I have a set of values $(v_1,v_2,..,v_n)$ positive integers. I want to assign probabilities depending on the value, with the lowest having the highest probability. The sum of probabilities should be $1$ for the whole set.

I can do it the other way around if I find $t = \sum v_i$ then $p(v_i) = \frac{v_i}{t}$ but the lowest value has the least probability.


For any finite set $\mathcal V = \{v_1, v_2, \ldots v_n\}$, you simply need to define $$p_i = g(v_i)$$ where $g(\cdot)$ is any function such that

  • $g(v) > 0$ for all $v \in \mathcal V$
  • $x < y$ implies that $g(x) > g(y)$ (monotonicity)

Then you can define the probability mass function $$p(v) = \begin{cases} \frac{v_i}{\sum_{u\in\mathcal V}g(u)}, & v = v_i \in \mathcal V \\ 0, & v \notin \mathcal V \end{cases}$$

Two simple functions with this property are $$g_1(v) = q^v, \ q \in (0,1)$$ and $$g_2(v) = v^{-s}, \ s > 0.$$ For $\mathcal V = \mathbb N$, these functions correspond to the geometric distribution and Zipfs law (when $s > 1$) respectively. The process described above truncates the distribution to a finite set of values.

Also note that $\mathcal V$ can be countably infinite by requiring

  • $\sum_{v\in\mathcal V}g(v) < \infty$

As a concrete example, suppose that the set of positive integers is $\{3, 5, 8, 16\}$ and you choose Zipfs law with $s=1$. Then $$p_t \propto \begin{cases} 1/3, &t=1 \\ 1/5, &t=2 \\ 1/8, &t=3 \\ 1/16, &t=4 \end{cases}$$ which gives probabilities $$P(V=v) = \begin{cases} 0.4624, &v=3 \\ 0.2775, &v=5 \\ 0.1734, &v=8 \\ 0.0867, &v=16 \\ 0, & \text{otherwise} \end{cases}$$

  • $\begingroup$ can the geometric distribution method be used on a set of random integers like the example for Zipfs law? $\endgroup$ – yolo expectz Nov 8 '20 at 20:19
  • $\begingroup$ Yes of course, you just need to select a parameter $p$. I can't give any advice on that without more information. In the example above, choosing $p=1/2$ gives probabilities $0.7804, 0.1951, 0.0244, 0.0001$. $\endgroup$ – knrumsey Nov 8 '20 at 20:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.