Mixing
Sum of Random Variables that are not in $L^p(mathbb{R})$
$begingroup$
let $Xsim P$ be a real-valued random variable that is not an element of $L^p(mathbb{R})$ for some $p>0$. Furthermore, let $X_1, dots, X_N$ represent $i.i.d.$ copies of $X$ and for any $iin{1,dots,N} $ let $ a_iinmathbb{R}$ such that there exist at least one $jin{1,dots,N} $ such that $a_j neq 0$.
My question is whether the assertion
begin{equation}
sum_{i=1}^{N} a_i X_i notin L^p(mathbb{R})
end{equation}
is true.
All the best,
Carl
probability probability-theory statistics
asked Jan 15 at 23:29
CarlCarl
9310
$endgroup$
add a comment |
$begingroup$
let $Xsim P$ be a real-valued random variable that is not an element of $L^p(mathbb{R})$ for some $p>0$. Furthermore, let $X_1, dots, X_N$ represent $i.i.d.$ copies of $X$ and for any $iin{1,dots,N} $ let $ a_iinmathbb{R}$ such that there exist at least one $jin{1,dots,N} $ such that $a_j neq 0$.
My question is whether the assertion
begin{equation}
sum_{i=1}^{N} a_i X_i notin L^p(mathbb{R})
end{equation}
is true.
All the best,
Carl
probability probability-theory statistics
asked Jan 15 at 23:29
CarlCarl
9310
$endgroup$
1
$begingroup$
For $p geq 1$ the answer is YES.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 23:39
$begingroup$
I believe the proof is highly non-trivial. Do you have a reference or a theorem/lemma which shows this?
$endgroup$
– Carl
Jan 15 at 23:49
add a comment |
$begingroup$
let $Xsim P$ be a real-valued random variable that is not an element of $L^p(mathbb{R})$ for some $p>0$. Furthermore, let $X_1, dots, X_N$ represent $i.i.d.$ copies of $X$ and for any $iin{1,dots,N} $ let $ a_iinmathbb{R}$ such that there exist at least one $jin{1,dots,N} $ such that $a_j neq 0$.
My question is whether the assertion
begin{equation}
sum_{i=1}^{N} a_i X_i notin L^p(mathbb{R})
end{equation}
is true.
All the best,
Carl
probability probability-theory statistics
asked Jan 15 at 23:29
CarlCarl
9310
$endgroup$
let $Xsim P$ be a real-valued random variable that is not an element of $L^p(mathbb{R})$ for some $p>0$. Furthermore, let $X_1, dots, X_N$ represent $i.i.d.$ copies of $X$ and for any $iin{1,dots,N} $ let $ a_iinmathbb{R}$ such that there exist at least one $jin{1,dots,N} $ such that $a_j neq 0$.
My question is whether the assertion
begin{equation}
sum_{i=1}^{N} a_i X_i notin L^p(mathbb{R})
end{equation}
is true.
All the best,
Carl
probability probability-theory statistics
probability probability-theory statistics
asked Jan 15 at 23:29
CarlCarl
9310
asked Jan 15 at 23:29
CarlCarl
9310
asked Jan 15 at 23:29
CarlCarl
9310
asked Jan 15 at 23:29
CarlCarl
9310
asked Jan 15 at 23:29
CarlCarl
9310
9310
1
$begingroup$
For $p geq 1$ the answer is YES.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 23:39
$begingroup$
I believe the proof is highly non-trivial. Do you have a reference or a theorem/lemma which shows this?
$endgroup$
– Carl
Jan 15 at 23:49
add a comment |
1
$begingroup$
For $p geq 1$ the answer is YES.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 23:39
$begingroup$
I believe the proof is highly non-trivial. Do you have a reference or a theorem/lemma which shows this?
$endgroup$
– Carl
Jan 15 at 23:49
1
1
$begingroup$
For $p geq 1$ the answer is YES.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 23:39
$begingroup$
For $p geq 1$ the answer is YES.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 23:39
$begingroup$
I believe the proof is highly non-trivial. Do you have a reference or a theorem/lemma which shows this?
$endgroup$
– Carl
Jan 15 at 23:49
$begingroup$
I believe the proof is highly non-trivial. Do you have a reference or a theorem/lemma which shows this?
$endgroup$
– Carl
Jan 15 at 23:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Partial answer: let us show that the result is true for $p geq 1$. For $p=1$ this is a trivial application of Fubini's Theorem, so let us assume that $p>1$.
Lemma
If $X$ and $Y$ are independent random variables with $EX=0=EY$ and $E|X+Y|^{p} <infty$ then $E|X|^{p} <infty$ and $E|Y|^{p} <infty$.
Proof: $E|X+Y|^{p} =int|x+y|^{p} dF_X(x)dF_Y(y) geq int |int (x+y)dF_Y(y)|^{p}dF_X(x)$ by Jensens's inequality applied to the convex function $x to x^{p}$. This gives $int |x|^{p}dF_X(x) <infty$ since $int ydF_Y(y)=0$. Simialrly, $int |y|^{p}dF_Y(y) <infty$
In our case we can always assume that the mean is $0$ since adding a constant does not change integrability. [Existence of $E|X_i|$ is immediate from an application of Fubini's Theorem since $E|sum_{i} a_i X_i| <infty$]. Hence the lemma tells us that $sum_{ineq j} a_i X_i in L^{p}$. If $a_i=0$ for all $i neq j$ the we already know that $X_j in L^{p}$ so there is nothing to prove. If some $a_i$ with $i neq j$ is non-zero we can repeat the argument.
answered Jan 16 at 0:03
Kavi Rama MurthyKavi Rama Murthy
61.5k42262
$endgroup$
$begingroup$
Thank you very much, Sir. Since we want to show that the sum is not $L^p$ I believe the statement holds only for $p>1$ and not for $pgeq 1$ since we need the existence of the first moment to use this Lemma.
$endgroup$
– Carl
Jan 16 at 0:38
1
$begingroup$
@Carl Goof comment. I have added one line to the answer to make it correct.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 5:23
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Partial answer: let us show that the result is true for $p geq 1$. For $p=1$ this is a trivial application of Fubini's Theorem, so let us assume that $p>1$.
Lemma
If $X$ and $Y$ are independent random variables with $EX=0=EY$ and $E|X+Y|^{p} <infty$ then $E|X|^{p} <infty$ and $E|Y|^{p} <infty$.
Proof: $E|X+Y|^{p} =int|x+y|^{p} dF_X(x)dF_Y(y) geq int |int (x+y)dF_Y(y)|^{p}dF_X(x)$ by Jensens's inequality applied to the convex function $x to x^{p}$. This gives $int |x|^{p}dF_X(x) <infty$ since $int ydF_Y(y)=0$. Simialrly, $int |y|^{p}dF_Y(y) <infty$
In our case we can always assume that the mean is $0$ since adding a constant does not change integrability. [Existence of $E|X_i|$ is immediate from an application of Fubini's Theorem since $E|sum_{i} a_i X_i| <infty$]. Hence the lemma tells us that $sum_{ineq j} a_i X_i in L^{p}$. If $a_i=0$ for all $i neq j$ the we already know that $X_j in L^{p}$ so there is nothing to prove. If some $a_i$ with $i neq j$ is non-zero we can repeat the argument.
answered Jan 16 at 0:03
Kavi Rama MurthyKavi Rama Murthy
61.5k42262
$endgroup$
$begingroup$
Thank you very much, Sir. Since we want to show that the sum is not $L^p$ I believe the statement holds only for $p>1$ and not for $pgeq 1$ since we need the existence of the first moment to use this Lemma.
$endgroup$
– Carl
Jan 16 at 0:38
1
$begingroup$
@Carl Goof comment. I have added one line to the answer to make it correct.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 5:23
add a comment |
$begingroup$
Partial answer: let us show that the result is true for $p geq 1$. For $p=1$ this is a trivial application of Fubini's Theorem, so let us assume that $p>1$.
Lemma
If $X$ and $Y$ are independent random variables with $EX=0=EY$ and $E|X+Y|^{p} <infty$ then $E|X|^{p} <infty$ and $E|Y|^{p} <infty$.
Proof: $E|X+Y|^{p} =int|x+y|^{p} dF_X(x)dF_Y(y) geq int |int (x+y)dF_Y(y)|^{p}dF_X(x)$ by Jensens's inequality applied to the convex function $x to x^{p}$. This gives $int |x|^{p}dF_X(x) <infty$ since $int ydF_Y(y)=0$. Simialrly, $int |y|^{p}dF_Y(y) <infty$
In our case we can always assume that the mean is $0$ since adding a constant does not change integrability. [Existence of $E|X_i|$ is immediate from an application of Fubini's Theorem since $E|sum_{i} a_i X_i| <infty$]. Hence the lemma tells us that $sum_{ineq j} a_i X_i in L^{p}$. If $a_i=0$ for all $i neq j$ the we already know that $X_j in L^{p}$ so there is nothing to prove. If some $a_i$ with $i neq j$ is non-zero we can repeat the argument.
answered Jan 16 at 0:03
Kavi Rama MurthyKavi Rama Murthy
61.5k42262
$endgroup$
$begingroup$
Thank you very much, Sir. Since we want to show that the sum is not $L^p$ I believe the statement holds only for $p>1$ and not for $pgeq 1$ since we need the existence of the first moment to use this Lemma.
$endgroup$
– Carl
Jan 16 at 0:38
1
$begingroup$
@Carl Goof comment. I have added one line to the answer to make it correct.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 5:23
add a comment |
$begingroup$
Partial answer: let us show that the result is true for $p geq 1$. For $p=1$ this is a trivial application of Fubini's Theorem, so let us assume that $p>1$.
Lemma
If $X$ and $Y$ are independent random variables with $EX=0=EY$ and $E|X+Y|^{p} <infty$ then $E|X|^{p} <infty$ and $E|Y|^{p} <infty$.
Proof: $E|X+Y|^{p} =int|x+y|^{p} dF_X(x)dF_Y(y) geq int |int (x+y)dF_Y(y)|^{p}dF_X(x)$ by Jensens's inequality applied to the convex function $x to x^{p}$. This gives $int |x|^{p}dF_X(x) <infty$ since $int ydF_Y(y)=0$. Simialrly, $int |y|^{p}dF_Y(y) <infty$
In our case we can always assume that the mean is $0$ since adding a constant does not change integrability. [Existence of $E|X_i|$ is immediate from an application of Fubini's Theorem since $E|sum_{i} a_i X_i| <infty$]. Hence the lemma tells us that $sum_{ineq j} a_i X_i in L^{p}$. If $a_i=0$ for all $i neq j$ the we already know that $X_j in L^{p}$ so there is nothing to prove. If some $a_i$ with $i neq j$ is non-zero we can repeat the argument.
answered Jan 16 at 0:03
Kavi Rama MurthyKavi Rama Murthy
61.5k42262
$endgroup$
Partial answer: let us show that the result is true for $p geq 1$. For $p=1$ this is a trivial application of Fubini's Theorem, so let us assume that $p>1$.
Lemma
If $X$ and $Y$ are independent random variables with $EX=0=EY$ and $E|X+Y|^{p} <infty$ then $E|X|^{p} <infty$ and $E|Y|^{p} <infty$.
Proof: $E|X+Y|^{p} =int|x+y|^{p} dF_X(x)dF_Y(y) geq int |int (x+y)dF_Y(y)|^{p}dF_X(x)$ by Jensens's inequality applied to the convex function $x to x^{p}$. This gives $int |x|^{p}dF_X(x) <infty$ since $int ydF_Y(y)=0$. Simialrly, $int |y|^{p}dF_Y(y) <infty$
In our case we can always assume that the mean is $0$ since adding a constant does not change integrability. [Existence of $E|X_i|$ is immediate from an application of Fubini's Theorem since $E|sum_{i} a_i X_i| <infty$]. Hence the lemma tells us that $sum_{ineq j} a_i X_i in L^{p}$. If $a_i=0$ for all $i neq j$ the we already know that $X_j in L^{p}$ so there is nothing to prove. If some $a_i$ with $i neq j$ is non-zero we can repeat the argument.
answered Jan 16 at 0:03
Kavi Rama MurthyKavi Rama Murthy
61.5k42262
edited Jan 16 at 5:22
answered Jan 16 at 0:03
Kavi Rama MurthyKavi Rama Murthy
61.5k42262
answered Jan 16 at 0:03
Kavi Rama MurthyKavi Rama Murthy
61.5k42262
answered Jan 16 at 0:03
Kavi Rama MurthyKavi Rama Murthy
61.5k42262
61.5k42262
$begingroup$
Thank you very much, Sir. Since we want to show that the sum is not $L^p$ I believe the statement holds only for $p>1$ and not for $pgeq 1$ since we need the existence of the first moment to use this Lemma.
$endgroup$
– Carl
Jan 16 at 0:38
1
$begingroup$
@Carl Goof comment. I have added one line to the answer to make it correct.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 5:23
add a comment |
$begingroup$
Thank you very much, Sir. Since we want to show that the sum is not $L^p$ I believe the statement holds only for $p>1$ and not for $pgeq 1$ since we need the existence of the first moment to use this Lemma.
$endgroup$
– Carl
Jan 16 at 0:38
1
$begingroup$
@Carl Goof comment. I have added one line to the answer to make it correct.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 5:23
$begingroup$
Thank you very much, Sir. Since we want to show that the sum is not $L^p$ I believe the statement holds only for $p>1$ and not for $pgeq 1$ since we need the existence of the first moment to use this Lemma.
$endgroup$
– Carl
Jan 16 at 0:38
$begingroup$
Thank you very much, Sir. Since we want to show that the sum is not $L^p$ I believe the statement holds only for $p>1$ and not for $pgeq 1$ since we need the existence of the first moment to use this Lemma.
$endgroup$
– Carl
Jan 16 at 0:38
1
1
$begingroup$
@Carl Goof comment. I have added one line to the answer to make it correct.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 5:23
$begingroup$
@Carl Goof comment. I have added one line to the answer to make it correct.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 5:23
add a comment |
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$begingroup$
For $p geq 1$ the answer is YES.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 23:39
$begingroup$
I believe the proof is highly non-trivial. Do you have a reference or a theorem/lemma which shows this?
$endgroup$
– Carl
Jan 15 at 23:49