# Does the law of total probability apply to hazards?

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

$$h(t|U=u)=\lim_{\Delta t\rightarrow0}\frac{P(tt,U=u)}{\Delta t}$$ 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:

$$h(t)=\lim_{\Delta t\rightarrow0}\frac{P(tt)}{\Delta t},$$

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 $$1/n$$):

$$h_n(t|U=u)=nP(tt,U=u),$$ 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(tt,U=u)f_U(u)du \\ &=\lim_{n\rightarrow\infty}\int_\mathcal{U}nP(tt,U=u)f_U(u)du \end{align}

which, by law of total probability, equals

$$\lim_{n\rightarrow\infty}nP(tt)=h(t).$$

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.

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

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

• 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? Dec 23, 2020 at 1:07
• 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. Dec 23, 2020 at 8:27