Do $sum_{x,y} f(x)g(x,y) = 1$ and $sum_{x} f(x) = 1$ for $f(x)>0$, $g(x,y)>0$ imply $sum_{y} g(x,y) =...












-2












$begingroup$


Do $sum_{x,y} f(x)g(x,y) = 1$ and $sum_{x} f(x) = 1$ for $f(x)>0$, $g(x,y)>0$ imply $sum_{y} g(x,y) = 1$? If so, how can I proof that?



Note: I am trying derive that $g(x,y)$ is a conditional probability without assuming it.



Previous mistake in the question removed.






probability probability-theory conditional-probability







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Of course not! Say there's only one value of $x$ and one value of $y$ under consideration, maybe $x=0$, $y=0$. Let $f(0)=2$, $g(0,0)=1/2$.
    $endgroup$
    – David C. Ullrich
    Dec 28 '18 at 18:06






  • 1




    $begingroup$
    $sum_{j=1}^2sum_{i=1}^3 frac{1}{2}cdot frac{1}{3} = 1$, but $sum_{i=1}^3 frac{1}{2} = 3/2$ and $sum_{i=1}^2 frac{1}{3} = 2/3$
    $endgroup$
    – Jakobian
    Dec 28 '18 at 18:06












  • $begingroup$
    What about the corrected version of the question? Anybody able to help?
    $endgroup$
    – J. Reinhard
    Dec 28 '18 at 18:41
















-2












$begingroup$


Do $sum_{x,y} f(x)g(x,y) = 1$ and $sum_{x} f(x) = 1$ for $f(x)>0$, $g(x,y)>0$ imply $sum_{y} g(x,y) = 1$? If so, how can I proof that?



Note: I am trying derive that $g(x,y)$ is a conditional probability without assuming it.



Previous mistake in the question removed.






probability probability-theory conditional-probability







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Of course not! Say there's only one value of $x$ and one value of $y$ under consideration, maybe $x=0$, $y=0$. Let $f(0)=2$, $g(0,0)=1/2$.
    $endgroup$
    – David C. Ullrich
    Dec 28 '18 at 18:06






  • 1




    $begingroup$
    $sum_{j=1}^2sum_{i=1}^3 frac{1}{2}cdot frac{1}{3} = 1$, but $sum_{i=1}^3 frac{1}{2} = 3/2$ and $sum_{i=1}^2 frac{1}{3} = 2/3$
    $endgroup$
    – Jakobian
    Dec 28 '18 at 18:06












  • $begingroup$
    What about the corrected version of the question? Anybody able to help?
    $endgroup$
    – J. Reinhard
    Dec 28 '18 at 18:41














-2












-2








-2





$begingroup$


Do $sum_{x,y} f(x)g(x,y) = 1$ and $sum_{x} f(x) = 1$ for $f(x)>0$, $g(x,y)>0$ imply $sum_{y} g(x,y) = 1$? If so, how can I proof that?



Note: I am trying derive that $g(x,y)$ is a conditional probability without assuming it.



Previous mistake in the question removed.






probability probability-theory conditional-probability







share|cite|improve this question











$endgroup$




Do $sum_{x,y} f(x)g(x,y) = 1$ and $sum_{x} f(x) = 1$ for $f(x)>0$, $g(x,y)>0$ imply $sum_{y} g(x,y) = 1$? If so, how can I proof that?



Note: I am trying derive that $g(x,y)$ is a conditional probability without assuming it.



Previous mistake in the question removed.






probability probability-theory conditional-probability




probability probability-theory conditional-probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 18:20







J. Reinhard

















asked Dec 28 '18 at 18:02









J. ReinhardJ. Reinhard

184




184








  • 2




    $begingroup$
    Of course not! Say there's only one value of $x$ and one value of $y$ under consideration, maybe $x=0$, $y=0$. Let $f(0)=2$, $g(0,0)=1/2$.
    $endgroup$
    – David C. Ullrich
    Dec 28 '18 at 18:06






  • 1




    $begingroup$
    $sum_{j=1}^2sum_{i=1}^3 frac{1}{2}cdot frac{1}{3} = 1$, but $sum_{i=1}^3 frac{1}{2} = 3/2$ and $sum_{i=1}^2 frac{1}{3} = 2/3$
    $endgroup$
    – Jakobian
    Dec 28 '18 at 18:06












  • $begingroup$
    What about the corrected version of the question? Anybody able to help?
    $endgroup$
    – J. Reinhard
    Dec 28 '18 at 18:41














  • 2




    $begingroup$
    Of course not! Say there's only one value of $x$ and one value of $y$ under consideration, maybe $x=0$, $y=0$. Let $f(0)=2$, $g(0,0)=1/2$.
    $endgroup$
    – David C. Ullrich
    Dec 28 '18 at 18:06






  • 1




    $begingroup$
    $sum_{j=1}^2sum_{i=1}^3 frac{1}{2}cdot frac{1}{3} = 1$, but $sum_{i=1}^3 frac{1}{2} = 3/2$ and $sum_{i=1}^2 frac{1}{3} = 2/3$
    $endgroup$
    – Jakobian
    Dec 28 '18 at 18:06












  • $begingroup$
    What about the corrected version of the question? Anybody able to help?
    $endgroup$
    – J. Reinhard
    Dec 28 '18 at 18:41








2




2




$begingroup$
Of course not! Say there's only one value of $x$ and one value of $y$ under consideration, maybe $x=0$, $y=0$. Let $f(0)=2$, $g(0,0)=1/2$.
$endgroup$
– David C. Ullrich
Dec 28 '18 at 18:06




$begingroup$
Of course not! Say there's only one value of $x$ and one value of $y$ under consideration, maybe $x=0$, $y=0$. Let $f(0)=2$, $g(0,0)=1/2$.
$endgroup$
– David C. Ullrich
Dec 28 '18 at 18:06




1




1




$begingroup$
$sum_{j=1}^2sum_{i=1}^3 frac{1}{2}cdot frac{1}{3} = 1$, but $sum_{i=1}^3 frac{1}{2} = 3/2$ and $sum_{i=1}^2 frac{1}{3} = 2/3$
$endgroup$
– Jakobian
Dec 28 '18 at 18:06






$begingroup$
$sum_{j=1}^2sum_{i=1}^3 frac{1}{2}cdot frac{1}{3} = 1$, but $sum_{i=1}^3 frac{1}{2} = 3/2$ and $sum_{i=1}^2 frac{1}{3} = 2/3$
$endgroup$
– Jakobian
Dec 28 '18 at 18:06














$begingroup$
What about the corrected version of the question? Anybody able to help?
$endgroup$
– J. Reinhard
Dec 28 '18 at 18:41




$begingroup$
What about the corrected version of the question? Anybody able to help?
$endgroup$
– J. Reinhard
Dec 28 '18 at 18:41










2 Answers
2






active

oldest

votes


















2












$begingroup$

There was already a very good finite counterexample given in the comments to the OP, but here is an infinite version:



Let $f(n)=frac{1}{n}$ and $g(n,m)=ncdot2^{-n}cdot2^{-m}$. Then the double sum is $1$ but the sum over $f$ is infinite.



EDIT: One of the conditions we wanted to prove has now been moved to a hypothesis, but this still doesn't result in a true statement.



Let $f(n)=frac{6}{pi^2n^2}$ and $g(n,m)=frac{pi^2n^2}{6}cdot2^{-n}cdot2^{-m}$. Then the double sum is $1$, and the sum over $f$ is $1$, but the double sum over $g$ is not, nor is the sum over any slice of $g$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What about the corrected version of the question? Anybody able to help?
    $endgroup$
    – J. Reinhard
    Dec 28 '18 at 18:41






  • 1




    $begingroup$
    @J.Reinhard I have made an edit to attempt to answer your question. It still appears that you cannot prove what you want.
    $endgroup$
    – DudeMan
    Dec 28 '18 at 18:46



















1












$begingroup$

Of course not.



If $f(x)=1$ then you're saying that $sum_{x,y} g(x,y) = 1$ implies that $sum_y g(x,y)=1$. But this is only possible if there is only one $x$.



Edit: This is an answer for the edited version. Still the answer is no.



Suppose there are two $x$'s and two $y$'s then:



Let $f(x_1)=frac{1}{2}$ and $f(x_2)=frac{1}{2}$ then $sum_x f(x) = 1$.



Now take $g(x_1,y_1) = 0.1, g(x_1 ,y_2) = 0.1, g(x_2,y_1) = 0.9, g(x_2,y_2)=0.9$



Then $sum_y g(x_1,y)=g(x_1,y_1)+g(x_1,y_2) = 0.2$ which is not $1$.



However $$sum_{x,y}f(x)g(x,y)= f(x_1)g(x_1,y_1) + f(x_1)g(x_1,y_2) + f(x_2)g(x_2,y_1)+f(x_2)g(x_2,y_2) =$$$$= frac{1}{2}cdot 0.2 + frac{1}{2}cdot 1.8 = 1$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What about the corrected version of the question? Anybody able to help?
    $endgroup$
    – J. Reinhard
    Dec 28 '18 at 18:41






  • 1




    $begingroup$
    @J.Reinhard Still no, look at my edited answer.
    $endgroup$
    – Yanko
    Dec 28 '18 at 18:48











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

There was already a very good finite counterexample given in the comments to the OP, but here is an infinite version:



Let $f(n)=frac{1}{n}$ and $g(n,m)=ncdot2^{-n}cdot2^{-m}$. Then the double sum is $1$ but the sum over $f$ is infinite.



EDIT: One of the conditions we wanted to prove has now been moved to a hypothesis, but this still doesn't result in a true statement.



Let $f(n)=frac{6}{pi^2n^2}$ and $g(n,m)=frac{pi^2n^2}{6}cdot2^{-n}cdot2^{-m}$. Then the double sum is $1$, and the sum over $f$ is $1$, but the double sum over $g$ is not, nor is the sum over any slice of $g$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What about the corrected version of the question? Anybody able to help?
    $endgroup$
    – J. Reinhard
    Dec 28 '18 at 18:41






  • 1




    $begingroup$
    @J.Reinhard I have made an edit to attempt to answer your question. It still appears that you cannot prove what you want.
    $endgroup$
    – DudeMan
    Dec 28 '18 at 18:46
















2












$begingroup$

There was already a very good finite counterexample given in the comments to the OP, but here is an infinite version:



Let $f(n)=frac{1}{n}$ and $g(n,m)=ncdot2^{-n}cdot2^{-m}$. Then the double sum is $1$ but the sum over $f$ is infinite.



EDIT: One of the conditions we wanted to prove has now been moved to a hypothesis, but this still doesn't result in a true statement.



Let $f(n)=frac{6}{pi^2n^2}$ and $g(n,m)=frac{pi^2n^2}{6}cdot2^{-n}cdot2^{-m}$. Then the double sum is $1$, and the sum over $f$ is $1$, but the double sum over $g$ is not, nor is the sum over any slice of $g$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What about the corrected version of the question? Anybody able to help?
    $endgroup$
    – J. Reinhard
    Dec 28 '18 at 18:41






  • 1




    $begingroup$
    @J.Reinhard I have made an edit to attempt to answer your question. It still appears that you cannot prove what you want.
    $endgroup$
    – DudeMan
    Dec 28 '18 at 18:46














2












2








2





$begingroup$

There was already a very good finite counterexample given in the comments to the OP, but here is an infinite version:



Let $f(n)=frac{1}{n}$ and $g(n,m)=ncdot2^{-n}cdot2^{-m}$. Then the double sum is $1$ but the sum over $f$ is infinite.



EDIT: One of the conditions we wanted to prove has now been moved to a hypothesis, but this still doesn't result in a true statement.



Let $f(n)=frac{6}{pi^2n^2}$ and $g(n,m)=frac{pi^2n^2}{6}cdot2^{-n}cdot2^{-m}$. Then the double sum is $1$, and the sum over $f$ is $1$, but the double sum over $g$ is not, nor is the sum over any slice of $g$.






share|cite|improve this answer











$endgroup$



There was already a very good finite counterexample given in the comments to the OP, but here is an infinite version:



Let $f(n)=frac{1}{n}$ and $g(n,m)=ncdot2^{-n}cdot2^{-m}$. Then the double sum is $1$ but the sum over $f$ is infinite.



EDIT: One of the conditions we wanted to prove has now been moved to a hypothesis, but this still doesn't result in a true statement.



Let $f(n)=frac{6}{pi^2n^2}$ and $g(n,m)=frac{pi^2n^2}{6}cdot2^{-n}cdot2^{-m}$. Then the double sum is $1$, and the sum over $f$ is $1$, but the double sum over $g$ is not, nor is the sum over any slice of $g$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 28 '18 at 18:45

























answered Dec 28 '18 at 18:12









DudeManDudeMan

1113




1113












  • $begingroup$
    What about the corrected version of the question? Anybody able to help?
    $endgroup$
    – J. Reinhard
    Dec 28 '18 at 18:41






  • 1




    $begingroup$
    @J.Reinhard I have made an edit to attempt to answer your question. It still appears that you cannot prove what you want.
    $endgroup$
    – DudeMan
    Dec 28 '18 at 18:46


















  • $begingroup$
    What about the corrected version of the question? Anybody able to help?
    $endgroup$
    – J. Reinhard
    Dec 28 '18 at 18:41






  • 1




    $begingroup$
    @J.Reinhard I have made an edit to attempt to answer your question. It still appears that you cannot prove what you want.
    $endgroup$
    – DudeMan
    Dec 28 '18 at 18:46
















$begingroup$
What about the corrected version of the question? Anybody able to help?
$endgroup$
– J. Reinhard
Dec 28 '18 at 18:41




$begingroup$
What about the corrected version of the question? Anybody able to help?
$endgroup$
– J. Reinhard
Dec 28 '18 at 18:41




1




1




$begingroup$
@J.Reinhard I have made an edit to attempt to answer your question. It still appears that you cannot prove what you want.
$endgroup$
– DudeMan
Dec 28 '18 at 18:46




$begingroup$
@J.Reinhard I have made an edit to attempt to answer your question. It still appears that you cannot prove what you want.
$endgroup$
– DudeMan
Dec 28 '18 at 18:46











1












$begingroup$

Of course not.



If $f(x)=1$ then you're saying that $sum_{x,y} g(x,y) = 1$ implies that $sum_y g(x,y)=1$. But this is only possible if there is only one $x$.



Edit: This is an answer for the edited version. Still the answer is no.



Suppose there are two $x$'s and two $y$'s then:



Let $f(x_1)=frac{1}{2}$ and $f(x_2)=frac{1}{2}$ then $sum_x f(x) = 1$.



Now take $g(x_1,y_1) = 0.1, g(x_1 ,y_2) = 0.1, g(x_2,y_1) = 0.9, g(x_2,y_2)=0.9$



Then $sum_y g(x_1,y)=g(x_1,y_1)+g(x_1,y_2) = 0.2$ which is not $1$.



However $$sum_{x,y}f(x)g(x,y)= f(x_1)g(x_1,y_1) + f(x_1)g(x_1,y_2) + f(x_2)g(x_2,y_1)+f(x_2)g(x_2,y_2) =$$$$= frac{1}{2}cdot 0.2 + frac{1}{2}cdot 1.8 = 1$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What about the corrected version of the question? Anybody able to help?
    $endgroup$
    – J. Reinhard
    Dec 28 '18 at 18:41






  • 1




    $begingroup$
    @J.Reinhard Still no, look at my edited answer.
    $endgroup$
    – Yanko
    Dec 28 '18 at 18:48
















1












$begingroup$

Of course not.



If $f(x)=1$ then you're saying that $sum_{x,y} g(x,y) = 1$ implies that $sum_y g(x,y)=1$. But this is only possible if there is only one $x$.



Edit: This is an answer for the edited version. Still the answer is no.



Suppose there are two $x$'s and two $y$'s then:



Let $f(x_1)=frac{1}{2}$ and $f(x_2)=frac{1}{2}$ then $sum_x f(x) = 1$.



Now take $g(x_1,y_1) = 0.1, g(x_1 ,y_2) = 0.1, g(x_2,y_1) = 0.9, g(x_2,y_2)=0.9$



Then $sum_y g(x_1,y)=g(x_1,y_1)+g(x_1,y_2) = 0.2$ which is not $1$.



However $$sum_{x,y}f(x)g(x,y)= f(x_1)g(x_1,y_1) + f(x_1)g(x_1,y_2) + f(x_2)g(x_2,y_1)+f(x_2)g(x_2,y_2) =$$$$= frac{1}{2}cdot 0.2 + frac{1}{2}cdot 1.8 = 1$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What about the corrected version of the question? Anybody able to help?
    $endgroup$
    – J. Reinhard
    Dec 28 '18 at 18:41






  • 1




    $begingroup$
    @J.Reinhard Still no, look at my edited answer.
    $endgroup$
    – Yanko
    Dec 28 '18 at 18:48














1












1








1





$begingroup$

Of course not.



If $f(x)=1$ then you're saying that $sum_{x,y} g(x,y) = 1$ implies that $sum_y g(x,y)=1$. But this is only possible if there is only one $x$.



Edit: This is an answer for the edited version. Still the answer is no.



Suppose there are two $x$'s and two $y$'s then:



Let $f(x_1)=frac{1}{2}$ and $f(x_2)=frac{1}{2}$ then $sum_x f(x) = 1$.



Now take $g(x_1,y_1) = 0.1, g(x_1 ,y_2) = 0.1, g(x_2,y_1) = 0.9, g(x_2,y_2)=0.9$



Then $sum_y g(x_1,y)=g(x_1,y_1)+g(x_1,y_2) = 0.2$ which is not $1$.



However $$sum_{x,y}f(x)g(x,y)= f(x_1)g(x_1,y_1) + f(x_1)g(x_1,y_2) + f(x_2)g(x_2,y_1)+f(x_2)g(x_2,y_2) =$$$$= frac{1}{2}cdot 0.2 + frac{1}{2}cdot 1.8 = 1$$






share|cite|improve this answer











$endgroup$



Of course not.



If $f(x)=1$ then you're saying that $sum_{x,y} g(x,y) = 1$ implies that $sum_y g(x,y)=1$. But this is only possible if there is only one $x$.



Edit: This is an answer for the edited version. Still the answer is no.



Suppose there are two $x$'s and two $y$'s then:



Let $f(x_1)=frac{1}{2}$ and $f(x_2)=frac{1}{2}$ then $sum_x f(x) = 1$.



Now take $g(x_1,y_1) = 0.1, g(x_1 ,y_2) = 0.1, g(x_2,y_1) = 0.9, g(x_2,y_2)=0.9$



Then $sum_y g(x_1,y)=g(x_1,y_1)+g(x_1,y_2) = 0.2$ which is not $1$.



However $$sum_{x,y}f(x)g(x,y)= f(x_1)g(x_1,y_1) + f(x_1)g(x_1,y_2) + f(x_2)g(x_2,y_1)+f(x_2)g(x_2,y_2) =$$$$= frac{1}{2}cdot 0.2 + frac{1}{2}cdot 1.8 = 1$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 28 '18 at 18:47

























answered Dec 28 '18 at 18:11









YankoYanko

7,0381629




7,0381629












  • $begingroup$
    What about the corrected version of the question? Anybody able to help?
    $endgroup$
    – J. Reinhard
    Dec 28 '18 at 18:41






  • 1




    $begingroup$
    @J.Reinhard Still no, look at my edited answer.
    $endgroup$
    – Yanko
    Dec 28 '18 at 18:48


















  • $begingroup$
    What about the corrected version of the question? Anybody able to help?
    $endgroup$
    – J. Reinhard
    Dec 28 '18 at 18:41






  • 1




    $begingroup$
    @J.Reinhard Still no, look at my edited answer.
    $endgroup$
    – Yanko
    Dec 28 '18 at 18:48
















$begingroup$
What about the corrected version of the question? Anybody able to help?
$endgroup$
– J. Reinhard
Dec 28 '18 at 18:41




$begingroup$
What about the corrected version of the question? Anybody able to help?
$endgroup$
– J. Reinhard
Dec 28 '18 at 18:41




1




1




$begingroup$
@J.Reinhard Still no, look at my edited answer.
$endgroup$
– Yanko
Dec 28 '18 at 18:48




$begingroup$
@J.Reinhard Still no, look at my edited answer.
$endgroup$
– Yanko
Dec 28 '18 at 18:48


















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