I'm reading up on reliability and I came across this question:

Show that if the hazard function is decreasing, the PDF, $f(t)$, is also a decreasing function and its mode must therefore occur at t = 0

I know that:

$h(t)\ = \frac{f(t)}{S(t)}$ where $h(t)$ is the hazard rate, $f(t)$ is the pdf and $S(t)$ is the reliability.

From the formula, we can infer that $f(t)$ increases with $h(t)$ so it's directly proportional and $s(t)$ decreases so it's indirectly proportional.

I also know that the mode of a probability density function is the point at which the maximum value occurs so I figure we're meant to derive some sort of formula and then set t=0 to get the maximum value, thereby proving that it is the mode.

What I'd like to know is how to derive a formula that proves that PDF decreases with the hazard function.

  • $\begingroup$ Could you state more specifically the statement you would like to prove? The relationship between $h$ and $f$ you give here is ordinarily taken to be the definition of $h,$ leaving nothing to prove. $\endgroup$ – whuber Mar 9 '19 at 18:38
  • $\begingroup$ @whuber I've updated the question $\endgroup$ – Jemima Mar 10 '19 at 21:45
  • $\begingroup$ Thank you. Because what you would like to prove isn't true, I suspect your reference must be making some additional assumptions about $f$: but what are they? $\endgroup$ – whuber Mar 10 '19 at 22:53
  • 1
    $\begingroup$ The pdf must be an asymptotically decreasing function regardless of the shape (constant, increasing, bath-tub etc or decreasing as in your case) of the hazard rate function. Perhaps you are asked to prove that if $h(t)$ is decreasing on $(0,\infty)$, then $f(t)$ is also decreasing on $(0,\infty)$ and hence has a mode at $0$? $\endgroup$ – Dilip Sarwate Mar 11 '19 at 3:29
  • $\begingroup$ Yes @DilipSarwate that's basically what I'm trying to prove $\endgroup$ – Jemima Mar 13 '19 at 23:06

It's tempting to show the derivative of $f$ must be non-positive: but there's no assurance $f$ is differentiable. Let's therefore attempt a direct comparison inspired by the finite difference: that is, to say that a function $f$ is decreasing literally means for all $t \ge 0$ and $\epsilon \gt 0,$

$$f(t+\epsilon) - f(t) \lt 0.$$

That's what we need to show. There's nothing available to try except to plug in the definitions. In the following, $S(t) = \int_t^\infty f(x)dx$ is the survival function, which we must assume to be nonzero. Let's write down the finite difference for $h,$ which we assume is negative:

$$\eqalign{ 0 \gt h(t+\epsilon) - h(t) &= \frac{f(t+\epsilon)}{S(t+\epsilon)} - \frac{f(t)}{S(t)}\\ &=\frac{f(t+\epsilon)S(t) - f(t)S(t+\epsilon)}{S(t+\epsilon)S(t)}. }$$

Clearly both $S(t)$ and $S(t+\epsilon)$ are positive, so we may ignore them when considering just the sign of the fraction. Let's focus on its numerator:

$$0 \gt f(t+\epsilon)S(t) - f(t)S(t+\epsilon) = \color{red}{(f(t+\epsilon)-f(t))S(t)} - \color{blue}{f(t)(S(t+\epsilon)-S(t))}.$$

This can be more simply written

$$\color{blue}{f(t)(S(t+\epsilon)-S(t))} \gt \color{red}{(f(t+\epsilon)-f(t))S(t)}.$$

Since $S(t+\epsilon) - S(t) = -\int_t^{t+\epsilon}f(x)dx \le 0$ and $f(t)\ge 0,$ the left hand side cannot be positive. The factor of $S(t)$ on the right hand side is positive (because it is nonzero). Therefore, the term it multiplies must be negative:

$$0 \gt f(t+\epsilon) - f(t),$$


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