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Let $$ \gamma = \frac{1}{\sum_{y}f(y)W(y)}, $$


$$ f(y) = 1 + e^{-|y|} $$

and $W(y)$ is a probability distribution (unknown) with $y \in \mathcal{Y}$ arbitrary but discrete, and $x \in \{0,1\}$. I want to calculate a lower bound on $\gamma$. I came up with one lower bound as follows: \begin{aligned} \gamma &= \frac{1}{\sum_{y}f(y)W(y)}\\ &\ge \frac{1}{\sum_{y}f(y)}, \because W(y) \le 1, \forall y \in \mathcal{Y}\\ \end{aligned} I wanted a tighter bond than this. Any ideas or references are appreciated. Thank you !

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$1 \lt f(y) \le 2$

so $1=\sum_{y}W(y) \lt \sum_{y}f(y)W(y) \le \sum_{y}2W(y) =2$

so $\frac12 \le \gamma \lt 1$.

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Thanks! I completely overlooked that! Also, can I ask a related question here or should I post it separately as a new question? – ubaabd Jun 19 '12 at 22:38
@Aitezaz It depends on the question, but probably best to ask a new one and link back to this one. – Henry Jun 19 '12 at 23:11

As $$f(y) = 1 + e^{-|y|}\leq 2$$ then $$E[f(y)]=\sum_{y} f(y)W(y)\leq \sum_{y} 2 ~W(y)=2 $$ Hence $$\gamma = \frac{1}{\sum_{y}f(y)W(y)}\geq \frac{1}{2}$$

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