independent variables implies integrability and same distribution












-1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















-1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















-1












-1








-1


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 25 at 20:38







mathex

















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




















  • $begingroup$
    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












1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 25 at 18:55

























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


















  • $begingroup$
    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
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– mathex
Jan 25 at 18:49














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I found on the internet this document:math.uni.wroc.pl/~pms/files/14.2/Article/14.2.8.pdf
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– 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














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but the proof of lemma 1 is not very logical!!
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– 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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