Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

One distance or discrimination of two probability distributions

$$ \textit{$P=(p_{1} ,...,p_{n} )$}$$ and $$\textit{$Q=(q_{1} ,...,q_{n} )$}$$ is the Variational distance defined by, $$ V(P,Q)=\sum _{i}|p_{i} -q_{i} | $$ It is fairly easy to see that V is a metric, and in particular that it satisfies the triangle inequality. I would like to introduce class priors, and define V', $$ V'(\alpha ,P,Q)=\sum _{i}|\alpha p_{i} -(1-\alpha )q_{i} | $$ and prove that V' (while not strictly a metric since the zero is not satisfied) satifies the triangle inequality,

$$V'(\alpha ,P,Q)+V'(\beta ,Q,R)\ge V'(\gamma ,P,R) $$ where $$\gamma =\frac{\alpha \beta }{\alpha \beta +(1-\alpha )(1-\beta )} $$

Any ideas?

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.