For independent RVs, does $E|X+Y|<infty$ imply $E|X|<infty$.
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Let $X,Y$ be independent random variables. Suppose $E|X+Y|$ is finite. Does it follow $E|X|<infty$?
I believe the answer is no and tried to come up with a counter-example. Below is my attempt where I couldn't complete. If it follows that $E|X|$ is finite, please ignore my attempt.
Attempt (Constants are ignored for readability.)
Let $X$ be defined on $mathbbN$ with distribution $P(X=k) = 1/k^2$. Note $EX = sum_k k/k^2=infty$.
Let $Y$ be identical to $X$. Distribution of $X+Y$ is given by;
beginequation
beginaligned
P(X+Y = z) &= sum_k=1^z-1 P(X=z-k)P(Y=k)\
&=sum_k=1^z-1 frac1(z-k)^2k^2
endaligned
endequation
Hence expected is given by;
beginequation
E(X+Y) = sum_z=2^inftysum_k=1^z-1 fracz(z-k)^2k^2
endequation
I couldn't argue that the sum is finite. Is above sum convergent?
sequences-and-series expectation convolution independence
edited Aug 11 at 1:25
Clement C.
47.2k33682
asked Aug 11 at 0:44
Jo'
1519
add a comment |Â
up vote
2
down vote
favorite
Let $X,Y$ be independent random variables. Suppose $E|X+Y|$ is finite. Does it follow $E|X|<infty$?
I believe the answer is no and tried to come up with a counter-example. Below is my attempt where I couldn't complete. If it follows that $E|X|$ is finite, please ignore my attempt.
Attempt (Constants are ignored for readability.)
Let $X$ be defined on $mathbbN$ with distribution $P(X=k) = 1/k^2$. Note $EX = sum_k k/k^2=infty$.
Let $Y$ be identical to $X$. Distribution of $X+Y$ is given by;
beginequation
beginaligned
P(X+Y = z) &= sum_k=1^z-1 P(X=z-k)P(Y=k)\
&=sum_k=1^z-1 frac1(z-k)^2k^2
endaligned
endequation
Hence expected is given by;
beginequation
E(X+Y) = sum_z=2^inftysum_k=1^z-1 fracz(z-k)^2k^2
endequation
I couldn't argue that the sum is finite. Is above sum convergent?
sequences-and-series expectation convolution independence
edited Aug 11 at 1:25
Clement C.
47.2k33682
asked Aug 11 at 0:44
Jo'
1519
1
You may want $Y$ to have the same distribution as $-X$ for your potential counterexample to have any chance of working. Otherwise, here, you have $X,Y>0$, and $X+Y > X$, so that $mathbbE[X+Y] geq mathbbE[X] = infty$.
– Clement C.
Aug 11 at 1:20
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $X,Y$ be independent random variables. Suppose $E|X+Y|$ is finite. Does it follow $E|X|<infty$?
I believe the answer is no and tried to come up with a counter-example. Below is my attempt where I couldn't complete. If it follows that $E|X|$ is finite, please ignore my attempt.
Attempt (Constants are ignored for readability.)
Let $X$ be defined on $mathbbN$ with distribution $P(X=k) = 1/k^2$. Note $EX = sum_k k/k^2=infty$.
Let $Y$ be identical to $X$. Distribution of $X+Y$ is given by;
beginequation
beginaligned
P(X+Y = z) &= sum_k=1^z-1 P(X=z-k)P(Y=k)\
&=sum_k=1^z-1 frac1(z-k)^2k^2
endaligned
endequation
Hence expected is given by;
beginequation
E(X+Y) = sum_z=2^inftysum_k=1^z-1 fracz(z-k)^2k^2
endequation
I couldn't argue that the sum is finite. Is above sum convergent?
sequences-and-series expectation convolution independence
edited Aug 11 at 1:25
Clement C.
47.2k33682
asked Aug 11 at 0:44
Jo'
1519
Let $X,Y$ be independent random variables. Suppose $E|X+Y|$ is finite. Does it follow $E|X|<infty$?
I believe the answer is no and tried to come up with a counter-example. Below is my attempt where I couldn't complete. If it follows that $E|X|$ is finite, please ignore my attempt.
Attempt (Constants are ignored for readability.)
Let $X$ be defined on $mathbbN$ with distribution $P(X=k) = 1/k^2$. Note $EX = sum_k k/k^2=infty$.
Let $Y$ be identical to $X$. Distribution of $X+Y$ is given by;
beginequation
beginaligned
P(X+Y = z) &= sum_k=1^z-1 P(X=z-k)P(Y=k)\
&=sum_k=1^z-1 frac1(z-k)^2k^2
endaligned
endequation
Hence expected is given by;
beginequation
E(X+Y) = sum_z=2^inftysum_k=1^z-1 fracz(z-k)^2k^2
endequation
I couldn't argue that the sum is finite. Is above sum convergent?
sequences-and-series expectation convolution independence
edited Aug 11 at 1:25
Clement C.
47.2k33682
asked Aug 11 at 0:44
Jo'
1519
edited Aug 11 at 1:25
Clement C.
47.2k33682
edited Aug 11 at 1:25
Clement C.
47.2k33682
edited Aug 11 at 1:25
Clement C.
47.2k33682
47.2k33682
asked Aug 11 at 0:44
Jo'
1519
asked Aug 11 at 0:44
Jo'
1519
asked Aug 11 at 0:44
Jo'
1519
1519
1
You may want $Y$ to have the same distribution as $-X$ for your potential counterexample to have any chance of working. Otherwise, here, you have $X,Y>0$, and $X+Y > X$, so that $mathbbE[X+Y] geq mathbbE[X] = infty$.
– Clement C.
Aug 11 at 1:20
add a comment |Â
1
You may want $Y$ to have the same distribution as $-X$ for your potential counterexample to have any chance of working. Otherwise, here, you have $X,Y>0$, and $X+Y > X$, so that $mathbbE[X+Y] geq mathbbE[X] = infty$.
– Clement C.
Aug 11 at 1:20
1
1
You may want $Y$ to have the same distribution as $-X$ for your potential counterexample to have any chance of working. Otherwise, here, you have $X,Y>0$, and $X+Y > X$, so that $mathbbE[X+Y] geq mathbbE[X] = infty$.
– Clement C.
Aug 11 at 1:20
You may want $Y$ to have the same distribution as $-X$ for your potential counterexample to have any chance of working. Otherwise, here, you have $X,Y>0$, and $X+Y > X$, so that $mathbbE[X+Y] geq mathbbE[X] = infty$.
– Clement C.
Aug 11 at 1:20
add a comment |Â
1 Answer
1
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oldest
votes
up vote
1
down vote
Since $|X| le |X+Y| + |Y|$, $E[|X|] = E[ |X| mid Y] le E[|X+Y| mid Y] + E[|Y| mid Y]$
and $E[|X|] le E[|X+Y|] + E[|Y|]$
Suppose $E[|X+Y|] = R < infty$. Now $|X+Y| ge |X| - |Y|$, so
$$R = E[|X+Y|] ge E[|X|-|Y|] = E[E[|X|-|Y| mid Y]] = E[E[|X|] - |Y|]$$
But that implies $textProb(E[|X|]-|Y| le R) > 0$ and thus $E[|X|] le R + s < infty$ for some $s$ where $textProb(|Y|<s) > 0$.
answered Aug 11 at 1:20
Robert Israel
305k22201443
1
Where do you use the conclusion of the first paragraph, in your argument?
– Clement C.
Aug 11 at 1:24
For the last equality(in equation environment), don't we require $E|X|$ to be finite for $E[|X| mid Y]$ to be defined?
– Jo'
Aug 11 at 2:09
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Since $|X| le |X+Y| + |Y|$, $E[|X|] = E[ |X| mid Y] le E[|X+Y| mid Y] + E[|Y| mid Y]$
and $E[|X|] le E[|X+Y|] + E[|Y|]$
Suppose $E[|X+Y|] = R < infty$. Now $|X+Y| ge |X| - |Y|$, so
$$R = E[|X+Y|] ge E[|X|-|Y|] = E[E[|X|-|Y| mid Y]] = E[E[|X|] - |Y|]$$
But that implies $textProb(E[|X|]-|Y| le R) > 0$ and thus $E[|X|] le R + s < infty$ for some $s$ where $textProb(|Y|<s) > 0$.
answered Aug 11 at 1:20
Robert Israel
305k22201443
1
Where do you use the conclusion of the first paragraph, in your argument?
– Clement C.
Aug 11 at 1:24
For the last equality(in equation environment), don't we require $E|X|$ to be finite for $E[|X| mid Y]$ to be defined?
– Jo'
Aug 11 at 2:09
add a comment |Â
up vote
1
down vote
Since $|X| le |X+Y| + |Y|$, $E[|X|] = E[ |X| mid Y] le E[|X+Y| mid Y] + E[|Y| mid Y]$
and $E[|X|] le E[|X+Y|] + E[|Y|]$
Suppose $E[|X+Y|] = R < infty$. Now $|X+Y| ge |X| - |Y|$, so
$$R = E[|X+Y|] ge E[|X|-|Y|] = E[E[|X|-|Y| mid Y]] = E[E[|X|] - |Y|]$$
But that implies $textProb(E[|X|]-|Y| le R) > 0$ and thus $E[|X|] le R + s < infty$ for some $s$ where $textProb(|Y|<s) > 0$.
answered Aug 11 at 1:20
Robert Israel
305k22201443
1
Where do you use the conclusion of the first paragraph, in your argument?
– Clement C.
Aug 11 at 1:24
For the last equality(in equation environment), don't we require $E|X|$ to be finite for $E[|X| mid Y]$ to be defined?
– Jo'
Aug 11 at 2:09
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Since $|X| le |X+Y| + |Y|$, $E[|X|] = E[ |X| mid Y] le E[|X+Y| mid Y] + E[|Y| mid Y]$
and $E[|X|] le E[|X+Y|] + E[|Y|]$
Suppose $E[|X+Y|] = R < infty$. Now $|X+Y| ge |X| - |Y|$, so
$$R = E[|X+Y|] ge E[|X|-|Y|] = E[E[|X|-|Y| mid Y]] = E[E[|X|] - |Y|]$$
But that implies $textProb(E[|X|]-|Y| le R) > 0$ and thus $E[|X|] le R + s < infty$ for some $s$ where $textProb(|Y|<s) > 0$.
answered Aug 11 at 1:20
Robert Israel
305k22201443
Since $|X| le |X+Y| + |Y|$, $E[|X|] = E[ |X| mid Y] le E[|X+Y| mid Y] + E[|Y| mid Y]$
and $E[|X|] le E[|X+Y|] + E[|Y|]$
Suppose $E[|X+Y|] = R < infty$. Now $|X+Y| ge |X| - |Y|$, so
$$R = E[|X+Y|] ge E[|X|-|Y|] = E[E[|X|-|Y| mid Y]] = E[E[|X|] - |Y|]$$
But that implies $textProb(E[|X|]-|Y| le R) > 0$ and thus $E[|X|] le R + s < infty$ for some $s$ where $textProb(|Y|<s) > 0$.
answered Aug 11 at 1:20
Robert Israel
305k22201443
answered Aug 11 at 1:20
Robert Israel
305k22201443
answered Aug 11 at 1:20
Robert Israel
305k22201443
answered Aug 11 at 1:20
Robert Israel
305k22201443
305k22201443
1
Where do you use the conclusion of the first paragraph, in your argument?
– Clement C.
Aug 11 at 1:24
For the last equality(in equation environment), don't we require $E|X|$ to be finite for $E[|X| mid Y]$ to be defined?
– Jo'
Aug 11 at 2:09
add a comment |Â
1
Where do you use the conclusion of the first paragraph, in your argument?
– Clement C.
Aug 11 at 1:24
For the last equality(in equation environment), don't we require $E|X|$ to be finite for $E[|X| mid Y]$ to be defined?
– Jo'
Aug 11 at 2:09
1
1
Where do you use the conclusion of the first paragraph, in your argument?
– Clement C.
Aug 11 at 1:24
Where do you use the conclusion of the first paragraph, in your argument?
– Clement C.
Aug 11 at 1:24
For the last equality(in equation environment), don't we require $E|X|$ to be finite for $E[|X| mid Y]$ to be defined?
– Jo'
Aug 11 at 2:09
For the last equality(in equation environment), don't we require $E|X|$ to be finite for $E[|X| mid Y]$ to be defined?
– Jo'
Aug 11 at 2:09
add a comment |Â
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You may want $Y$ to have the same distribution as $-X$ for your potential counterexample to have any chance of working. Otherwise, here, you have $X,Y>0$, and $X+Y > X$, so that $mathbbE[X+Y] geq mathbbE[X] = infty$.
– Clement C.
Aug 11 at 1:20