2c why is it not always true?
http://adv-ml-2017.wikidot.com/forum/t-2226047/2c-why-is-it-not-always-true
Posts in the discussion thread "2c why is it not always true?"Tue, 25 Jan 2022 17:17:06 +0000http://adv-ml-2017.wikidot.com/forum/t-2226047#post-2798266(no title)
http://adv-ml-2017.wikidot.com/forum/t-2226047/2c-why-is-it-not-always-true#post-2798266
Sat, 22 Apr 2017 11:43:41 +0000student
Not really, since the dependency on phi{x_i, x_j} together with the sum over x_j is actually m_{ji} which mu_i depends on, unless u mean it's m(t+1)_{ij} so the fix point is needed…
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http://adv-ml-2017.wikidot.com/forum/t-2226047#post-2794398Re: 2c why is it not always true?
http://adv-ml-2017.wikidot.com/forum/t-2226047/2c-why-is-it-not-always-true#post-2794398
Mon, 17 Apr 2017 22:30:46 +0000Amir Globerson2872628
When you sum mu_{ij} over x_j you get some function of x_i, which presumably depends on phi(x_i,x_j) Now you want to show that that is equal to mu_i, which does not explicitly depend on phi. Take for example the case where all messages are set to one. Then what you want to prove in 2c will not be correct. right?
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http://adv-ml-2017.wikidot.com/forum/t-2226047#post-27930822c why is it not always true?
http://adv-ml-2017.wikidot.com/forum/t-2226047/2c-why-is-it-not-always-true#post-2793082
Sun, 16 Apr 2017 08:16:19 +0000student
Sum_xj(uij) goes like Pie_k(mki). (Since it goes like sum_xj{phiij*pie(mki)*pie(mki)} … Unless I miss some "wrong" equisons) ui also goes like that pie, so the both r equal, no? Regardless the fix point
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