Timeline for Existence of solution for this set of polynomial equations
Current License: CC BY-SA 3.0
12 events
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| Dec 1, 2014 at 15:04 | comment | added | R B | And we may assume that $p_r>0$ (updated in the question). | |
| Dec 1, 2014 at 15:01 | comment | added | R B | Thanks Pietro. Indeed, the positively values $t_i$ correspond to symmetric equilibrium strategies in a game I'm studying, and $p_if(t_i)$ is the utility of playing strategy $i$ when others play the symmetric strategy, this is why I require that $p_if(t_i)$ will be constant for the strategies in the support of the equilibrium, but if I could identify when a solution exist I'll be able to find an equilibrium (or equivalently, the support). In short, what I'd really like is to characterize is which $t_i$'s are positive, given an input vector $p$ (what is the smallest $j$ such that $t_j=0$). | |
| Dec 1, 2014 at 14:17 | comment | added | Pietro Majer | (I edited and made the minimal changes needed to get a non-incorrect statement. A complete discussion could be made starting from the preceding comment). | |
| Dec 1, 2014 at 14:11 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Dec 1, 2014 at 14:08 | comment | added | Pietro Majer | So the complete discussion should take into account that some $t_{i_0}$ can be zero. It seems this has a double effect: it makes the ${i_0}$-th equation (and in case $i_0>0$, the ${i_0-1}$-th equation too) automatically satisfied. Moreover, $p_i f(t_i)$ must be constant only in any interval of $i$ corresponding to non-vanishing $t_i$. | |
| Dec 1, 2014 at 13:47 | comment | added | Pietro Majer | You are right: writing the equations $p_{i}f(t_i) =p_{i+1 }f(t_{i+1}) $ excludes the case where some $t_i$ vanishes. | |
| Dec 1, 2014 at 13:06 | comment | added | R B | Actually, consider the following input: $p_1=p_2=\frac{4}{11}$, $p_3=\frac{3}{11}$ and $n=2$. This gives us $t_1=t_2=0.5$ and $t_3=0$, despite the fact that $p_3\geq \frac{p_1}{2}$. What am I missing? | |
| Dec 1, 2014 at 12:40 | comment | added | R B | I think that $p_r\geq \frac{p_1}{n}$ is what makes sure $t_r$ doesn't vanish, doesn't it? It might also suggest that a solution is still unique, where $t_i=0$ for all $i$ such that $p_i\geq \frac{p_1}{n}$ . | |
| Dec 1, 2014 at 11:26 | comment | added | Pietro Majer | Note that assuming $p_r>0$ (hence $p_i>0$ for all $i$) implies that no $t_i$ can vanish, so we didn't loose solutions. If some $p_i$ vanishes, a small discussion is required, but in general uniqueness drops (e.g., if all $p_i$ vanish) . | |
| Dec 1, 2014 at 11:12 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Dec 1, 2014 at 11:03 | history | edited | Pietro Majer | CC BY-SA 3.0 |
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| Dec 1, 2014 at 10:40 | history | answered | Pietro Majer | CC BY-SA 3.0 |