How would one go about solving the following given that the function h(x) isn’t provided in the question? I’m at a loss on where to begin.

Suppose h(x) is such that h(x) > 0 for x = 1,2,3,...,I. Argue that $p(x) = h(x)/ \sum_{i=1}^I h(i)$ is a valid pmf


1 Answer 1


The fact that $h(x)$ isn't provided is a strong hint: it doesn't matter what $h(x)$ is except that it's always positive ($\geq 0$ would also be ok).

First, why is $p(x)>0$ always true? Second, what's the other property that a pmf has to have?

  • $\begingroup$ Thank you! However I’m still stuck on the second portion. The second property is that the sum of all of the probabilities where the function is defined is 1 but I’m unsure how to show this without knowing the function. $\endgroup$
    – Ronjondon
    Sep 6, 2020 at 15:24
  • $\begingroup$ Ok. The sum of $h(x)$ is something, call it $H$. $p(x)=h(x)/H$, so what's the sum of $p(x)$. $\endgroup$ Sep 8, 2020 at 2:36

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