4
$\begingroup$

I am working on Problem 1.13 from Statistical Models and Methods for Lifetime Data, 2nd ed., by Jerald Lawless, and I have been stuck on part (b) for some time now; I would greatly appreciate any assistance. I will summarize the problem below.

Suppose that given $\lambda>0$, $T$ is an exponential random variable with hazard rate $\lambda$ so that the pdf $f(t\mid \lambda)$ is of the form $f(t\mid \lambda)=\lambda e^{-\lambda t}$ for $t\geq 0$. Consider a general setting where the distribution of $T\mid \lambda$ is exponential with rate $\lambda$ where $\lambda$ has a continuous distribution on $(0,\infty)$. We must show that the hazard function for the marginal distribution of $T$ is monotone decreasing.

Here is what I have done. First of all, we have by the Law of Total Probability that $$\operatorname P(T=t)=\int_0^\infty \operatorname P(T=t\mid \Lambda=\lambda)\operatorname P(\Lambda=\lambda)\,d\lambda=\int_0^\infty \lambda e^{-\lambda t}f(\lambda)\,d\lambda$$

We can also find that $$\operatorname P(T\geq t\mid \Lambda=\lambda)=1-\int_0^t \lambda e^{-\lambda w}\,dw=e^{-\lambda t}$$

Therefore, again by the Law of Total Probability, this leads to the survival function $$\operatorname P(T\geq t)=\int_0^\infty \operatorname P(T\geq t\mid \Lambda=\lambda)\operatorname P(\Lambda=\lambda)\,d\lambda=\int_0^\infty e^{-\lambda t}f(\lambda)\,d\lambda$$

We now define the hazard function as $h(t)=\operatorname P(T=t)/\operatorname P(T\geq t)$, and therefore $$h(t)=\frac{\int_0^\infty \lambda e^{-\lambda t}f(\lambda)\,d\lambda}{\int_0^\infty e^{-\lambda t}f(\lambda)\,d\lambda}$$

I assume everything up until this point has been correct, but here is where I start to become uncertain. My instinct now is to take the derivative wrt $t$ and show that it is always negative. Using the quotient rule, I found the following result:

$$h'(t)=\frac{-\left(\int_0^\infty e^{-\lambda t}f(\lambda)\,d\lambda\right)\left(\int_0^\infty \lambda^2 e^{-\lambda t}f(\lambda)\,d\lambda\right)+\left(\int_0^\infty \lambda e^{-\lambda t}f(\lambda)\,d\lambda\right)^2}{\left(\int_0^\infty e^{-\lambda t}f(\lambda)\,d\lambda\right)^2}$$

I believe that this is as far as I can simplify this expression. But it is unclear to me that this would necessarily be negative, as we have a difference of terms on the numerator. I do know that each of the squared integrals should be positive, and that $\lambda>0$; thus, the only way for the whole expression to be negative is if the magnitude of the first multiplied terms in the numerator (with the negative sign in front) is greater than the second term in the numerator. Is there a way I can conclusively show that $h'(t)<0$ for all $t$?

$\endgroup$
1
$\begingroup$

You have made quite a bit of progress; all that remains is to show that the numerator of your derivative $h'$ is negative. Expanding the products of integrals into double-integrals you get:

$$\begin{equation} \begin{aligned} \text{NUM} &\equiv \left(\int_0^\infty \lambda e^{-\lambda t}f(\lambda) \ d\lambda\right)^2 -\left(\int_0^\infty e^{-\lambda t}f(\lambda) \ d\lambda\right) \left(\int_0^\infty \lambda^2 e^{-\lambda t}f(\lambda) \ d\lambda\right) \\[6pt] &= \int_0^\infty \int_0^\infty \lambda \phi e^{-(\lambda+\phi) t} f(\lambda) f(\phi) \ d\lambda \ d\phi - \int_0^\infty \int_0^\infty \lambda^2 e^{-(\lambda+\phi) t} f(\lambda) f(\phi) \ d\lambda \ d\phi \\[6pt] &= \int_0^\infty \int_0^\infty \lambda (\phi - \lambda) e^{-(\lambda+\phi) t} f(\lambda) f(\phi) \ d\lambda \ d\phi. \\[6pt] \end{aligned} \end{equation}$$

Now, since the term $e^{-(\lambda+\phi) t} f(\lambda) f(\phi)$ is symmetric in the parameters, we can switch the two parameters in the remaining terms in the integrand to get:

$$\int_0^\infty \int_0^\infty \lambda (\phi - \lambda) e^{-(\lambda+\phi) t} f(\lambda) f(\phi) \ d\lambda \ d\phi = \int_0^\infty \int_0^\infty \phi (\lambda - \phi) e^{-(\lambda+\phi) t} f(\lambda) f(\phi) \ d\lambda \ d\phi.$$

Using this equation we then have:

$$\begin{equation} \begin{aligned} \text{NUM} &= \int_0^\infty \int_0^\infty \lambda (\phi - \lambda) e^{-(\lambda+\phi) t} f(\lambda) f(\phi) \ d\lambda \ d\phi \\[6pt] &= \frac{1}{2} \Bigg[ \int_0^\infty \int_0^\infty \lambda (\phi - \lambda) e^{-(\lambda+\phi) t} f(\lambda) f(\phi) \ d\lambda \ d\phi \\[6pt] &\quad \quad \quad - \int_0^\infty \int_0^\infty \phi (\phi - \lambda) e^{-(\lambda+\phi) t} f(\lambda) f(\phi) \ d\lambda \ d\phi \Bigg] \\[6pt] &= \frac{1}{2} \int_0^\infty \int_0^\infty (\lambda - \phi) (\phi - \lambda) e^{-(\lambda+\phi) t} f(\lambda) f(\phi) \ d\lambda \ d\phi \\[6pt] &= - \frac{1}{2} \int_0^\infty \int_0^\infty (\phi - \lambda)^2 e^{-(\lambda+\phi) t} f(\lambda) f(\phi) \ d\lambda \ d\phi < 0. \end{aligned} \end{equation}$$

This establishes that $h'(t) < 0$ for all $t > 0$ which shows that the hazard function is strictly decreasing.

$\endgroup$
-1
$\begingroup$

I like to use the eta function proposed by Ronald Glaser (1980) in his paper Bathtub and related Failure Rate Characterizations to understand the shape of hazard functions, because the hazard usually has complicated derivatives. The eta function uses only the density of $T$ and its derivative.

If $\eta '(t)<0$ for all $t$, then $h'(t)<0$ for all $t$, which proves that $h(t)$ is decreasing, where $$\eta(t)=-\frac{f'(t)}{f(t)}$$

In the case of $T \sim \mathrm{Exp}(\lambda)$ with $\lambda>0$, $$f(t)=\lambda e^{-\lambda t}$$ for $t\geq0$, and $$f'(t)=-\lambda^2e^{-\lambda t}$$ for $t\geq 0$ and so Glaser's eta function can be written as $$\eta (t)=\frac{1}{\lambda}$$ and its derivative is then $$\eta'(t)= -\frac{1}{\lambda^2}$$ which is negative, showing that $h(t)$ is monotone decreasing, as required. The proof that the eta function sufficiently shows this property of the hazard function is in Glaser's paper named above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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