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Let $W(y/x)$ be a conditional probability distribution where $x \in \{0, 1\}$ and $y$ is arbitrary but discrete, then Bhattacharyya coefficient is given by $$ B(W) = \sum_{y} \sqrt{W(y/0)W(y/1)} $$ Suppose I want to form another distribution $D(y/x) = \gamma(x) f(y/x)W(y/x)$ where $\gamma(x) \in \mathbb{R}$ and $f: \mathbb{R} \to \mathbb{R}$. I introduced $\gamma(x)$ to make $D(y/x)$ a valid distribution by making $$ \gamma(x) = \frac{1}{ \sum_{y} f(y/x)W(y/x)} $$

Under what conditions on $f(y/x)$ and/or $\gamma(x)$ the following will be true?

$$ B(D) \ge B(W) $$

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Surely you realize that $\gamma$ does depend on $x$, in general. Hence the definition of $D$ is odd... –  Did Jun 19 '12 at 17:48
    
you are right. I forgot to mention that $f$ also depends on $x$, i.e., $W(y/0)$ is multiplied by $\gamma(0) f(y/0)$ and $W(y/1)$ is multiplied by $\gamma(1) f(y/1)$ –  ubaabd Jun 19 '12 at 18:34
    
I do not follow: $D(y\mid x)=\gamma(x)f(y)W(y\mid x)$ makes sense, hence one does not have to introduce a dependence of $f$ on $x$. // About the question itself: since the transformation $W\mapsto D$ is reversible (at least for nonzero values of the $f(y)$s), some additional hypothesis is needed to get $B(D)\geqslant B(W)$. You might want to explain the kind you have in mind. –  Did Jun 20 '12 at 6:17
    
I do not follow why do we need further hypothesis on $f(y)$ because I can form another distribution with reversible mapping, e.g., $D(y/x)=W(y+x/x)$ and in this case $B(D) \ge B(W)$. I am thinking in terms of overlap between the two distributions $W(y/0)$ and $W(y/1)$. If I perform the above mentioned transformation, the overlap between the two distributions $D(y/0), D(y/1)$ will increase and so will the Bhattacharyya coefficent. But, just to be clearly explicit, the function I have in my mind is $$ f(y) = \frac{1}{1+e^{-|y|}}. $$ –  ubaabd Jun 20 '12 at 16:16
    
Also, the dependence on $x$ was to phrase the fact that $W(y/0)$ is multiplied by a function other than one that is multiplied by $W(y/1)$. In plain words, if I multiply $W(y/0)$ and $W(y/1)$ by different functions$f(y/0), f(y/1)$, respectively, what are the conditions on $f(y/x)$ under which Bhattacharyya coefficent will increase. I phrased the problem in a general setting, even though, as described above I am multiplying both distributions by the same function. –  ubaabd Jun 20 '12 at 16:22

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