Mixing
Is $sum_{k=1}^infty sum_{ell=1}^infty a_ib_j1_{{X=a_i}}1_{{Y=b_j}}=sum_{ell=1}^infty sum_{k=1}^infty $ true?
Let $(Omega ,mathcal F,mathbb P)$ a probability space. Let $X,Y$ random variables. We suppose $Omega $ discrete. We write $$X=sum_{i=1}^infty a_iboldsymbol 1_{{X=a_i}}$$
and $$Y=sum_{k=1}^infty b_k boldsymbol 1_{{Y=b_k}}.$$
We want to write $X+Y$ as a sum of $sum_{k}x_kboldsymbol 1_{A_k}$. At a moment in the proof, they do :
$$sum_{k=1}^infty a_kboldsymbol 1_{{X=a_k}}sum_{j=1}^infty boldsymbol 1_{{Y=b_j}}=sum_{k=1}^infty sum_{j=1}^infty a_kb_jboldsymbol 1_{{X=a_k,Y=b_j}}=sum_{j=1}^infty sum_{k=1}^infty a_kb_jboldsymbol 1_{{X=a_k,Y=b_j}}.$$
In the last equality they permute the two sum. Don't we need condition to do that ? Like $sum_{i,j}a_ib_j$ converge uniformly ? Because nothing has been mentioned at this subject.
probability measure-theory
asked Nov 21 '18 at 18:25
user617786user617786
234
add a comment |
Let $(Omega ,mathcal F,mathbb P)$ a probability space. Let $X,Y$ random variables. We suppose $Omega $ discrete. We write $$X=sum_{i=1}^infty a_iboldsymbol 1_{{X=a_i}}$$
and $$Y=sum_{k=1}^infty b_k boldsymbol 1_{{Y=b_k}}.$$
We want to write $X+Y$ as a sum of $sum_{k}x_kboldsymbol 1_{A_k}$. At a moment in the proof, they do :
$$sum_{k=1}^infty a_kboldsymbol 1_{{X=a_k}}sum_{j=1}^infty boldsymbol 1_{{Y=b_j}}=sum_{k=1}^infty sum_{j=1}^infty a_kb_jboldsymbol 1_{{X=a_k,Y=b_j}}=sum_{j=1}^infty sum_{k=1}^infty a_kb_jboldsymbol 1_{{X=a_k,Y=b_j}}.$$
In the last equality they permute the two sum. Don't we need condition to do that ? Like $sum_{i,j}a_ib_j$ converge uniformly ? Because nothing has been mentioned at this subject.
probability measure-theory
asked Nov 21 '18 at 18:25
user617786user617786
234
@WillM. why do you need that?
– Federico
Nov 21 '18 at 18:37
My bad, I read non-negative random variables.
– Will M.
Nov 21 '18 at 18:40
I don't understand you question, @Surb
– Will M.
Nov 21 '18 at 18:42
Lebuesgue-Fubini-Tonelli theorem says iterated series of nonnegative functions can be commuted at will; all will diverge or all will converge to the same value (which is exactly the question OP is asking).
– Will M.
Nov 21 '18 at 18:45
"At a moment in the proof, they do" Which proof in what book/notes are you referring to?
– user587192
Nov 21 '18 at 18:53
add a comment |
Let $(Omega ,mathcal F,mathbb P)$ a probability space. Let $X,Y$ random variables. We suppose $Omega $ discrete. We write $$X=sum_{i=1}^infty a_iboldsymbol 1_{{X=a_i}}$$
and $$Y=sum_{k=1}^infty b_k boldsymbol 1_{{Y=b_k}}.$$
We want to write $X+Y$ as a sum of $sum_{k}x_kboldsymbol 1_{A_k}$. At a moment in the proof, they do :
$$sum_{k=1}^infty a_kboldsymbol 1_{{X=a_k}}sum_{j=1}^infty boldsymbol 1_{{Y=b_j}}=sum_{k=1}^infty sum_{j=1}^infty a_kb_jboldsymbol 1_{{X=a_k,Y=b_j}}=sum_{j=1}^infty sum_{k=1}^infty a_kb_jboldsymbol 1_{{X=a_k,Y=b_j}}.$$
In the last equality they permute the two sum. Don't we need condition to do that ? Like $sum_{i,j}a_ib_j$ converge uniformly ? Because nothing has been mentioned at this subject.
probability measure-theory
asked Nov 21 '18 at 18:25
user617786user617786
234
Let $(Omega ,mathcal F,mathbb P)$ a probability space. Let $X,Y$ random variables. We suppose $Omega $ discrete. We write $$X=sum_{i=1}^infty a_iboldsymbol 1_{{X=a_i}}$$
and $$Y=sum_{k=1}^infty b_k boldsymbol 1_{{Y=b_k}}.$$
We want to write $X+Y$ as a sum of $sum_{k}x_kboldsymbol 1_{A_k}$. At a moment in the proof, they do :
$$sum_{k=1}^infty a_kboldsymbol 1_{{X=a_k}}sum_{j=1}^infty boldsymbol 1_{{Y=b_j}}=sum_{k=1}^infty sum_{j=1}^infty a_kb_jboldsymbol 1_{{X=a_k,Y=b_j}}=sum_{j=1}^infty sum_{k=1}^infty a_kb_jboldsymbol 1_{{X=a_k,Y=b_j}}.$$
In the last equality they permute the two sum. Don't we need condition to do that ? Like $sum_{i,j}a_ib_j$ converge uniformly ? Because nothing has been mentioned at this subject.
probability measure-theory
probability measure-theory
asked Nov 21 '18 at 18:25
user617786user617786
234
asked Nov 21 '18 at 18:25
user617786user617786
234
asked Nov 21 '18 at 18:25
user617786user617786
234
asked Nov 21 '18 at 18:25
user617786user617786
234
asked Nov 21 '18 at 18:25
user617786user617786
234
234
@WillM. why do you need that?
– Federico
Nov 21 '18 at 18:37
My bad, I read non-negative random variables.
– Will M.
Nov 21 '18 at 18:40
I don't understand you question, @Surb
– Will M.
Nov 21 '18 at 18:42
Lebuesgue-Fubini-Tonelli theorem says iterated series of nonnegative functions can be commuted at will; all will diverge or all will converge to the same value (which is exactly the question OP is asking).
– Will M.
Nov 21 '18 at 18:45
"At a moment in the proof, they do" Which proof in what book/notes are you referring to?
– user587192
Nov 21 '18 at 18:53
add a comment |
@WillM. why do you need that?
– Federico
Nov 21 '18 at 18:37
My bad, I read non-negative random variables.
– Will M.
Nov 21 '18 at 18:40
I don't understand you question, @Surb
– Will M.
Nov 21 '18 at 18:42
Lebuesgue-Fubini-Tonelli theorem says iterated series of nonnegative functions can be commuted at will; all will diverge or all will converge to the same value (which is exactly the question OP is asking).
– Will M.
Nov 21 '18 at 18:45
"At a moment in the proof, they do" Which proof in what book/notes are you referring to?
– user587192
Nov 21 '18 at 18:53
@WillM. why do you need that?
– Federico
Nov 21 '18 at 18:37
@WillM. why do you need that?
– Federico
Nov 21 '18 at 18:37
My bad, I read non-negative random variables.
– Will M.
Nov 21 '18 at 18:40
My bad, I read non-negative random variables.
– Will M.
Nov 21 '18 at 18:40
I don't understand you question, @Surb
– Will M.
Nov 21 '18 at 18:42
I don't understand you question, @Surb
– Will M.
Nov 21 '18 at 18:42
Lebuesgue-Fubini-Tonelli theorem says iterated series of nonnegative functions can be commuted at will; all will diverge or all will converge to the same value (which is exactly the question OP is asking).
– Will M.
Nov 21 '18 at 18:45
Lebuesgue-Fubini-Tonelli theorem says iterated series of nonnegative functions can be commuted at will; all will diverge or all will converge to the same value (which is exactly the question OP is asking).
– Will M.
Nov 21 '18 at 18:45
"At a moment in the proof, they do" Which proof in what book/notes are you referring to?
– user587192
Nov 21 '18 at 18:53
"At a moment in the proof, they do" Which proof in what book/notes are you referring to?
– user587192
Nov 21 '18 at 18:53
add a comment |
1 Answer
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For a fix $omega in Omega $, $$sum_{j=1}^infty sum_{i=1}^infty a_ib_jboldsymbol 1_{{X=a_i, Y=b_j}}(omega ),$$
is a finite sum.
answered Nov 21 '18 at 18:32
SurbSurb
37.4k94375
It seems to me that the sum contains exactly one non-zero term if the ${a_i}$ and ${b_i}$ are distinct (i.e $a_i=a_j$ implies $i=j$)?
– irchans
Nov 21 '18 at 18:40
@irchans: yes exactly.
– Surb
Nov 21 '18 at 18:40
add a comment |
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1 Answer
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1 Answer
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For a fix $omega in Omega $, $$sum_{j=1}^infty sum_{i=1}^infty a_ib_jboldsymbol 1_{{X=a_i, Y=b_j}}(omega ),$$
is a finite sum.
answered Nov 21 '18 at 18:32
SurbSurb
37.4k94375
It seems to me that the sum contains exactly one non-zero term if the ${a_i}$ and ${b_i}$ are distinct (i.e $a_i=a_j$ implies $i=j$)?
– irchans
Nov 21 '18 at 18:40
@irchans: yes exactly.
– Surb
Nov 21 '18 at 18:40
add a comment |
For a fix $omega in Omega $, $$sum_{j=1}^infty sum_{i=1}^infty a_ib_jboldsymbol 1_{{X=a_i, Y=b_j}}(omega ),$$
is a finite sum.
answered Nov 21 '18 at 18:32
SurbSurb
37.4k94375
It seems to me that the sum contains exactly one non-zero term if the ${a_i}$ and ${b_i}$ are distinct (i.e $a_i=a_j$ implies $i=j$)?
– irchans
Nov 21 '18 at 18:40
@irchans: yes exactly.
– Surb
Nov 21 '18 at 18:40
add a comment |
For a fix $omega in Omega $, $$sum_{j=1}^infty sum_{i=1}^infty a_ib_jboldsymbol 1_{{X=a_i, Y=b_j}}(omega ),$$
is a finite sum.
answered Nov 21 '18 at 18:32
SurbSurb
37.4k94375
For a fix $omega in Omega $, $$sum_{j=1}^infty sum_{i=1}^infty a_ib_jboldsymbol 1_{{X=a_i, Y=b_j}}(omega ),$$
is a finite sum.
answered Nov 21 '18 at 18:32
SurbSurb
37.4k94375
answered Nov 21 '18 at 18:32
SurbSurb
37.4k94375
answered Nov 21 '18 at 18:32
SurbSurb
37.4k94375
answered Nov 21 '18 at 18:32
SurbSurb
37.4k94375
37.4k94375
It seems to me that the sum contains exactly one non-zero term if the ${a_i}$ and ${b_i}$ are distinct (i.e $a_i=a_j$ implies $i=j$)?
– irchans
Nov 21 '18 at 18:40
@irchans: yes exactly.
– Surb
Nov 21 '18 at 18:40
add a comment |
It seems to me that the sum contains exactly one non-zero term if the ${a_i}$ and ${b_i}$ are distinct (i.e $a_i=a_j$ implies $i=j$)?
– irchans
Nov 21 '18 at 18:40
@irchans: yes exactly.
– Surb
Nov 21 '18 at 18:40
It seems to me that the sum contains exactly one non-zero term if the ${a_i}$ and ${b_i}$ are distinct (i.e $a_i=a_j$ implies $i=j$)?
– irchans
Nov 21 '18 at 18:40
It seems to me that the sum contains exactly one non-zero term if the ${a_i}$ and ${b_i}$ are distinct (i.e $a_i=a_j$ implies $i=j$)?
– irchans
Nov 21 '18 at 18:40
@irchans: yes exactly.
– Surb
Nov 21 '18 at 18:40
@irchans: yes exactly.
– Surb
Nov 21 '18 at 18:40
add a comment |
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@WillM. why do you need that?
– Federico
Nov 21 '18 at 18:37
My bad, I read non-negative random variables.
– Will M.
Nov 21 '18 at 18:40
I don't understand you question, @Surb
– Will M.
Nov 21 '18 at 18:42
Lebuesgue-Fubini-Tonelli theorem says iterated series of nonnegative functions can be commuted at will; all will diverge or all will converge to the same value (which is exactly the question OP is asking).
– Will M.
Nov 21 '18 at 18:45
"At a moment in the proof, they do" Which proof in what book/notes are you referring to?
– user587192
Nov 21 '18 at 18:53