1
$\begingroup$

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

$\endgroup$

1 Answer 1

4
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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