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Dieter Kadelka
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A necessary assumption will be that $f_m > 0$. Further, assume that each $f_m$ is constant. Then there is nothing to optimize and you have to compare $\alpha_2$, which may go to $-\infty$ and $\alpha_1 > 0$. So without any essential additional assumptions there is no bound. Of course, if you assume that each $f_m > \beta_m$ (not necessarily constant) for some $\beta_m > 0$ then you have $|\alpha_1 - \alpha_2| \leq \sum_{m=1}^M (\log(1+\beta_m) - \log(\beta_m)) < \infty$.

A necessary assumption will be that $f_m > 0$. Further, assume that each $f_m$ is constant. Then there is nothing to optimize and you have to compare $\alpha_2$, which may go to $-\infty$ and $\alpha_1 > 0$. So without any essential additional assumptions there is no bound.

A necessary assumption will be that $f_m > 0$. Further, assume that each $f_m$ is constant. Then there is nothing to optimize and you have to compare $\alpha_2$, which may go to $-\infty$ and $\alpha_1 > 0$. So without any essential additional assumptions there is no bound. Of course, if you assume that each $f_m > \beta_m$ (not necessarily constant) for some $\beta_m > 0$ then you have $|\alpha_1 - \alpha_2| \leq \sum_{m=1}^M (\log(1+\beta_m) - \log(\beta_m)) < \infty$.

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Dieter Kadelka
  • 2.1k
  • 2
  • 11
  • 15

A necessary assumption will be that $f_m > 0$. Further, assume that each $f_m$ is constant. Then there is nothing to optimize and you have to compare $\alpha_2$, which may go to $-\infty$ and $\alpha_1 > 0$. So without any essential additional assumptions there is no bound.