Suppose that we have a probability distribution on the positive integers. I want to relate monotonicity of its hazard rate to convexity of a transformation of its generating function. I have a specific claim below. I am fairly certain that it is true, and if it is, it feels like there should be an elementary proof, but so far it has eluded me.

Let $F$ be a cdf on the positive integers, and let $f(k) = F(k) - F(k-1)$ be the associated probability function.

Define the hazard rate at k to be $\frac{f(k)}{1-F(k-1)}$ (this is the probability that the random variable is exactly k, given that it is at least k).

Given $F$, define the function $g: [0,1] \rightarrow \mathbb{R}$ by $g(x) = \sum_{k \geq 0} x^k (1-F(k))$

Claim: The function $1/g(x)$ is convex if the hazard rate of $F$ is increasing, and is concave if the hazard rate of $F$ is decreasing.

I mentioned generating functions, because $1 -(1-x) g(x) = \sum_{k \geq 0} x^k f(k)$ (the right side is the standard probability generating function).

The claim "f1 is convex if f2 is increasing, and f1 is concave if f2 is decreasing" basically makes it sound like f2 = f1', yet in this case the domain of $g$ is [0,1] and the domain of the hazard rate is the integers, so the connection isn't so clear. Any help appreciated - thanks!

  • $\begingroup$ You can format math using LaTeX, i.e. putting dollar signs around math. $\endgroup$ – usul Oct 26 '16 at 5:32

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