Why is this sequence of random variables pairwise independent?












2












$begingroup$


I have a sequence $(X_n: Omega to mathbb{R})_{n=1}^infty$ of pairwise independent random variables.



Define for $n geq 1: X_n' := X_n I_{{X_n leq n}}$ where $I_A$ is the indicator function on $A$. Is it true that $(X_n')_{n=1}^infty$ is a sequence of pairwise independent random variables?



Intuitively, this seems true. I know that pairwise independence is preserved under a Borel transformation $g: mathbb{R} to mathbb{R}$ so I tried to write



$$X_n' = g circ X_n$$



for some suitable $g$ but did not come up with anything useful. Any hints?






probability-theory measure-theory independence







share|cite|improve this question











$endgroup$












  • $begingroup$
    It should follow from the definition of independent random variables. About your third paragraph, are you sure that this result holds? I think $g$ needs to preserve measure or something like that.
    $endgroup$
    – Yanko
    Dec 15 '18 at 15:13










  • $begingroup$
    $g$ must be measurable, which is what I meant with "Borel".
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:14










  • $begingroup$
    If $g$ is a constant you sure the claim holds?
    $endgroup$
    – Yanko
    Dec 15 '18 at 15:16










  • $begingroup$
    I'm thinking about it. Give me a second please. Thanks for your useful comment.
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:16












  • $begingroup$
    Fix Borel sets $A,B$. We have to prove that $P(gcirc X in A, g circ Y in B) = P(g circ X in A) P(g circ Y in B)$ where $g=1$ is the constant 1 function. If $1 in A cap B$, then we have $1 = 1$. Otherwise, we have $0=0$, so I don't think there is a contradiction. What do you think?
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:23


















2












$begingroup$


I have a sequence $(X_n: Omega to mathbb{R})_{n=1}^infty$ of pairwise independent random variables.



Define for $n geq 1: X_n' := X_n I_{{X_n leq n}}$ where $I_A$ is the indicator function on $A$. Is it true that $(X_n')_{n=1}^infty$ is a sequence of pairwise independent random variables?



Intuitively, this seems true. I know that pairwise independence is preserved under a Borel transformation $g: mathbb{R} to mathbb{R}$ so I tried to write



$$X_n' = g circ X_n$$



for some suitable $g$ but did not come up with anything useful. Any hints?






probability-theory measure-theory independence







share|cite|improve this question











$endgroup$












  • $begingroup$
    It should follow from the definition of independent random variables. About your third paragraph, are you sure that this result holds? I think $g$ needs to preserve measure or something like that.
    $endgroup$
    – Yanko
    Dec 15 '18 at 15:13










  • $begingroup$
    $g$ must be measurable, which is what I meant with "Borel".
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:14










  • $begingroup$
    If $g$ is a constant you sure the claim holds?
    $endgroup$
    – Yanko
    Dec 15 '18 at 15:16










  • $begingroup$
    I'm thinking about it. Give me a second please. Thanks for your useful comment.
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:16












  • $begingroup$
    Fix Borel sets $A,B$. We have to prove that $P(gcirc X in A, g circ Y in B) = P(g circ X in A) P(g circ Y in B)$ where $g=1$ is the constant 1 function. If $1 in A cap B$, then we have $1 = 1$. Otherwise, we have $0=0$, so I don't think there is a contradiction. What do you think?
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:23
















2












2








2





$begingroup$


I have a sequence $(X_n: Omega to mathbb{R})_{n=1}^infty$ of pairwise independent random variables.



Define for $n geq 1: X_n' := X_n I_{{X_n leq n}}$ where $I_A$ is the indicator function on $A$. Is it true that $(X_n')_{n=1}^infty$ is a sequence of pairwise independent random variables?



Intuitively, this seems true. I know that pairwise independence is preserved under a Borel transformation $g: mathbb{R} to mathbb{R}$ so I tried to write



$$X_n' = g circ X_n$$



for some suitable $g$ but did not come up with anything useful. Any hints?






probability-theory measure-theory independence







share|cite|improve this question











$endgroup$




I have a sequence $(X_n: Omega to mathbb{R})_{n=1}^infty$ of pairwise independent random variables.



Define for $n geq 1: X_n' := X_n I_{{X_n leq n}}$ where $I_A$ is the indicator function on $A$. Is it true that $(X_n')_{n=1}^infty$ is a sequence of pairwise independent random variables?



Intuitively, this seems true. I know that pairwise independence is preserved under a Borel transformation $g: mathbb{R} to mathbb{R}$ so I tried to write



$$X_n' = g circ X_n$$



for some suitable $g$ but did not come up with anything useful. Any hints?






probability-theory measure-theory independence




probability-theory measure-theory independence






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 15:13







Math_QED

















asked Dec 15 '18 at 15:08









Math_QEDMath_QED

7,58331452




7,58331452












  • $begingroup$
    It should follow from the definition of independent random variables. About your third paragraph, are you sure that this result holds? I think $g$ needs to preserve measure or something like that.
    $endgroup$
    – Yanko
    Dec 15 '18 at 15:13










  • $begingroup$
    $g$ must be measurable, which is what I meant with "Borel".
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:14










  • $begingroup$
    If $g$ is a constant you sure the claim holds?
    $endgroup$
    – Yanko
    Dec 15 '18 at 15:16










  • $begingroup$
    I'm thinking about it. Give me a second please. Thanks for your useful comment.
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:16












  • $begingroup$
    Fix Borel sets $A,B$. We have to prove that $P(gcirc X in A, g circ Y in B) = P(g circ X in A) P(g circ Y in B)$ where $g=1$ is the constant 1 function. If $1 in A cap B$, then we have $1 = 1$. Otherwise, we have $0=0$, so I don't think there is a contradiction. What do you think?
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:23




















  • $begingroup$
    It should follow from the definition of independent random variables. About your third paragraph, are you sure that this result holds? I think $g$ needs to preserve measure or something like that.
    $endgroup$
    – Yanko
    Dec 15 '18 at 15:13










  • $begingroup$
    $g$ must be measurable, which is what I meant with "Borel".
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:14










  • $begingroup$
    If $g$ is a constant you sure the claim holds?
    $endgroup$
    – Yanko
    Dec 15 '18 at 15:16










  • $begingroup$
    I'm thinking about it. Give me a second please. Thanks for your useful comment.
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:16












  • $begingroup$
    Fix Borel sets $A,B$. We have to prove that $P(gcirc X in A, g circ Y in B) = P(g circ X in A) P(g circ Y in B)$ where $g=1$ is the constant 1 function. If $1 in A cap B$, then we have $1 = 1$. Otherwise, we have $0=0$, so I don't think there is a contradiction. What do you think?
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:23


















$begingroup$
It should follow from the definition of independent random variables. About your third paragraph, are you sure that this result holds? I think $g$ needs to preserve measure or something like that.
$endgroup$
– Yanko
Dec 15 '18 at 15:13




$begingroup$
It should follow from the definition of independent random variables. About your third paragraph, are you sure that this result holds? I think $g$ needs to preserve measure or something like that.
$endgroup$
– Yanko
Dec 15 '18 at 15:13












$begingroup$
$g$ must be measurable, which is what I meant with "Borel".
$endgroup$
– Math_QED
Dec 15 '18 at 15:14




$begingroup$
$g$ must be measurable, which is what I meant with "Borel".
$endgroup$
– Math_QED
Dec 15 '18 at 15:14












$begingroup$
If $g$ is a constant you sure the claim holds?
$endgroup$
– Yanko
Dec 15 '18 at 15:16




$begingroup$
If $g$ is a constant you sure the claim holds?
$endgroup$
– Yanko
Dec 15 '18 at 15:16












$begingroup$
I'm thinking about it. Give me a second please. Thanks for your useful comment.
$endgroup$
– Math_QED
Dec 15 '18 at 15:16






$begingroup$
I'm thinking about it. Give me a second please. Thanks for your useful comment.
$endgroup$
– Math_QED
Dec 15 '18 at 15:16














$begingroup$
Fix Borel sets $A,B$. We have to prove that $P(gcirc X in A, g circ Y in B) = P(g circ X in A) P(g circ Y in B)$ where $g=1$ is the constant 1 function. If $1 in A cap B$, then we have $1 = 1$. Otherwise, we have $0=0$, so I don't think there is a contradiction. What do you think?
$endgroup$
– Math_QED
Dec 15 '18 at 15:23






$begingroup$
Fix Borel sets $A,B$. We have to prove that $P(gcirc X in A, g circ Y in B) = P(g circ X in A) P(g circ Y in B)$ where $g=1$ is the constant 1 function. If $1 in A cap B$, then we have $1 = 1$. Otherwise, we have $0=0$, so I don't think there is a contradiction. What do you think?
$endgroup$
– Math_QED
Dec 15 '18 at 15:23












2 Answers
2






active

oldest

votes


















2












$begingroup$

Try
$$
g_ncolon xmapsto xmathbf 1_{left[-infty,nright]}(x).
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Don't you mean $xmapsto xmathbf1_{(-infty,n]}(x)$?
    $endgroup$
    – drhab
    Dec 15 '18 at 15:36












  • $begingroup$
    Yeah I noticed it too but the idea remains the same.
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:36



















1












$begingroup$

It is immediate that $Xmathbf1_{X_nleq n}$ is measurable wrt $sigma(X_n)$ because it is the product of two random variables that are both measurable wrt $sigma(X_n)$.



That is enough to conclude that also the $X_n'$ are pairwise disjoint.



There is indeed a measurable function $g:mathbb Rtomathbb R$ such that $X_n'=gcirc X_n$.



Let $h:mathbb R^2tomathbb R$ denote the function prescribed by $(x,y)mapsto xy$ and let $k_n:mathbb Rtomathbb R^2$ denote the function that is prescribed by $xmapsto(x,mathbf1_{(-infty,n]}(x))$.



Both functions can be shown to be measurable so that also their composition is measurable.



Then function $g=hcirc k_n$ will do the job.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer. Gives more insight.
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:37










  • $begingroup$
    You are very welcome.
    $endgroup$
    – drhab
    Dec 15 '18 at 15:40











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Try
$$
g_ncolon xmapsto xmathbf 1_{left[-infty,nright]}(x).
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Don't you mean $xmapsto xmathbf1_{(-infty,n]}(x)$?
    $endgroup$
    – drhab
    Dec 15 '18 at 15:36












  • $begingroup$
    Yeah I noticed it too but the idea remains the same.
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:36
















2












$begingroup$

Try
$$
g_ncolon xmapsto xmathbf 1_{left[-infty,nright]}(x).
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Don't you mean $xmapsto xmathbf1_{(-infty,n]}(x)$?
    $endgroup$
    – drhab
    Dec 15 '18 at 15:36












  • $begingroup$
    Yeah I noticed it too but the idea remains the same.
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:36














2












2








2





$begingroup$

Try
$$
g_ncolon xmapsto xmathbf 1_{left[-infty,nright]}(x).
$$






share|cite|improve this answer











$endgroup$



Try
$$
g_ncolon xmapsto xmathbf 1_{left[-infty,nright]}(x).
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 18 '18 at 13:21

























answered Dec 15 '18 at 15:14









Davide GiraudoDavide Giraudo

126k16151263




126k16151263












  • $begingroup$
    Don't you mean $xmapsto xmathbf1_{(-infty,n]}(x)$?
    $endgroup$
    – drhab
    Dec 15 '18 at 15:36












  • $begingroup$
    Yeah I noticed it too but the idea remains the same.
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:36


















  • $begingroup$
    Don't you mean $xmapsto xmathbf1_{(-infty,n]}(x)$?
    $endgroup$
    – drhab
    Dec 15 '18 at 15:36












  • $begingroup$
    Yeah I noticed it too but the idea remains the same.
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:36
















$begingroup$
Don't you mean $xmapsto xmathbf1_{(-infty,n]}(x)$?
$endgroup$
– drhab
Dec 15 '18 at 15:36






$begingroup$
Don't you mean $xmapsto xmathbf1_{(-infty,n]}(x)$?
$endgroup$
– drhab
Dec 15 '18 at 15:36














$begingroup$
Yeah I noticed it too but the idea remains the same.
$endgroup$
– Math_QED
Dec 15 '18 at 15:36




$begingroup$
Yeah I noticed it too but the idea remains the same.
$endgroup$
– Math_QED
Dec 15 '18 at 15:36











1












$begingroup$

It is immediate that $Xmathbf1_{X_nleq n}$ is measurable wrt $sigma(X_n)$ because it is the product of two random variables that are both measurable wrt $sigma(X_n)$.



That is enough to conclude that also the $X_n'$ are pairwise disjoint.



There is indeed a measurable function $g:mathbb Rtomathbb R$ such that $X_n'=gcirc X_n$.



Let $h:mathbb R^2tomathbb R$ denote the function prescribed by $(x,y)mapsto xy$ and let $k_n:mathbb Rtomathbb R^2$ denote the function that is prescribed by $xmapsto(x,mathbf1_{(-infty,n]}(x))$.



Both functions can be shown to be measurable so that also their composition is measurable.



Then function $g=hcirc k_n$ will do the job.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer. Gives more insight.
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:37










  • $begingroup$
    You are very welcome.
    $endgroup$
    – drhab
    Dec 15 '18 at 15:40
















1












$begingroup$

It is immediate that $Xmathbf1_{X_nleq n}$ is measurable wrt $sigma(X_n)$ because it is the product of two random variables that are both measurable wrt $sigma(X_n)$.



That is enough to conclude that also the $X_n'$ are pairwise disjoint.



There is indeed a measurable function $g:mathbb Rtomathbb R$ such that $X_n'=gcirc X_n$.



Let $h:mathbb R^2tomathbb R$ denote the function prescribed by $(x,y)mapsto xy$ and let $k_n:mathbb Rtomathbb R^2$ denote the function that is prescribed by $xmapsto(x,mathbf1_{(-infty,n]}(x))$.



Both functions can be shown to be measurable so that also their composition is measurable.



Then function $g=hcirc k_n$ will do the job.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer. Gives more insight.
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:37










  • $begingroup$
    You are very welcome.
    $endgroup$
    – drhab
    Dec 15 '18 at 15:40














1












1








1





$begingroup$

It is immediate that $Xmathbf1_{X_nleq n}$ is measurable wrt $sigma(X_n)$ because it is the product of two random variables that are both measurable wrt $sigma(X_n)$.



That is enough to conclude that also the $X_n'$ are pairwise disjoint.



There is indeed a measurable function $g:mathbb Rtomathbb R$ such that $X_n'=gcirc X_n$.



Let $h:mathbb R^2tomathbb R$ denote the function prescribed by $(x,y)mapsto xy$ and let $k_n:mathbb Rtomathbb R^2$ denote the function that is prescribed by $xmapsto(x,mathbf1_{(-infty,n]}(x))$.



Both functions can be shown to be measurable so that also their composition is measurable.



Then function $g=hcirc k_n$ will do the job.






share|cite|improve this answer









$endgroup$



It is immediate that $Xmathbf1_{X_nleq n}$ is measurable wrt $sigma(X_n)$ because it is the product of two random variables that are both measurable wrt $sigma(X_n)$.



That is enough to conclude that also the $X_n'$ are pairwise disjoint.



There is indeed a measurable function $g:mathbb Rtomathbb R$ such that $X_n'=gcirc X_n$.



Let $h:mathbb R^2tomathbb R$ denote the function prescribed by $(x,y)mapsto xy$ and let $k_n:mathbb Rtomathbb R^2$ denote the function that is prescribed by $xmapsto(x,mathbf1_{(-infty,n]}(x))$.



Both functions can be shown to be measurable so that also their composition is measurable.



Then function $g=hcirc k_n$ will do the job.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 15 '18 at 15:33









drhabdrhab

101k545136




101k545136












  • $begingroup$
    Thank you for your answer. Gives more insight.
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:37










  • $begingroup$
    You are very welcome.
    $endgroup$
    – drhab
    Dec 15 '18 at 15:40


















  • $begingroup$
    Thank you for your answer. Gives more insight.
    $endgroup$
    – Math_QED
    Dec 15 '18 at 15:37










  • $begingroup$
    You are very welcome.
    $endgroup$
    – drhab
    Dec 15 '18 at 15:40
















$begingroup$
Thank you for your answer. Gives more insight.
$endgroup$
– Math_QED
Dec 15 '18 at 15:37




$begingroup$
Thank you for your answer. Gives more insight.
$endgroup$
– Math_QED
Dec 15 '18 at 15:37












$begingroup$
You are very welcome.
$endgroup$
– drhab
Dec 15 '18 at 15:40




$begingroup$
You are very welcome.
$endgroup$
– drhab
Dec 15 '18 at 15:40


















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