Transform a probability distribution to a special "reversed form"

Question

Suppose we have a distribution $p(x_i),i\in[1,N]$. I want to transform it into a "reversed" distribution $q(x_i)$. By reverse, I mean if $p(x_i)$ is high, then $q(x_i)$ is low. I think it can be formally expressed as: $$\begin{aligned} \sum_{i=1}^N p(x_i)&=1\\ q(x_i)&=f(p(x_i))\\ \sum_{i=1}^N q(x_i)&=1\\ p(x_a)\ge p(x_b) &\Rightarrow q(x_a)\le q(x_b) \end{aligned}$$ Is there any existing transform of $f$. If not, how can I find the $f$?

Answer 1

I am not aware if there are "usual" such transformation but you can build one quite easily as follow: reverse the probability (without caring about the sum equal 1). and then normalize to get a sum equal 1.

For example you can define $q'(x_i)=(1-p(x_i))$.

Thus $S = \sum_i (1-p(x_i))$ will be used for normalization.

And you obtain $q(x_i)=q'(x_i)/S=(1-p(x_i))/S$ that fulfil your conditions.