independent variables implies integrability and same distribution
$begingroup$
I want to solve this difficult problem, so I need your help,
I want to prove that if $X$ and $Y$ are independent real variables such that $X+Y$ and $X-Y$ are independent, then $X^2$ and $Y^2$ are integrable and $X-E(X)$ and $Y-E(Y)$ are identically distributed .
I don't know exactly how to begin.
Thanks for your answers.
probability-theory measure-theory probability-distributions probability-limit-theorems
asked Jan 25 at 16:27
mathexmathex
887
$endgroup$
add a comment |
$begingroup$
I want to solve this difficult problem, so I need your help,
I want to prove that if $X$ and $Y$ are independent real variables such that $X+Y$ and $X-Y$ are independent, then $X^2$ and $Y^2$ are integrable and $X-E(X)$ and $Y-E(Y)$ are identically distributed .
I don't know exactly how to begin.
Thanks for your answers.
probability-theory measure-theory probability-distributions probability-limit-theorems
asked Jan 25 at 16:27
mathexmathex
887
$endgroup$
$begingroup$
this is not true
$endgroup$
– Joey Doey
Jan 25 at 16:52
$begingroup$
yes you're wright
$endgroup$
– mathex
Jan 25 at 16:57
$begingroup$
Just take $X$ and $Y$ to be distinct constants for a counterexample.
$endgroup$
– Mike Earnest
Jan 25 at 16:58
$begingroup$
I corrected it, I think it is true now
$endgroup$
– mathex
Jan 25 at 17:00
$begingroup$
it is a problem about a characterisation of normal distribution, the first question is as above, the second is to deduce from the central limit theorem that X and Y are normally distributed
$endgroup$
– mathex
Jan 25 at 17:02
add a comment |
$begingroup$
I want to solve this difficult problem, so I need your help,
I want to prove that if $X$ and $Y$ are independent real variables such that $X+Y$ and $X-Y$ are independent, then $X^2$ and $Y^2$ are integrable and $X-E(X)$ and $Y-E(Y)$ are identically distributed .
I don't know exactly how to begin.
Thanks for your answers.
probability-theory measure-theory probability-distributions probability-limit-theorems
asked Jan 25 at 16:27
mathexmathex
887
$endgroup$
I want to solve this difficult problem, so I need your help,
I want to prove that if $X$ and $Y$ are independent real variables such that $X+Y$ and $X-Y$ are independent, then $X^2$ and $Y^2$ are integrable and $X-E(X)$ and $Y-E(Y)$ are identically distributed .
I don't know exactly how to begin.
Thanks for your answers.
probability-theory measure-theory probability-distributions probability-limit-theorems
probability-theory measure-theory probability-distributions probability-limit-theorems
asked Jan 25 at 16:27
mathexmathex
887
asked Jan 25 at 16:27
mathexmathex
887
edited Jan 25 at 20:38
mathex
asked Jan 25 at 16:27
mathexmathex
887
asked Jan 25 at 16:27
mathexmathex
887
asked Jan 25 at 16:27
mathexmathex
887
887
$begingroup$
this is not true
$endgroup$
– Joey Doey
Jan 25 at 16:52
$begingroup$
yes you're wright
$endgroup$
– mathex
Jan 25 at 16:57
$begingroup$
Just take $X$ and $Y$ to be distinct constants for a counterexample.
$endgroup$
– Mike Earnest
Jan 25 at 16:58
$begingroup$
I corrected it, I think it is true now
$endgroup$
– mathex
Jan 25 at 17:00
$begingroup$
it is a problem about a characterisation of normal distribution, the first question is as above, the second is to deduce from the central limit theorem that X and Y are normally distributed
$endgroup$
– mathex
Jan 25 at 17:02
add a comment |
$begingroup$
this is not true
$endgroup$
– Joey Doey
Jan 25 at 16:52
$begingroup$
yes you're wright
$endgroup$
– mathex
Jan 25 at 16:57
$begingroup$
Just take $X$ and $Y$ to be distinct constants for a counterexample.
$endgroup$
– Mike Earnest
Jan 25 at 16:58
$begingroup$
I corrected it, I think it is true now
$endgroup$
– mathex
Jan 25 at 17:00
$begingroup$
it is a problem about a characterisation of normal distribution, the first question is as above, the second is to deduce from the central limit theorem that X and Y are normally distributed
$endgroup$
– mathex
Jan 25 at 17:02
$begingroup$
this is not true
$endgroup$
– Joey Doey
Jan 25 at 16:52
$begingroup$
this is not true
$endgroup$
– Joey Doey
Jan 25 at 16:52
$begingroup$
yes you're wright
$endgroup$
– mathex
Jan 25 at 16:57
$begingroup$
yes you're wright
$endgroup$
– mathex
Jan 25 at 16:57
$begingroup$
Just take $X$ and $Y$ to be distinct constants for a counterexample.
$endgroup$
– Mike Earnest
Jan 25 at 16:58
$begingroup$
Just take $X$ and $Y$ to be distinct constants for a counterexample.
$endgroup$
– Mike Earnest
Jan 25 at 16:58
$begingroup$
I corrected it, I think it is true now
$endgroup$
– mathex
Jan 25 at 17:00
$begingroup$
I corrected it, I think it is true now
$endgroup$
– mathex
Jan 25 at 17:00
$begingroup$
it is a problem about a characterisation of normal distribution, the first question is as above, the second is to deduce from the central limit theorem that X and Y are normally distributed
$endgroup$
– mathex
Jan 25 at 17:02
$begingroup$
it is a problem about a characterisation of normal distribution, the first question is as above, the second is to deduce from the central limit theorem that X and Y are normally distributed
$endgroup$
– mathex
Jan 25 at 17:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Assume that $X$ and $Y$ have zero mean. Let us consider the characteristic functions $f$ of $X$, $g$ of $Y$, $h$ of $X+Y$, $j$ of $X-Y$.
Now, $h(s)j(t)$ is the average of $e^{i(s+t)X+i(s-t)Y}$, thus $h(s)j(t)=f(s+t)g(s-t)$. In particular, $h(s)=f(s)g(s)$, $j(t)=f(t)g(-t)$.
Thus $f(s)f(t)g(s)g(-t)=f(s+t)g(s-t)$. (E)
Now write $s=u+v$, $t=u-v$, then $f(u+v)f(u+(-v))g(u-(-v))g(v-u)=f(2u)g(2v)$.
Applying (E) to the left-hand side yields:
$f(2u)g(2v)=f(u)f(v)g(v)g(-u)f(u)f(-v)g(u)g(v)=f(u)^2g(v)^2|f(v)|^2|g(u)|^2$.
Thus $f(2u)=c_1f(u)^2|g(u)|^2$, $g(2v)=c_2|f(v)|^2g(v)^2$, for some constants $c_1,c_2$. Hence $c_1=c_2=1$ (with $u=v=0$).
In particular, $|f|=|g|$, and the set of zeroes of $fg$ is stable under division by $2$ and closed. Since it does not contain $0$, $|f|$ does not vanish.
In particular, there exists a unique differentiable function $theta$ such that $f=ge^{itheta}$, and $theta(0)=0$.
The equations and uniqueness yield that $theta(2t)=2theta(t)$, which entails $theta(t)=theta’(0)t$. Now since $X$ and $Y$ have zero mean, $theta’(0)=0$ hence $f=g$.
Edit: I realize I did not address the last part of the question. I am adding it now.
$X+Y$ and $X-Y$ are independent and integrable thus so is their product $X^2-Y^2$, thus, for some large $R>0$, so is $|X^2-Y^2|1(|Y| leq R) geq X^21(|Y| leq R) -R^2$. Therefore, $X^21(|Y| leq R)$ is $L^1$, hence, if $|Y| leq R$ has positive probability, $X^2$ is integrable.
answered Jan 25 at 17:17
MindlackMindlack
4,945211
$endgroup$
$begingroup$
but why $X$ and $Y$ are integrable??
$endgroup$
– mathex
Jan 25 at 18:48
$begingroup$
in fact $X^k$ is integrable for all nonnegative integer k
$endgroup$
– mathex
Jan 25 at 18:49
$begingroup$
I found on the internet this document:math.uni.wroc.pl/~pms/files/14.2/Article/14.2.8.pdf
$endgroup$
– mathex
Jan 25 at 18:50
$begingroup$
but the proof of lemma 1 is not very logical!!
$endgroup$
– mathex
Jan 25 at 18:51
$begingroup$
I know that if $X$ and $Y$ are independent, then $X+Y$ is integrable if and only if $X$ and $Y$ are (very easy to prove by Fubini...), by I didn't comprehend how $X+Y$ is integrable??
$endgroup$
– mathex
Jan 25 at 19:02
|
show 2 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assume that $X$ and $Y$ have zero mean. Let us consider the characteristic functions $f$ of $X$, $g$ of $Y$, $h$ of $X+Y$, $j$ of $X-Y$.
Now, $h(s)j(t)$ is the average of $e^{i(s+t)X+i(s-t)Y}$, thus $h(s)j(t)=f(s+t)g(s-t)$. In particular, $h(s)=f(s)g(s)$, $j(t)=f(t)g(-t)$.
Thus $f(s)f(t)g(s)g(-t)=f(s+t)g(s-t)$. (E)
Now write $s=u+v$, $t=u-v$, then $f(u+v)f(u+(-v))g(u-(-v))g(v-u)=f(2u)g(2v)$.
Applying (E) to the left-hand side yields:
$f(2u)g(2v)=f(u)f(v)g(v)g(-u)f(u)f(-v)g(u)g(v)=f(u)^2g(v)^2|f(v)|^2|g(u)|^2$.
Thus $f(2u)=c_1f(u)^2|g(u)|^2$, $g(2v)=c_2|f(v)|^2g(v)^2$, for some constants $c_1,c_2$. Hence $c_1=c_2=1$ (with $u=v=0$).
In particular, $|f|=|g|$, and the set of zeroes of $fg$ is stable under division by $2$ and closed. Since it does not contain $0$, $|f|$ does not vanish.
In particular, there exists a unique differentiable function $theta$ such that $f=ge^{itheta}$, and $theta(0)=0$.
The equations and uniqueness yield that $theta(2t)=2theta(t)$, which entails $theta(t)=theta’(0)t$. Now since $X$ and $Y$ have zero mean, $theta’(0)=0$ hence $f=g$.
Edit: I realize I did not address the last part of the question. I am adding it now.
$X+Y$ and $X-Y$ are independent and integrable thus so is their product $X^2-Y^2$, thus, for some large $R>0$, so is $|X^2-Y^2|1(|Y| leq R) geq X^21(|Y| leq R) -R^2$. Therefore, $X^21(|Y| leq R)$ is $L^1$, hence, if $|Y| leq R$ has positive probability, $X^2$ is integrable.
answered Jan 25 at 17:17
MindlackMindlack
4,945211
$endgroup$
$begingroup$
but why $X$ and $Y$ are integrable??
$endgroup$
– mathex
Jan 25 at 18:48
$begingroup$
in fact $X^k$ is integrable for all nonnegative integer k
$endgroup$
– mathex
Jan 25 at 18:49
$begingroup$
I found on the internet this document:math.uni.wroc.pl/~pms/files/14.2/Article/14.2.8.pdf
$endgroup$
– mathex
Jan 25 at 18:50
$begingroup$
but the proof of lemma 1 is not very logical!!
$endgroup$
– mathex
Jan 25 at 18:51
$begingroup$
I know that if $X$ and $Y$ are independent, then $X+Y$ is integrable if and only if $X$ and $Y$ are (very easy to prove by Fubini...), by I didn't comprehend how $X+Y$ is integrable??
$endgroup$
– mathex
Jan 25 at 19:02
|
show 2 more comments
$begingroup$
Assume that $X$ and $Y$ have zero mean. Let us consider the characteristic functions $f$ of $X$, $g$ of $Y$, $h$ of $X+Y$, $j$ of $X-Y$.
Now, $h(s)j(t)$ is the average of $e^{i(s+t)X+i(s-t)Y}$, thus $h(s)j(t)=f(s+t)g(s-t)$. In particular, $h(s)=f(s)g(s)$, $j(t)=f(t)g(-t)$.
Thus $f(s)f(t)g(s)g(-t)=f(s+t)g(s-t)$. (E)
Now write $s=u+v$, $t=u-v$, then $f(u+v)f(u+(-v))g(u-(-v))g(v-u)=f(2u)g(2v)$.
Applying (E) to the left-hand side yields:
$f(2u)g(2v)=f(u)f(v)g(v)g(-u)f(u)f(-v)g(u)g(v)=f(u)^2g(v)^2|f(v)|^2|g(u)|^2$.
Thus $f(2u)=c_1f(u)^2|g(u)|^2$, $g(2v)=c_2|f(v)|^2g(v)^2$, for some constants $c_1,c_2$. Hence $c_1=c_2=1$ (with $u=v=0$).
In particular, $|f|=|g|$, and the set of zeroes of $fg$ is stable under division by $2$ and closed. Since it does not contain $0$, $|f|$ does not vanish.
In particular, there exists a unique differentiable function $theta$ such that $f=ge^{itheta}$, and $theta(0)=0$.
The equations and uniqueness yield that $theta(2t)=2theta(t)$, which entails $theta(t)=theta’(0)t$. Now since $X$ and $Y$ have zero mean, $theta’(0)=0$ hence $f=g$.
Edit: I realize I did not address the last part of the question. I am adding it now.
$X+Y$ and $X-Y$ are independent and integrable thus so is their product $X^2-Y^2$, thus, for some large $R>0$, so is $|X^2-Y^2|1(|Y| leq R) geq X^21(|Y| leq R) -R^2$. Therefore, $X^21(|Y| leq R)$ is $L^1$, hence, if $|Y| leq R$ has positive probability, $X^2$ is integrable.
answered Jan 25 at 17:17
MindlackMindlack
4,945211
$endgroup$
$begingroup$
but why $X$ and $Y$ are integrable??
$endgroup$
– mathex
Jan 25 at 18:48
$begingroup$
in fact $X^k$ is integrable for all nonnegative integer k
$endgroup$
– mathex
Jan 25 at 18:49
$begingroup$
I found on the internet this document:math.uni.wroc.pl/~pms/files/14.2/Article/14.2.8.pdf
$endgroup$
– mathex
Jan 25 at 18:50
$begingroup$
but the proof of lemma 1 is not very logical!!
$endgroup$
– mathex
Jan 25 at 18:51
$begingroup$
I know that if $X$ and $Y$ are independent, then $X+Y$ is integrable if and only if $X$ and $Y$ are (very easy to prove by Fubini...), by I didn't comprehend how $X+Y$ is integrable??
$endgroup$
– mathex
Jan 25 at 19:02
|
show 2 more comments
$begingroup$
Assume that $X$ and $Y$ have zero mean. Let us consider the characteristic functions $f$ of $X$, $g$ of $Y$, $h$ of $X+Y$, $j$ of $X-Y$.
Now, $h(s)j(t)$ is the average of $e^{i(s+t)X+i(s-t)Y}$, thus $h(s)j(t)=f(s+t)g(s-t)$. In particular, $h(s)=f(s)g(s)$, $j(t)=f(t)g(-t)$.
Thus $f(s)f(t)g(s)g(-t)=f(s+t)g(s-t)$. (E)
Now write $s=u+v$, $t=u-v$, then $f(u+v)f(u+(-v))g(u-(-v))g(v-u)=f(2u)g(2v)$.
Applying (E) to the left-hand side yields:
$f(2u)g(2v)=f(u)f(v)g(v)g(-u)f(u)f(-v)g(u)g(v)=f(u)^2g(v)^2|f(v)|^2|g(u)|^2$.
Thus $f(2u)=c_1f(u)^2|g(u)|^2$, $g(2v)=c_2|f(v)|^2g(v)^2$, for some constants $c_1,c_2$. Hence $c_1=c_2=1$ (with $u=v=0$).
In particular, $|f|=|g|$, and the set of zeroes of $fg$ is stable under division by $2$ and closed. Since it does not contain $0$, $|f|$ does not vanish.
In particular, there exists a unique differentiable function $theta$ such that $f=ge^{itheta}$, and $theta(0)=0$.
The equations and uniqueness yield that $theta(2t)=2theta(t)$, which entails $theta(t)=theta’(0)t$. Now since $X$ and $Y$ have zero mean, $theta’(0)=0$ hence $f=g$.
Edit: I realize I did not address the last part of the question. I am adding it now.
$X+Y$ and $X-Y$ are independent and integrable thus so is their product $X^2-Y^2$, thus, for some large $R>0$, so is $|X^2-Y^2|1(|Y| leq R) geq X^21(|Y| leq R) -R^2$. Therefore, $X^21(|Y| leq R)$ is $L^1$, hence, if $|Y| leq R$ has positive probability, $X^2$ is integrable.
answered Jan 25 at 17:17
MindlackMindlack
4,945211
$endgroup$
Assume that $X$ and $Y$ have zero mean. Let us consider the characteristic functions $f$ of $X$, $g$ of $Y$, $h$ of $X+Y$, $j$ of $X-Y$.
Now, $h(s)j(t)$ is the average of $e^{i(s+t)X+i(s-t)Y}$, thus $h(s)j(t)=f(s+t)g(s-t)$. In particular, $h(s)=f(s)g(s)$, $j(t)=f(t)g(-t)$.
Thus $f(s)f(t)g(s)g(-t)=f(s+t)g(s-t)$. (E)
Now write $s=u+v$, $t=u-v$, then $f(u+v)f(u+(-v))g(u-(-v))g(v-u)=f(2u)g(2v)$.
Applying (E) to the left-hand side yields:
$f(2u)g(2v)=f(u)f(v)g(v)g(-u)f(u)f(-v)g(u)g(v)=f(u)^2g(v)^2|f(v)|^2|g(u)|^2$.
Thus $f(2u)=c_1f(u)^2|g(u)|^2$, $g(2v)=c_2|f(v)|^2g(v)^2$, for some constants $c_1,c_2$. Hence $c_1=c_2=1$ (with $u=v=0$).
In particular, $|f|=|g|$, and the set of zeroes of $fg$ is stable under division by $2$ and closed. Since it does not contain $0$, $|f|$ does not vanish.
In particular, there exists a unique differentiable function $theta$ such that $f=ge^{itheta}$, and $theta(0)=0$.
The equations and uniqueness yield that $theta(2t)=2theta(t)$, which entails $theta(t)=theta’(0)t$. Now since $X$ and $Y$ have zero mean, $theta’(0)=0$ hence $f=g$.
Edit: I realize I did not address the last part of the question. I am adding it now.
$X+Y$ and $X-Y$ are independent and integrable thus so is their product $X^2-Y^2$, thus, for some large $R>0$, so is $|X^2-Y^2|1(|Y| leq R) geq X^21(|Y| leq R) -R^2$. Therefore, $X^21(|Y| leq R)$ is $L^1$, hence, if $|Y| leq R$ has positive probability, $X^2$ is integrable.
answered Jan 25 at 17:17
MindlackMindlack
4,945211
edited Jan 25 at 18:55
answered Jan 25 at 17:17
MindlackMindlack
4,945211
answered Jan 25 at 17:17
MindlackMindlack
4,945211
answered Jan 25 at 17:17
MindlackMindlack
4,945211
4,945211
$begingroup$
but why $X$ and $Y$ are integrable??
$endgroup$
– mathex
Jan 25 at 18:48
$begingroup$
in fact $X^k$ is integrable for all nonnegative integer k
$endgroup$
– mathex
Jan 25 at 18:49
$begingroup$
I found on the internet this document:math.uni.wroc.pl/~pms/files/14.2/Article/14.2.8.pdf
$endgroup$
– mathex
Jan 25 at 18:50
$begingroup$
but the proof of lemma 1 is not very logical!!
$endgroup$
– mathex
Jan 25 at 18:51
$begingroup$
I know that if $X$ and $Y$ are independent, then $X+Y$ is integrable if and only if $X$ and $Y$ are (very easy to prove by Fubini...), by I didn't comprehend how $X+Y$ is integrable??
$endgroup$
– mathex
Jan 25 at 19:02
|
show 2 more comments
$begingroup$
but why $X$ and $Y$ are integrable??
$endgroup$
– mathex
Jan 25 at 18:48
$begingroup$
in fact $X^k$ is integrable for all nonnegative integer k
$endgroup$
– mathex
Jan 25 at 18:49
$begingroup$
I found on the internet this document:math.uni.wroc.pl/~pms/files/14.2/Article/14.2.8.pdf
$endgroup$
– mathex
Jan 25 at 18:50
$begingroup$
but the proof of lemma 1 is not very logical!!
$endgroup$
– mathex
Jan 25 at 18:51
$begingroup$
I know that if $X$ and $Y$ are independent, then $X+Y$ is integrable if and only if $X$ and $Y$ are (very easy to prove by Fubini...), by I didn't comprehend how $X+Y$ is integrable??
$endgroup$
– mathex
Jan 25 at 19:02
$begingroup$
but why $X$ and $Y$ are integrable??
$endgroup$
– mathex
Jan 25 at 18:48
$begingroup$
but why $X$ and $Y$ are integrable??
$endgroup$
– mathex
Jan 25 at 18:48
$begingroup$
in fact $X^k$ is integrable for all nonnegative integer k
$endgroup$
– mathex
Jan 25 at 18:49
$begingroup$
in fact $X^k$ is integrable for all nonnegative integer k
$endgroup$
– mathex
Jan 25 at 18:49
$begingroup$
I found on the internet this document:math.uni.wroc.pl/~pms/files/14.2/Article/14.2.8.pdf
$endgroup$
– mathex
Jan 25 at 18:50
$begingroup$
I found on the internet this document:math.uni.wroc.pl/~pms/files/14.2/Article/14.2.8.pdf
$endgroup$
– mathex
Jan 25 at 18:50
$begingroup$
but the proof of lemma 1 is not very logical!!
$endgroup$
– mathex
Jan 25 at 18:51
$begingroup$
but the proof of lemma 1 is not very logical!!
$endgroup$
– mathex
Jan 25 at 18:51
$begingroup$
I know that if $X$ and $Y$ are independent, then $X+Y$ is integrable if and only if $X$ and $Y$ are (very easy to prove by Fubini...), by I didn't comprehend how $X+Y$ is integrable??
$endgroup$
– mathex
Jan 25 at 19:02
$begingroup$
I know that if $X$ and $Y$ are independent, then $X+Y$ is integrable if and only if $X$ and $Y$ are (very easy to prove by Fubini...), by I didn't comprehend how $X+Y$ is integrable??
$endgroup$
– mathex
Jan 25 at 19:02
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show 2 more comments
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$begingroup$
this is not true
$endgroup$
– Joey Doey
Jan 25 at 16:52
$begingroup$
yes you're wright
$endgroup$
– mathex
Jan 25 at 16:57
$begingroup$
Just take $X$ and $Y$ to be distinct constants for a counterexample.
$endgroup$
– Mike Earnest
Jan 25 at 16:58
$begingroup$
I corrected it, I think it is true now
$endgroup$
– mathex
Jan 25 at 17:00
$begingroup$
it is a problem about a characterisation of normal distribution, the first question is as above, the second is to deduce from the central limit theorem that X and Y are normally distributed
$endgroup$
– mathex
Jan 25 at 17:02