If $mathbb{E} e^{it(X+Y)} = mathbb{E} e^{itX} mathbb{E} e^{itY}$ then $X$ and $Y$ are uncorrelated












2












$begingroup$


Let $X$ and $Y$ be random variables such that
$mathbb{E} e^{it(X+Y)} = mathbb{E} e^{itX} mathbb{E} e^{itY}$ and covariance exists.



I want to show that they are uncorrelated, i.e., $mathbb{E}(X -mathbb{E}X) (Y - mathbb{E}Y) = 0$.
According to this paper https://www.tandfonline.com/doi/pdf/10.1198/tast.2009.09051?needAccess=true, this is true.
I can prove it when both have finite second moments, but I do not know how I should proceed for general general random variables. Any hints or reference are appreciated.



Differentiating $varphi_{X+Y}(t) = varphi_X(t) varphi_Y(t) $ twice and letting $t=0$, begin{equation} varphi_X ''(0) varphi_Y(0) + 2 varphi_X'(0) varphi_Y'(0) + varphi_X(0) varphi_Y''(0) = varphi_{X+Y}''(0) end{equation}
Since begin{equation} varphi_X^{(k)}(t) = mathbb{E} (( iX)^k e^{itX})end{equation} holds for $X in L^k$,
begin{equation} mathbb{E} ( iX)^2 + 2 mathbb{E} ( iX) mathbb{E} ( iY)+mathbb{E} ( iY)^2 = mathbb{E} (i^2(X+Y)^2) end{equation}






probability probability-theory characteristic-functions







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$endgroup$












  • $begingroup$
    What about $X = Y sim operatorname{Cauchy}(0,1)$?
    $endgroup$
    – Rhys Steele
    Dec 2 '18 at 19:38












  • $begingroup$
    @RhysSteele One instance of subindependence is when a random variable X is Cauchy with mean 0 and scale s and another random variable Y=X. Then X+Y is also Cauchy but with scale 2s. The characteristic function of either X or Y in t is then exp(-s·|t|), and the characteristic function of X+Y is exp(-2s·|t|). wikiwand.com/en/Subindependence
    $endgroup$
    – GouldBach
    Dec 3 '18 at 1:49










  • $begingroup$
    Yes, that was my point. The Cauchy random variable does not have finite second moment and in particular for $X=Y sim operatorname{Cauchy}(0,1)$ we do not have $mathbb{E}[(X - mathbb{E}X)^2] = 0$ (i.e. that $X$ and $Y$ are uncorrelated) but we do have subindependence. Is this not a counterexample to the general claim?
    $endgroup$
    – Rhys Steele
    Dec 3 '18 at 15:19










  • $begingroup$
    @RhysSteele Yes. Thank you. I will edit my claim.
    $endgroup$
    – GouldBach
    Dec 4 '18 at 0:14
















2












$begingroup$


Let $X$ and $Y$ be random variables such that
$mathbb{E} e^{it(X+Y)} = mathbb{E} e^{itX} mathbb{E} e^{itY}$ and covariance exists.



I want to show that they are uncorrelated, i.e., $mathbb{E}(X -mathbb{E}X) (Y - mathbb{E}Y) = 0$.
According to this paper https://www.tandfonline.com/doi/pdf/10.1198/tast.2009.09051?needAccess=true, this is true.
I can prove it when both have finite second moments, but I do not know how I should proceed for general general random variables. Any hints or reference are appreciated.



Differentiating $varphi_{X+Y}(t) = varphi_X(t) varphi_Y(t) $ twice and letting $t=0$, begin{equation} varphi_X ''(0) varphi_Y(0) + 2 varphi_X'(0) varphi_Y'(0) + varphi_X(0) varphi_Y''(0) = varphi_{X+Y}''(0) end{equation}
Since begin{equation} varphi_X^{(k)}(t) = mathbb{E} (( iX)^k e^{itX})end{equation} holds for $X in L^k$,
begin{equation} mathbb{E} ( iX)^2 + 2 mathbb{E} ( iX) mathbb{E} ( iY)+mathbb{E} ( iY)^2 = mathbb{E} (i^2(X+Y)^2) end{equation}






probability probability-theory characteristic-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    What about $X = Y sim operatorname{Cauchy}(0,1)$?
    $endgroup$
    – Rhys Steele
    Dec 2 '18 at 19:38












  • $begingroup$
    @RhysSteele One instance of subindependence is when a random variable X is Cauchy with mean 0 and scale s and another random variable Y=X. Then X+Y is also Cauchy but with scale 2s. The characteristic function of either X or Y in t is then exp(-s·|t|), and the characteristic function of X+Y is exp(-2s·|t|). wikiwand.com/en/Subindependence
    $endgroup$
    – GouldBach
    Dec 3 '18 at 1:49










  • $begingroup$
    Yes, that was my point. The Cauchy random variable does not have finite second moment and in particular for $X=Y sim operatorname{Cauchy}(0,1)$ we do not have $mathbb{E}[(X - mathbb{E}X)^2] = 0$ (i.e. that $X$ and $Y$ are uncorrelated) but we do have subindependence. Is this not a counterexample to the general claim?
    $endgroup$
    – Rhys Steele
    Dec 3 '18 at 15:19










  • $begingroup$
    @RhysSteele Yes. Thank you. I will edit my claim.
    $endgroup$
    – GouldBach
    Dec 4 '18 at 0:14














2












2








2





$begingroup$


Let $X$ and $Y$ be random variables such that
$mathbb{E} e^{it(X+Y)} = mathbb{E} e^{itX} mathbb{E} e^{itY}$ and covariance exists.



I want to show that they are uncorrelated, i.e., $mathbb{E}(X -mathbb{E}X) (Y - mathbb{E}Y) = 0$.
According to this paper https://www.tandfonline.com/doi/pdf/10.1198/tast.2009.09051?needAccess=true, this is true.
I can prove it when both have finite second moments, but I do not know how I should proceed for general general random variables. Any hints or reference are appreciated.



Differentiating $varphi_{X+Y}(t) = varphi_X(t) varphi_Y(t) $ twice and letting $t=0$, begin{equation} varphi_X ''(0) varphi_Y(0) + 2 varphi_X'(0) varphi_Y'(0) + varphi_X(0) varphi_Y''(0) = varphi_{X+Y}''(0) end{equation}
Since begin{equation} varphi_X^{(k)}(t) = mathbb{E} (( iX)^k e^{itX})end{equation} holds for $X in L^k$,
begin{equation} mathbb{E} ( iX)^2 + 2 mathbb{E} ( iX) mathbb{E} ( iY)+mathbb{E} ( iY)^2 = mathbb{E} (i^2(X+Y)^2) end{equation}






probability probability-theory characteristic-functions







share|cite|improve this question











$endgroup$




Let $X$ and $Y$ be random variables such that
$mathbb{E} e^{it(X+Y)} = mathbb{E} e^{itX} mathbb{E} e^{itY}$ and covariance exists.



I want to show that they are uncorrelated, i.e., $mathbb{E}(X -mathbb{E}X) (Y - mathbb{E}Y) = 0$.
According to this paper https://www.tandfonline.com/doi/pdf/10.1198/tast.2009.09051?needAccess=true, this is true.
I can prove it when both have finite second moments, but I do not know how I should proceed for general general random variables. Any hints or reference are appreciated.



Differentiating $varphi_{X+Y}(t) = varphi_X(t) varphi_Y(t) $ twice and letting $t=0$, begin{equation} varphi_X ''(0) varphi_Y(0) + 2 varphi_X'(0) varphi_Y'(0) + varphi_X(0) varphi_Y''(0) = varphi_{X+Y}''(0) end{equation}
Since begin{equation} varphi_X^{(k)}(t) = mathbb{E} (( iX)^k e^{itX})end{equation} holds for $X in L^k$,
begin{equation} mathbb{E} ( iX)^2 + 2 mathbb{E} ( iX) mathbb{E} ( iY)+mathbb{E} ( iY)^2 = mathbb{E} (i^2(X+Y)^2) end{equation}






probability probability-theory characteristic-functions




probability probability-theory characteristic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 0:14







GouldBach

















asked Dec 2 '18 at 18:08









GouldBachGouldBach

43918




43918












  • $begingroup$
    What about $X = Y sim operatorname{Cauchy}(0,1)$?
    $endgroup$
    – Rhys Steele
    Dec 2 '18 at 19:38












  • $begingroup$
    @RhysSteele One instance of subindependence is when a random variable X is Cauchy with mean 0 and scale s and another random variable Y=X. Then X+Y is also Cauchy but with scale 2s. The characteristic function of either X or Y in t is then exp(-s·|t|), and the characteristic function of X+Y is exp(-2s·|t|). wikiwand.com/en/Subindependence
    $endgroup$
    – GouldBach
    Dec 3 '18 at 1:49










  • $begingroup$
    Yes, that was my point. The Cauchy random variable does not have finite second moment and in particular for $X=Y sim operatorname{Cauchy}(0,1)$ we do not have $mathbb{E}[(X - mathbb{E}X)^2] = 0$ (i.e. that $X$ and $Y$ are uncorrelated) but we do have subindependence. Is this not a counterexample to the general claim?
    $endgroup$
    – Rhys Steele
    Dec 3 '18 at 15:19










  • $begingroup$
    @RhysSteele Yes. Thank you. I will edit my claim.
    $endgroup$
    – GouldBach
    Dec 4 '18 at 0:14


















  • $begingroup$
    What about $X = Y sim operatorname{Cauchy}(0,1)$?
    $endgroup$
    – Rhys Steele
    Dec 2 '18 at 19:38












  • $begingroup$
    @RhysSteele One instance of subindependence is when a random variable X is Cauchy with mean 0 and scale s and another random variable Y=X. Then X+Y is also Cauchy but with scale 2s. The characteristic function of either X or Y in t is then exp(-s·|t|), and the characteristic function of X+Y is exp(-2s·|t|). wikiwand.com/en/Subindependence
    $endgroup$
    – GouldBach
    Dec 3 '18 at 1:49










  • $begingroup$
    Yes, that was my point. The Cauchy random variable does not have finite second moment and in particular for $X=Y sim operatorname{Cauchy}(0,1)$ we do not have $mathbb{E}[(X - mathbb{E}X)^2] = 0$ (i.e. that $X$ and $Y$ are uncorrelated) but we do have subindependence. Is this not a counterexample to the general claim?
    $endgroup$
    – Rhys Steele
    Dec 3 '18 at 15:19










  • $begingroup$
    @RhysSteele Yes. Thank you. I will edit my claim.
    $endgroup$
    – GouldBach
    Dec 4 '18 at 0:14
















$begingroup$
What about $X = Y sim operatorname{Cauchy}(0,1)$?
$endgroup$
– Rhys Steele
Dec 2 '18 at 19:38






$begingroup$
What about $X = Y sim operatorname{Cauchy}(0,1)$?
$endgroup$
– Rhys Steele
Dec 2 '18 at 19:38














$begingroup$
@RhysSteele One instance of subindependence is when a random variable X is Cauchy with mean 0 and scale s and another random variable Y=X. Then X+Y is also Cauchy but with scale 2s. The characteristic function of either X or Y in t is then exp(-s·|t|), and the characteristic function of X+Y is exp(-2s·|t|). wikiwand.com/en/Subindependence
$endgroup$
– GouldBach
Dec 3 '18 at 1:49




$begingroup$
@RhysSteele One instance of subindependence is when a random variable X is Cauchy with mean 0 and scale s and another random variable Y=X. Then X+Y is also Cauchy but with scale 2s. The characteristic function of either X or Y in t is then exp(-s·|t|), and the characteristic function of X+Y is exp(-2s·|t|). wikiwand.com/en/Subindependence
$endgroup$
– GouldBach
Dec 3 '18 at 1:49












$begingroup$
Yes, that was my point. The Cauchy random variable does not have finite second moment and in particular for $X=Y sim operatorname{Cauchy}(0,1)$ we do not have $mathbb{E}[(X - mathbb{E}X)^2] = 0$ (i.e. that $X$ and $Y$ are uncorrelated) but we do have subindependence. Is this not a counterexample to the general claim?
$endgroup$
– Rhys Steele
Dec 3 '18 at 15:19




$begingroup$
Yes, that was my point. The Cauchy random variable does not have finite second moment and in particular for $X=Y sim operatorname{Cauchy}(0,1)$ we do not have $mathbb{E}[(X - mathbb{E}X)^2] = 0$ (i.e. that $X$ and $Y$ are uncorrelated) but we do have subindependence. Is this not a counterexample to the general claim?
$endgroup$
– Rhys Steele
Dec 3 '18 at 15:19












$begingroup$
@RhysSteele Yes. Thank you. I will edit my claim.
$endgroup$
– GouldBach
Dec 4 '18 at 0:14




$begingroup$
@RhysSteele Yes. Thank you. I will edit my claim.
$endgroup$
– GouldBach
Dec 4 '18 at 0:14










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