Why is the Hazard function not a pdf?

Question

I am trying to understand why the hazard function is not a PDF. For a random variable T, people often define the PDF of this random variable as:

$$f(t)=\lim_{\delta \to 0} \frac{P(t\leqslant T <t+\delta)}{\delta}$$

By this definition, the hazard should also be a conditional PDF.

$$\lambda(t)=\lim_{\delta \to 0} \frac{P(t\leqslant T < t+\delta \mid T\geqslant t)}{\delta}$$

Seems like the two functions share the same type of definition! They are the limit of a probability. So why is one a PDF and the other not?

I guess is the reason this is not a PDF because the conditioning is not on a single event $T=t$ but rather on $T\geqslant$? If it were on a single event $T=t$, would this be a PDF?

Answer 1

The argument of a conditional pdf cannot depend on the conditioning event in any way, shape or form. In $$f_{T\mid A}(t\mid A) = \lim_{\delta\to 0} \frac{P\{t < T \leq t+\delta\mid A\}}{\delta},$$ $A$ can be a fixed event such as $\{T>5\}$ but not something that depends on $t$ such as $\{T > t\}$.

Another important reason why a hazard function $h(t)$ (or any scalar submultiple thereof) cannot possibly be a pdf is that $$\int_0^\infty h(t)\, \mathrm dt = \infty$$ whereas pdfs of lifetimes have more mundane values for their integrals: $$\int_0^\infty f_T(t)\, \mathrm dt = 1.$$

Answer 2

I think the counterexample Björn suggested is a enough to answer the question. Let me write it out in more detail:

Björn's Counterexample

Let $T\stackrel{d}{=}\mathrm{Exp}(\lambda)$ for any $\lambda > 0$ then $f_T(t) = \lambda e^{-\lambda t}$ which implies $S(t) = \mathcal{P}(T>t) = 1-(1-e^{-\lambda t}) = e^{-\lambda t}$.

The hazard function can be calculated as $\lambda(t) = \dfrac{f_T(t)}{S(t)} = \dfrac{\lambda e^{-\lambda t}}{e^{-\lambda t}} = \lambda$. (hence the notation $\lambda(t)$ for a hazard function).

Clearly this implies that $\lambda(t)$ is not a probability density function, since $\int_0^{+\infty} \lambda(u) \mathrm{d}u \not = 1$.

Back to your question(s)

The first part of your question was:

`Why is one a pdf and the other not'

As far as I remember, $f_T(t)$ is defined as a pdf if certain conditions hold. One of these conditions is $\int_\Omega f_T(t) \mathrm{d} t =1$. (Which can clearly be violated in the case of $\lambda(t)$).

The density function $f_T(t)$ can be viewed as the limit of a cumulative distribution function in a narrow timeslot, which is what the notation $f_T(t) = \lim_{\delta \to 0} \frac{\mathcal{P}(t\leqslant T<t+\delta)}{\delta}$ suggests. I would not call this the definition since it lacks the conditions.

I guess is the reason this is not a pdf because the conditioning is not on a single event $T=t$ but rather on $T\geqslant t$? If it were on a single event $T=t$, would this be a pdf?

Notice how:

$$ \mathcal{P}(t\leqslant T<t+\delta \mid T = t) = 1 $$ which implies that conditioning on $T=t$ isn't that usefull.

A conditional probability function $f_{Y\mid X}(y\mid X=x)$ for continious r.v. is defined as $$ f_{Y\mid X}(y\mid X=x) = \dfrac{f_{Y,X}(Y=y,X=x)}{f_X(x)} $$ and the hazard function does not fit here. You need (at least) two different random variables. You could look at something like this:

$$ f_{T \mid Z}(t\mid Z = 1) $$ for a certain $Z$ as an indicator of treatment or control. Then cleary this is a probability function.

Answer 3

The Hazard function can be equivalently seen as the ratio of the pdf $f(t)$ to the survival function $S(t)$:

$$\lambda(t)=\lim_{\delta \to 0} \frac{P(t\leqslant T < t+\delta \mid T\geqslant t)}{\delta}$$ by definition of conditional probability: $$=\lim_{\delta \to 0} \frac{P(t\leqslant T < t+\delta \cap T\geqslant t) / P(T \ge t)}{\delta}$$ $$=\lim_{\delta \to 0} \frac{P(t\leqslant T < t+\delta) / P(T \ge t)}{\delta}$$ by definition of $f(t):$ $$= \frac{f(t)}{P(T \ge t)}$$ by definition of survival function: $$ =\frac{f(t)}{S(t)}$$

Thus since the hazard function is the pdf scaled by the survival function, it will not generally integrate to $1$, and hence is not a pdf.