Is it possible to find four real numbers, for which $a+b

Question

For those interested, this question arises because in markov random fields one encounteres a lot of algorithms which look vor variables that minimize the potential of multiple cliques. So if a variable $X$ gives the clique potentials $a$ and $b$ these algorithms would choose this over a variable $Y$ that gives clique potentials $c$ and $d$, if $a+b<c+d$.

From a probabilistic point of view, it would make more sense to choose $Y$ if $ {\rm e}^{-a} + {\rm e}^{-b} <{\rm e}^{-c} + {\rm e}^{-d} $

There are more examples but for any given $a,b,$ you can choose any $c < -\log_e(e^{-a}+e^{-b})$ and any $d>a+b-c$ — Henry, Feb 28 '22 at 15:16
Thank you all very much for your answers, it really helped me ! — kolo kolo, Feb 28 '22 at 15:39

score 2 · Accepted Answer · answered Feb 28 '22 at 15:07

2

Certainly. You just need $c$ small and $d$ large compared to $a,b$. Try $a=b=1,c=0,d=3$

answered Feb 28 '22 at 15:07

Ross Millikan

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Is it possible to find four real numbers, for which $a+b

1 Answers1