# The limit of discrete hazard functions

Suppose there's a discrete hazard function $$\displaystyle h(t) = \frac{f(t)}{S(t)}$$ that is defined on a discrete domain $$\mathcal{T} = \{t_1 < t_2 < ... < t_n\}$$, where $$f(t)$$ is the PMF $$P(T_i = t)$$ and $$S(t)$$ is the complementary CDF $$P(T_i \geq t)$$.

It's easy to derive that if $$n<\infty$$, $$\displaystyle \frac{f(t_n)}{S(t_n)} = 1$$. But what if $$n \rightarrow \infty$$, is $$\displaystyle \frac{f(t_n)}{S(t_n)}$$ still 1?

• Hint: Consider $S(t)=C\exp(-t^2/2)$ where $(t_n)=(1,2,3,\ldots)$ and $C$ is a normalization constant (equal approximately to $1/0.753314$) that makes $S$ a CCDF.
– whuber
Commented Jun 7, 2020 at 15:32
• Thanks! I tried to take the derivative of $- S(t)$ as $f(t)$, and got the ratio as $t$, which will diverge to infinity as $t \rightarrow \infty$. But I'm not sure whether I could take the derivative since it is a discrete distribution. Commented Jun 7, 2020 at 19:28
• Taking the derivative is suggestive, isn't it? There are many other ways to show a sequence diverges, though. Why not compute $f(n),$ construct the ratio $f(n)/S(n),$ and examine that series?
– whuber
Commented Jun 7, 2020 at 20:09

For insight, consider a continuous distribution supported on the positive numbers with a nonzero differentiable survival function (complementary CDF) $$S$$ so that $$f(t) = -S^\prime (t)$$ for all $$t\gt 0.$$ In this case the condition

$$1 = \lim_{t\to \infty} \frac{f(t)}{S(t)} = -\lim_{t\to\infty} \frac{\mathrm{d}}{\mathrm{d}t} \log(S(t))$$

indicates that asymptotically (for large $$t$$) the survival function behaves exponentially.

Consequently, when the survival function decays at a different rate, we should expect different limiting behavior. Changing the rate can readily be achieved by scaling $$t,$$ suggesting we consider $$S(t)$$ proportional to $$\exp(-\lambda t)$$ for $$\lambda \gt 0.$$ Here,

$$f(t) \ \propto\ -\frac{\mathrm{d}}{\mathrm{d}t} \exp(-\lambda t)\ \propto\ \exp(-\lambda t).$$

Let us therefore examine its discrete analog given by

$$f(n) = C\, \exp(-\lambda n) = C\, a^n$$

for $$n = 1, 2, 3, \ldots$$ where $$a = \exp(-\lambda) \lt 1$$ and $$C$$ is a normalizing constant (which we won't have to compute since it will disappear in the ratio $$f/S;$$ all we need to check is that it's not infinite).

The survival function is easily computed as the sum of a geometric series

$$S(n) = \sum_{i=n}^\infty f(i) = C\sum_{i=n}^\infty a^i = \frac{Ca^n}{1-a},$$

whence

$$\frac{f(n)}{S(n)} = 1-a \lt 1.$$

Finally, since it is necessary that $$1=S(1)$$ we find $$C = 1-a,$$ verifying $$S$$ really defines a genuine probability distribution.

This exhibits explicit discrete distributions with constant hazard functions $$f/S$$ for any possible positive constant. In particular, their limiting values are not $$1.$$

Further reasoning along these lines will help you generate sequences of probabilities $$f(n),$$ $$n=1,2,3,\ldots,$$ for which the ratios $$f(n)/S(n)$$ have any specified limit on the interval $$[0,1]$$ or do not reach a limit at all. In particular,

• $$f(n) = 1/n-1/(n+1)$$ gives $$S(n)=1/n,$$ whence $$\lim_{n\to\infty} \frac{f(n)}{S(n)} = \lim_{n\to\infty} 1 - \frac{n}{n+1} = 0.$$

• $$f(n) = 1/(e\,(n-1)!)$$ (a shifted Poisson distribution) defines a distribution with countable support having a limiting value of $$1$$ for its hazard function.

• Create the sequence $$f(n)$$ by concatenating constant sequences of length $$2^i-1$$ and values $$2^{-\binom{i}{2}}$$ for $$i=1,2,3,\ldots.$$ This sequence begins $$f(n) = \left(\frac{1}{2}, \frac{1}{8}, \frac{1}{8}, \frac{1}{8}, \frac{1}{64}, \ldots, \frac{1}{64}, \frac{1}{1024}, \ldots\right)$$ where the constant subsequences have lengths $$1,$$ $$3,$$ $$7,$$ and so on. It sums to $$1,$$ making it a probability mass function. You may compute that $$f(2^i-i-1)/S(2^i-i-1) = 1/2$$ for $$i=2, 3, 4, \ldots$$ while $$f(2^i-i)/S(2^i-i) = 2^{-i}$$ grows arbitrarily small. Consequently $$f(n)/S(n)$$ has no limit.