$ and $Y$ are square integrable, then $XY$ is integrable.
$endgroup$
– Will M.
Dec 18 '18 at 1:36
$begingroup$
@WillM. math.stackexchange.com/a/1520480/6179
$endgroup$
– Did
Dec 18 '18 at 7:20
|
show 1 more comment
$begingroup$
I have a question and hope you can help me. The problem is about stochastic variables $X,Y$, which are square integrable, independent and identically distributed on $(Omega, mathcal{S}, P)$. Furthermore $mathcal{A} subsetmathcal{S}$ is a sub-sigma-algebra.
Now I have to show three statements:
a) $mathbb{V}(X) = mathbb{E}(X|mathcal{A}) + mathbb{E}[X-mathbb{E}(X|mathcal{A})]^2 Rightarrow mathbb{V}(mathbb{E}(X|mathcal{A})) leq mathbb{V}(X)$
b) $mathbb{E}X^2 = mathbb{E}Y^2$ and $mathbb{E}(Y|mathcal{A}) = X$ P-a.s. $Rightarrow X=Y$ P.-a.s.
c) $mathbb{E}(X|Y) = Y$ P-a.s. and $mathbb{E}(Y|X) = X$ P-a.s. $Rightarrow X=Y$ P-a.s.
I already managed to proof a) by using variance-decomposition-theorem and a little bit of calculation.
Furthermore I think I have an idea how to solve c):
$mathbb{E}[XY|X] = X mathbb{E}[Y|X] = X X = X^2$ P.-almost sure, where the second equal-sign uses the assumption. Similarly there one can conclude $mathbb{E}[XY|Y] = Y^2$ P-a.s.
Applying $mathbb{E}$ on both sides gives me:
$mathbb{E}[X^2]=mathbb{E}[mathbb{E}[XY|X]] = mathbb{E}[XY] = mathbb{E}[mathbb{E}[XY|Y]] = mathbb{E}[Y^2]$ P-a.s. Now, by using this result, I can conclude $mathbb{E}[(X-Y)^2] = 0$ to get the required result.
Sadly I don't have a clue how to prove b) to finish this task.
I would be very glad if anyone could help me solving b) too!
Thanks in advance for your help!
measure-theory conditional-expectation conditional-probability variance
asked Dec 17 '18 at 20:57
TNickyTNicky
85
$endgroup$
I have a question and hope you can help me. The problem is about stochastic variables $X,Y$, which are square integrable, independent and identically distributed on $(Omega, mathcal{S}, P)$. Furthermore $mathcal{A} subsetmathcal{S}$ is a sub-sigma-algebra.
Now I have to show three statements:
a) $mathbb{V}(X) = mathbb{E}(X|mathcal{A}) + mathbb{E}[X-mathbb{E}(X|mathcal{A})]^2 Rightarrow mathbb{V}(mathbb{E}(X|mathcal{A})) leq mathbb{V}(X)$
b) $mathbb{E}X^2 = mathbb{E}Y^2$ and $mathbb{E}(Y|mathcal{A}) = X$ P-a.s. $Rightarrow X=Y$ P.-a.s.
c) $mathbb{E}(X|Y) = Y$ P-a.s. and $mathbb{E}(Y|X) = X$ P-a.s. $Rightarrow X=Y$ P-a.s.
I already managed to proof a) by using variance-decomposition-theorem and a little bit of calculation.
Furthermore I think I have an idea how to solve c):
$mathbb{E}[XY|X] = X mathbb{E}[Y|X] = X X = X^2$ P.-almost sure, where the second equal-sign uses the assumption. Similarly there one can conclude $mathbb{E}[XY|Y] = Y^2$ P-a.s.
Applying $mathbb{E}$ on both sides gives me:
$mathbb{E}[X^2]=mathbb{E}[mathbb{E}[XY|X]] = mathbb{E}[XY] = mathbb{E}[mathbb{E}[XY|Y]] = mathbb{E}[Y^2]$ P-a.s. Now, by using this result, I can conclude $mathbb{E}[(X-Y)^2] = 0$ to get the required result.
Sadly I don't have a clue how to prove b) to finish this task.
I would be very glad if anyone could help me solving b) too!
Thanks in advance for your help!
measure-theory conditional-expectation conditional-probability variance
measure-theory conditional-expectation conditional-probability variance
asked Dec 17 '18 at 20:57
TNickyTNicky
85
asked Dec 17 '18 at 20:57
TNickyTNicky
85
edited Dec 17 '18 at 22:48
TNicky
asked Dec 17 '18 at 20:57
TNickyTNicky
85
asked Dec 17 '18 at 20:57
TNickyTNicky
85
asked Dec 17 '18 at 20:57
TNickyTNicky
85
85
$begingroup$
Implication c) holds for integrable random variables X and Y not necessarily square integrable, as shown in answers to several questions on this site.
$endgroup$
– Did
Dec 17 '18 at 22:19
$begingroup$
@Did I believe c) it to be true in general, but I failed to provide a proof. Do you have any of this links or a proof-sketch?
$endgroup$
– Will M.
Dec 17 '18 at 22:37
$begingroup$
I thought, that my proof for c) (given above) is valid. Is there anything wrong with it? The only step I need square integrability is (at least I think so) $mathbb{E}[XY|X] = X mathbb{E}[Y|X]$. With Hölder's inequality I can conclude, that if X,Y are square-integrable then XY is integrable: $int XY dP leq int |XY| dp stackrel{Hölder}{=} left( int |X|^2 dP right)^{frac{1}{2}} left( int |Y|^2 dP right)^{frac{1}{2}} < infty$. Or am I wrong?
$endgroup$
– TNicky
Dec 17 '18 at 22:42
$begingroup$
It is true that if $X$ and $Y$ are square integrable, then $XY$ is integrable.
$endgroup$
– Will M.
Dec 18 '18 at 1:36
$begingroup$
@WillM. math.stackexchange.com/a/1520480/6179
$endgroup$
– Did
Dec 18 '18 at 7:20
|
show 1 more comment
$begingroup$
Implication c) holds for integrable random variables X and Y not necessarily square integrable, as shown in answers to several questions on this site.
$endgroup$
– Did
Dec 17 '18 at 22:19
$begingroup$
@Did I believe c) it to be true in general, but I failed to provide a proof. Do you have any of this links or a proof-sketch?
$endgroup$
– Will M.
Dec 17 '18 at 22:37
$begingroup$
I thought, that my proof for c) (given above) is valid. Is there anything wrong with it? The only step I need square integrability is (at least I think so) $mathbb{E}[XY|X] = X mathbb{E}[Y|X]$. With Hölder's inequality I can conclude, that if X,Y are square-integrable then XY is integrable: $int XY dP leq int |XY| dp stackrel{Hölder}{=} left( int |X|^2 dP right)^{frac{1}{2}} left( int |Y|^2 dP right)^{frac{1}{2}} < infty$. Or am I wrong?
$endgroup$
– TNicky
Dec 17 '18 at 22:42
$begingroup$
It is true that if $X$ and $Y$ are square integrable, then $XY$ is integrable.
$endgroup$
– Will M.
Dec 18 '18 at 1:36
$begingroup$
@WillM. math.stackexchange.com/a/1520480/6179
$endgroup$
– Did
Dec 18 '18 at 7:20
$begingroup$
Implication c) holds for integrable random variables X and Y not necessarily square integrable, as shown in answers to several questions on this site.
$endgroup$
– Did
Dec 17 '18 at 22:19
$begingroup$
Implication c) holds for integrable random variables X and Y not necessarily square integrable, as shown in answers to several questions on this site.
$endgroup$
– Did
Dec 17 '18 at 22:19
$begingroup$
@Did I believe c) it to be true in general, but I failed to provide a proof. Do you have any of this links or a proof-sketch?
$endgroup$
– Will M.
Dec 17 '18 at 22:37
$begingroup$
@Did I believe c) it to be true in general, but I failed to provide a proof. Do you have any of this links or a proof-sketch?
$endgroup$
– Will M.
Dec 17 '18 at 22:37
$begingroup$
I thought, that my proof for c) (given above) is valid. Is there anything wrong with it? The only step I need square integrability is (at least I think so) $mathbb{E}[XY|X] = X mathbb{E}[Y|X]$. With Hölder's inequality I can conclude, that if X,Y are square-integrable then XY is integrable: $int XY dP leq int |XY| dp stackrel{Hölder}{=} left( int |X|^2 dP right)^{frac{1}{2}} left( int |Y|^2 dP right)^{frac{1}{2}} < infty$. Or am I wrong?
$endgroup$
– TNicky
Dec 17 '18 at 22:42
$begingroup$
I thought, that my proof for c) (given above) is valid. Is there anything wrong with it? The only step I need square integrability is (at least I think so) $mathbb{E}[XY|X] = X mathbb{E}[Y|X]$. With Hölder's inequality I can conclude, that if X,Y are square-integrable then XY is integrable: $int XY dP leq int |XY| dp stackrel{Hölder}{=} left( int |X|^2 dP right)^{frac{1}{2}} left( int |Y|^2 dP right)^{frac{1}{2}} < infty$. Or am I wrong?
$endgroup$
– TNicky
Dec 17 '18 at 22:42
$begingroup$
It is true that if $X$ and $Y$ are square integrable, then $XY$ is integrable.
$endgroup$
– Will M.
Dec 18 '18 at 1:36
$begingroup$
It is true that if $X$ and $Y$ are square integrable, then $XY$ is integrable.
$endgroup$
– Will M.
Dec 18 '18 at 1:36
$begingroup$
@WillM. math.stackexchange.com/a/1520480/6179
$endgroup$
– Did
Dec 18 '18 at 7:20
$begingroup$
@WillM. math.stackexchange.com/a/1520480/6179
$endgroup$
– Did
Dec 18 '18 at 7:20
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Partial solution.
Since you have a), I give a proof of b). Observe the hypothesis $mathbf{E}(Y mid mathscr{A}) = X$ implies $X$ to be measurable with respect to $mathscr{A}.$ Bearing this in mind,
$$begin{align*}
mathbf{E}((Y-X)^2) &= mathbf{E}big( mathbf{E}((Y-X)^2 mid mathscr{A}) big) \
&= mathbf{E}(mathbf{E}(Y^2 mid mathscr{A}) - 2mathbf{E}(Y mid mathscr{A}) X+ X^2) \
&= mathbf{E}(mathbf{E}(Y^2 mid mathscr{A}) - X^2) = 0.
end{align*}$$
As for c), if $X$ and $Y$ were square integrable, you have $mathbf{E}(XY) = E(X^2) = E(Y^2),$ so b) finishes the proof upon applying it with $mathscr{A} = sigma(X).$ For the general case, I am not sure how to tackle it.
answered Dec 17 '18 at 22:11
Will M.Will M.
2,865315
$endgroup$
$begingroup$
Hi and thanks for your quick respond! I have a question about the second equal-sign. Why is it $X^2$ and not $mathbb{E}[X|mathcal{A}]$? I think it doesn't make much difference in the result, due to if i write it this way I get zero as well, but maybe I missed something so my way might be faulty (?)
$endgroup$
– TNicky
Dec 17 '18 at 22:46
$begingroup$
You would need $mathbf{E}(X^2 mid mathscr{A}) = X^2$ since $X$ is $mathscr{A}$-measurable.
$endgroup$
– Will M.
Dec 18 '18 at 1:35
$begingroup$
Ok! But I don't see, why $X$ is $mathcal{A}$-measurable. We also need this for the step with the term $mathbb{E}[XY|mathcal{A}]$.
$endgroup$
– TNicky
Dec 18 '18 at 7:52
$begingroup$
If $X = mathbf{E}(Y mid mathscr{A})$ then $X$ is $mathscr{A}$-measurable. If this is not clear, you need to read again the definition of conditional expectation.
$endgroup$
– Will M.
Dec 18 '18 at 7:53
$begingroup$
AAh sure it is clear! Stupid me! Thank you very much!
$endgroup$
– TNicky
Dec 18 '18 at 7:55
|
show 2 more comments
Your Answer
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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oldest
votes
$begingroup$
Partial solution.
Since you have a), I give a proof of b). Observe the hypothesis $mathbf{E}(Y mid mathscr{A}) = X$ implies $X$ to be measurable with respect to $mathscr{A}.$ Bearing this in mind,
$$begin{align*}
mathbf{E}((Y-X)^2) &= mathbf{E}big( mathbf{E}((Y-X)^2 mid mathscr{A}) big) \
&= mathbf{E}(mathbf{E}(Y^2 mid mathscr{A}) - 2mathbf{E}(Y mid mathscr{A}) X+ X^2) \
&= mathbf{E}(mathbf{E}(Y^2 mid mathscr{A}) - X^2) = 0.
end{align*}$$
As for c), if $X$ and $Y$ were square integrable, you have $mathbf{E}(XY) = E(X^2) = E(Y^2),$ so b) finishes the proof upon applying it with $mathscr{A} = sigma(X).$ For the general case, I am not sure how to tackle it.
answered Dec 17 '18 at 22:11
Will M.Will M.
2,865315
$endgroup$
$begingroup$
Hi and thanks for your quick respond! I have a question about the second equal-sign. Why is it $X^2$ and not $mathbb{E}[X|mathcal{A}]$? I think it doesn't make much difference in the result, due to if i write it this way I get zero as well, but maybe I missed something so my way might be faulty (?)
$endgroup$
– TNicky
Dec 17 '18 at 22:46
$begingroup$
You would need $mathbf{E}(X^2 mid mathscr{A}) = X^2$ since $X$ is $mathscr{A}$-measurable.
$endgroup$
– Will M.
Dec 18 '18 at 1:35
$begingroup$
Ok! But I don't see, why $X$ is $mathcal{A}$-measurable. We also need this for the step with the term $mathbb{E}[XY|mathcal{A}]$.
$endgroup$
– TNicky
Dec 18 '18 at 7:52
$begingroup$
If $X = mathbf{E}(Y mid mathscr{A})$ then $X$ is $mathscr{A}$-measurable. If this is not clear, you need to read again the definition of conditional expectation.
$endgroup$
– Will M.
Dec 18 '18 at 7:53
$begingroup$
AAh sure it is clear! Stupid me! Thank you very much!
$endgroup$
– TNicky
Dec 18 '18 at 7:55
|
show 2 more comments
$begingroup$
Partial solution.
Since you have a), I give a proof of b). Observe the hypothesis $mathbf{E}(Y mid mathscr{A}) = X$ implies $X$ to be measurable with respect to $mathscr{A}.$ Bearing this in mind,
$$begin{align*}
mathbf{E}((Y-X)^2) &= mathbf{E}big( mathbf{E}((Y-X)^2 mid mathscr{A}) big) \
&= mathbf{E}(mathbf{E}(Y^2 mid mathscr{A}) - 2mathbf{E}(Y mid mathscr{A}) X+ X^2) \
&= mathbf{E}(mathbf{E}(Y^2 mid mathscr{A}) - X^2) = 0.
end{align*}$$
As for c), if $X$ and $Y$ were square integrable, you have $mathbf{E}(XY) = E(X^2) = E(Y^2),$ so b) finishes the proof upon applying it with $mathscr{A} = sigma(X).$ For the general case, I am not sure how to tackle it.
answered Dec 17 '18 at 22:11
Will M.Will M.
2,865315
$endgroup$
$begingroup$
Hi and thanks for your quick respond! I have a question about the second equal-sign. Why is it $X^2$ and not $mathbb{E}[X|mathcal{A}]$? I think it doesn't make much difference in the result, due to if i write it this way I get zero as well, but maybe I missed something so my way might be faulty (?)
$endgroup$
– TNicky
Dec 17 '18 at 22:46
$begingroup$
You would need $mathbf{E}(X^2 mid mathscr{A}) = X^2$ since $X$ is $mathscr{A}$-measurable.
$endgroup$
– Will M.
Dec 18 '18 at 1:35
$begingroup$
Ok! But I don't see, why $X$ is $mathcal{A}$-measurable. We also need this for the step with the term $mathbb{E}[XY|mathcal{A}]$.
$endgroup$
– TNicky
Dec 18 '18 at 7:52
$begingroup$
If $X = mathbf{E}(Y mid mathscr{A})$ then $X$ is $mathscr{A}$-measurable. If this is not clear, you need to read again the definition of conditional expectation.
$endgroup$
– Will M.
Dec 18 '18 at 7:53
$begingroup$
AAh sure it is clear! Stupid me! Thank you very much!
$endgroup$
– TNicky
Dec 18 '18 at 7:55
|
show 2 more comments
$begingroup$
Partial solution.
Since you have a), I give a proof of b). Observe the hypothesis $mathbf{E}(Y mid mathscr{A}) = X$ implies $X$ to be measurable with respect to $mathscr{A}.$ Bearing this in mind,
$$begin{align*}
mathbf{E}((Y-X)^2) &= mathbf{E}big( mathbf{E}((Y-X)^2 mid mathscr{A}) big) \
&= mathbf{E}(mathbf{E}(Y^2 mid mathscr{A}) - 2mathbf{E}(Y mid mathscr{A}) X+ X^2) \
&= mathbf{E}(mathbf{E}(Y^2 mid mathscr{A}) - X^2) = 0.
end{align*}$$
As for c), if $X$ and $Y$ were square integrable, you have $mathbf{E}(XY) = E(X^2) = E(Y^2),$ so b) finishes the proof upon applying it with $mathscr{A} = sigma(X).$ For the general case, I am not sure how to tackle it.
answered Dec 17 '18 at 22:11
Will M.Will M.
2,865315
$endgroup$
Partial solution.
Since you have a), I give a proof of b). Observe the hypothesis $mathbf{E}(Y mid mathscr{A}) = X$ implies $X$ to be measurable with respect to $mathscr{A}.$ Bearing this in mind,
$$begin{align*}
mathbf{E}((Y-X)^2) &= mathbf{E}big( mathbf{E}((Y-X)^2 mid mathscr{A}) big) \
&= mathbf{E}(mathbf{E}(Y^2 mid mathscr{A}) - 2mathbf{E}(Y mid mathscr{A}) X+ X^2) \
&= mathbf{E}(mathbf{E}(Y^2 mid mathscr{A}) - X^2) = 0.
end{align*}$$
As for c), if $X$ and $Y$ were square integrable, you have $mathbf{E}(XY) = E(X^2) = E(Y^2),$ so b) finishes the proof upon applying it with $mathscr{A} = sigma(X).$ For the general case, I am not sure how to tackle it.
answered Dec 17 '18 at 22:11
Will M.Will M.
2,865315
answered Dec 17 '18 at 22:11
Will M.Will M.
2,865315
answered Dec 17 '18 at 22:11
Will M.Will M.
2,865315
answered Dec 17 '18 at 22:11
Will M.Will M.
2,865315
2,865315
$begingroup$
Hi and thanks for your quick respond! I have a question about the second equal-sign. Why is it $X^2$ and not $mathbb{E}[X|mathcal{A}]$? I think it doesn't make much difference in the result, due to if i write it this way I get zero as well, but maybe I missed something so my way might be faulty (?)
$endgroup$
– TNicky
Dec 17 '18 at 22:46
$begingroup$
You would need $mathbf{E}(X^2 mid mathscr{A}) = X^2$ since $X$ is $mathscr{A}$-measurable.
$endgroup$
– Will M.
Dec 18 '18 at 1:35
$begingroup$
Ok! But I don't see, why $X$ is $mathcal{A}$-measurable. We also need this for the step with the term $mathbb{E}[XY|mathcal{A}]$.
$endgroup$
– TNicky
Dec 18 '18 at 7:52
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If $X = mathbf{E}(Y mid mathscr{A})$ then $X$ is $mathscr{A}$-measurable. If this is not clear, you need to read again the definition of conditional expectation.
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– Will M.
Dec 18 '18 at 7:53
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AAh sure it is clear! Stupid me! Thank you very much!
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– TNicky
Dec 18 '18 at 7:55
|
show 2 more comments
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Hi and thanks for your quick respond! I have a question about the second equal-sign. Why is it $X^2$ and not $mathbb{E}[X|mathcal{A}]$? I think it doesn't make much difference in the result, due to if i write it this way I get zero as well, but maybe I missed something so my way might be faulty (?)
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– TNicky
Dec 17 '18 at 22:46
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You would need $mathbf{E}(X^2 mid mathscr{A}) = X^2$ since $X$ is $mathscr{A}$-measurable.
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– Will M.
Dec 18 '18 at 1:35
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Ok! But I don't see, why $X$ is $mathcal{A}$-measurable. We also need this for the step with the term $mathbb{E}[XY|mathcal{A}]$.
$endgroup$
– TNicky
Dec 18 '18 at 7:52
$begingroup$
If $X = mathbf{E}(Y mid mathscr{A})$ then $X$ is $mathscr{A}$-measurable. If this is not clear, you need to read again the definition of conditional expectation.
$endgroup$
– Will M.
Dec 18 '18 at 7:53
$begingroup$
AAh sure it is clear! Stupid me! Thank you very much!
$endgroup$
– TNicky
Dec 18 '18 at 7:55
$begingroup$
Hi and thanks for your quick respond! I have a question about the second equal-sign. Why is it $X^2$ and not $mathbb{E}[X|mathcal{A}]$? I think it doesn't make much difference in the result, due to if i write it this way I get zero as well, but maybe I missed something so my way might be faulty (?)
$endgroup$
– TNicky
Dec 17 '18 at 22:46
$begingroup$
Hi and thanks for your quick respond! I have a question about the second equal-sign. Why is it $X^2$ and not $mathbb{E}[X|mathcal{A}]$? I think it doesn't make much difference in the result, due to if i write it this way I get zero as well, but maybe I missed something so my way might be faulty (?)
$endgroup$
– TNicky
Dec 17 '18 at 22:46
$begingroup$
You would need $mathbf{E}(X^2 mid mathscr{A}) = X^2$ since $X$ is $mathscr{A}$-measurable.
$endgroup$
– Will M.
Dec 18 '18 at 1:35
$begingroup$
You would need $mathbf{E}(X^2 mid mathscr{A}) = X^2$ since $X$ is $mathscr{A}$-measurable.
$endgroup$
– Will M.
Dec 18 '18 at 1:35
$begingroup$
Ok! But I don't see, why $X$ is $mathcal{A}$-measurable. We also need this for the step with the term $mathbb{E}[XY|mathcal{A}]$.
$endgroup$
– TNicky
Dec 18 '18 at 7:52
$begingroup$
Ok! But I don't see, why $X$ is $mathcal{A}$-measurable. We also need this for the step with the term $mathbb{E}[XY|mathcal{A}]$.
$endgroup$
– TNicky
Dec 18 '18 at 7:52
$begingroup$
If $X = mathbf{E}(Y mid mathscr{A})$ then $X$ is $mathscr{A}$-measurable. If this is not clear, you need to read again the definition of conditional expectation.
$endgroup$
– Will M.
Dec 18 '18 at 7:53
$begingroup$
If $X = mathbf{E}(Y mid mathscr{A})$ then $X$ is $mathscr{A}$-measurable. If this is not clear, you need to read again the definition of conditional expectation.
$endgroup$
– Will M.
Dec 18 '18 at 7:53
$begingroup$
AAh sure it is clear! Stupid me! Thank you very much!
$endgroup$
– TNicky
Dec 18 '18 at 7:55
$begingroup$
AAh sure it is clear! Stupid me! Thank you very much!
$endgroup$
– TNicky
Dec 18 '18 at 7:55
|
show 2 more comments
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$begingroup$
Implication c) holds for integrable random variables X and Y not necessarily square integrable, as shown in answers to several questions on this site.
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– Did
Dec 17 '18 at 22:19
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@Did I believe c) it to be true in general, but I failed to provide a proof. Do you have any of this links or a proof-sketch?
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– Will M.
Dec 17 '18 at 22:37
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I thought, that my proof for c) (given above) is valid. Is there anything wrong with it? The only step I need square integrability is (at least I think so) $mathbb{E}[XY|X] = X mathbb{E}[Y|X]$. With Hölder's inequality I can conclude, that if X,Y are square-integrable then XY is integrable: $int XY dP leq int |XY| dp stackrel{Hölder}{=} left( int |X|^2 dP right)^{frac{1}{2}} left( int |Y|^2 dP right)^{frac{1}{2}} < infty$. Or am I wrong?
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– TNicky
Dec 17 '18 at 22:42
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It is true that if $X$ and $Y$ are square integrable, then $XY$ is integrable.
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– Will M.
Dec 18 '18 at 1:36
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@WillM. math.stackexchange.com/a/1520480/6179
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– Did
Dec 18 '18 at 7:20