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I'm really struggling to answer myself the following question!

Is there an absolutely continuous probability distribution that is supported on a bounded interval [a,b], for which the hazard function $\frac{f(x)}{1-F(x)}$ is monotonically increasing but does not diverge to infinity, when $x$ approaches $b$.

I have had no luck with my attempts so far and would greatly appreciate an answer.

Clearly, the hazard function of the exponential distribution does not diverge to infinity, as it is constant, but its support is not bounded from above.

And if no such distribution can exist, how could that be proved.

Thanks in advance!

Projectsral
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  • Hi, so I tried some functions of the form $f(x)=ab - ax, a \leq x \leq b$ before and clearly for some $a,b$, this is a probability density function, for example for $a=2, b=1$, but I could not find any that satisfies the properties specified above. Did you have some specific choice for $a$ and $b$ that solves the problem, which I do not see at the moment. – Projectsral Oct 09 '23 at 09:45
  • Sorry, my example doesn't work. When $a=0,b=1$ the hazard function converges to $1$ but it is not increasing. – geetha290krm Oct 09 '23 at 09:58

1 Answers1

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Such a function can't exist. Already the finite limit condition

$$\lim_{x\to b} \frac{f(x)}{1-F(x)} < \infty$$ cannot be fulfilled.

Since $\lim_{x\to b} F(x) = 1$, the doniminator in the limit condition tends to $0$, so the enumarator $f(x)$ has to as well. Since $f$ is continous, we have $f(b)=0$.

For each $n \ge 1$, let $x_n$ be the largest $x \in [a,b]$ such that $f(x_n)=\frac1n$. They may not exist for each $n$, but since $f$ is continuous, non-negative and $f(b)=0$, but cannot be identical to $0$ on the whole interval $[a,b]$, they have to exist for all $n \ge N$ for some $N$. Our argument will never use any specific or "small" values of $n$, so that is enough.

Since $f(b)=0$ and $f$ is continuous, we have $f(x) \le \frac1n$ for $x\in [x_n,b]$ (equality only for $x=x_n$).

We have

$$1-F(x_n)=\int_{x_n}^b f(x)dx \le \frac1n (b-x_n) = f(x_n) (b-x_n)$$

and therefore

$$\frac{f(x_n)}{1-F(x_n)} \ge \frac1{b-x_n}.$$

By definition, the $x_n$ are increasing and bounded by $b$, so tend to a limit: $\lim_{n\to\infty}x_n=u.$

Now if $u < b$, that means your distribution isn't really supported on $[a,b]$, as $f$ can only be zero on the inverval $[u,b]$, by definition of $x_n$.

But $u = b$ implies that $ \frac1{b-x_n}$ and therefore $\frac{f(x_n)}{1-F(x_n)}$ will be unbounded.

Ingix
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