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














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




















  • 2




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






  • 1




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












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
















-2














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




















  • 2




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






  • 1




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












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














-2












-2








-2







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















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

11




11








  • 2




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






  • 1




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












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














  • 2




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






  • 1




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












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








2




2




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




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




1




1




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






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














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




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










2 Answers
2






active

oldest

votes


















2














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























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






  • 1




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



















1














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























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






  • 1




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











Your Answer





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2 Answers
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2 Answers
2






active

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active

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active

oldest

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2














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























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






  • 1




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
















2














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























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






  • 1




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














2












2








2






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














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












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






  • 1




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


















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






  • 1




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
















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




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




1




1




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




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











1














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























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






  • 1




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
















1














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























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






  • 1




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














1












1








1






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














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

6,469727




6,469727












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






  • 1




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


















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






  • 1




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
















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




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




1




1




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




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


















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Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
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