Consider the hazard function for a random variable $T$, conditional on some other random variable $U$:

$$ h(t|U=u)=\lim_{\Delta t\rightarrow0}\frac{P(t<T<t+\Delta t|T>t,U=u)}{\Delta t} $$ where $U$ has probability density function $f_U(u)$. Suppose, given $h(t|U=u)$, we would like to derive the hazard function irrespective of $U$, what we could call the marginal hazard function:

$$ h(t)=\lim_{\Delta t\rightarrow0}\frac{P(t<T<t+\Delta t|T>t)}{\Delta t}, $$

My question is, can we apply the law of total probability in this setting? In other words, is it the case that $h(t)=E_u[h(t|U=u)]=\int_\mathcal{U}h(t|U=u)f_U(u)du?$

Here is my best attempt to answer the question. I'm not very familiar with measure theory so please forgive me if I get this wrong. Essentially, we are trying to bring a limit outside of a integral, which implies Lebesgue's dominated convergence theorem (DCT) might be useful. DCT states that, for a sequence of functions $f_n(x)$ with $n\in\mathbb{N}$, $\lim_{n\rightarrow\infty}\int f_n(x)dx=\int \lim_{n\rightarrow\infty} f_n(x)dx$, so long as there exists some function $g(x)$ such that $g(x)\geq|f_n(x)|$ for all $x, n$, where $|.|$ indicates absolute value.

To apply DCT, we can re-write the conditional hazard by first defining a sequence of functions (replacing $\Delta t$ with $1/n$):

$$ h_n(t|U=u)=nP(t<T<t+1/n|T>t,U=u), $$ for $n\in\mathbb{N}$. Then the conditional hazard is defined as the pointwise limit of this sequence:

$$ h(t|U=u)=\lim_{n\rightarrow\infty}h_n(t|U=u) $$

Then, the DCT indicates that, so long as some function $g(u)$ exists such that $g(u)\geq|h_n(t|U=u)|$ for all $n$ and $u$,

\begin{align} E_u[h(t|U=u)] &=\int_\mathcal{U}h(t|U=u)f_U(u)du \\ &=\int_\mathcal{U}\lim_{n\rightarrow\infty}nP(t<T<t+1/n|T>t,U=u)f_U(u)du \\ &=\lim_{n\rightarrow\infty}\int_\mathcal{U}nP(t<T<t+1/n|T>t,U=u)f_U(u)du \end{align}

which, by law of total probability, equals

$$ \lim_{n\rightarrow\infty}nP(t<T<t+1/n|T>t)=h(t). $$

It seems reasonable to assume that such a $g(u)$ exists - if I understand correctly, this would just require that the conditional hazard is finite.


1 Answer 1


One case where this implication fails: suppose $U$ is uniformly distributed on the open interval $(0, 1)$, and $T = U$. Let $t = 0$. Then conditional on any particular value $U = u$, the hazard is zero, since $P(T < v) = 0$ for all $v < u$. But the unconditional hazard $h(0) = 1$, since for all $0 \le d \le 1$, $\frac{P(T < d)}{d} = \frac{P(U < d)}{d} = 1$.

Admittedly your dominating function $g$ doesn't exist in this example, since as $u$ goes to zero the conditional hazard goes to infinity near zero. But for $t = 0$ the conditional hazard is never actually infinite.

I think your argument that the implication holds when $g$ exists is promising, with 2 possible gaps.

First, for DCT to apply, $g(u)$ has to dominate not just $h_n(u)$, but also $h_n(u) f_U(u)$. When the pdf of $U$ is unbounded this isn't a trivial difference.

Second, in a formal write-up, it might be worth explaining why the existence of the limit

$$\lim_{n \to \infty} nP(t < T < t + \frac{1}{n} | T > t)$$

implies the existence of the limit $h(t)$. For example, note that the existence of $h(t)$ is not implied by the existence of the limit

$$\lim_{n \to \infty} 2^nP(t < T < t + \frac{1}{2^n} | T > t).$$

  • $\begingroup$ Thanks for your answer. Could you help me see why the existence of $h(t)$ is not implied by the existence of your second limit? $\endgroup$ Dec 23, 2020 at 1:07
  • 1
    $\begingroup$ Consider the probability mass function $p(x) = x$ whenever $x = 2^{-n}$ for $n \in \mathbb N$. Then $\lim_{n \to \infty} \frac{2^n}{c} P(0 < X < 2^{-n}c) = \frac{1}{2c}$ for $c \in (0.5, 1]$. So $\frac{$P(0 < X < d)}{d}$ takes every value between 0.5 and 1 on any open interval containing 0, so the limit as $d$ goes to 0 isn't defined. There are also continuous counterexamples. $\endgroup$
    – fblundun
    Dec 23, 2020 at 8:27

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