Difference between hazard function and intensity function? I'm reviewing a paper whereby the nonhomogeneous poisson process is explored. I'm familiar with the intensity function, but the idea of the hazard function is throwing me off. Especially, I cannot figure out when they two are the same? Literature is sparse and I'm just becoming confused. To me, it seems like the intensity function deals with reoccurring failure, while the hazard function deals only with the time to first failure?
 A: From the theory on counting processes and survival analysis, we find (informally) a simple relation between the hazard rate and the intensity function (refer to [1]).
Definitions
Let $X_1,\cdots, X_n$ be a sample of $n$ uncensored continuously distributed survival times (e.g. time of failure after repair) with density function $f$ and distribution function $F$.
Hazard rate $\alpha$
The hazard rate $\alpha(t)$ is the probability that the event $X_i$ happens at time $t$ given that it has not happened before $t$. Loosely, this means: $P(X_i \text{ happens at time }t \text{ given the past})=P(X_i\in [t,t+dt)\vert X_i\geq t) =\alpha(t)dt $ and $\alpha$ can be expressed as $\alpha = \frac{f}{1-F}$.
Event-counting process $N(t)$
Now, let's define $N=\left(N(t)\right)_{t\geq 0}$ as the event-counting process related to $(X_i)_i$, i.e. $N$ counts the number of events $X_i$ that occurred before or at time $t$:
$$N(t)=\# \{ i : X_i\leq t \} $$
Intensity process $\lambda$
Then, the intensity process $\lambda$ of $N$ is "a measure of the rate of change of its predictable part" [2]. Loosely put, $\lambda(t)$ is the instantaneous ($dt\rightarrow 0$) expected number of events counted in the interval $[t,t+dt)$ given the past (which is represented by the $\sigma$-algebra $\mathcal{F}_{t^-}$). This yields $$\lambda(t)dt=\mathbb{E}\left(N(t+dt)-N(t)\vert \mathcal{F}_{t^-}\right)$$
Expected number of observations in $[t, t+dt)$
In terms of $\alpha$
Looking informally at the expected number of events to be observed  in the interval $[t,t+dt)$ (with $dt$ small) given the past, we have:
$$\mathbb{E}\left(\# \{ i:X_i\in[t,t+dt)\}\vert \mathcal{F}_{t^-}\right)=\#\{i:X_i\geq t \}\cdot \alpha(t)dt \qquad (1)$$
In terms of $\lambda$
The left part of equation $(1)$, $\mathbb{E}\left(\# \{ i:X_i\in[t,t+dt)\}\vert \mathcal{F}_{t^-}\right)$, is the expected number of events to be observed  in the interval $[t,t+dt)$. Thus it is equivalent to $\mathbb{E}\left(N(t+dt)-N(t)\right)$ by definition of $N(t)$, which counts events prior to $t$. So $N(t+dt)-N(t)$ count events in the interval $[t,t+dt)$ and we know that this equals $\lambda(t) dt$.
Hence we have
$$\mathbb{E}\left(\# \{ i:X_i\in[t,t+dt)\}\vert \mathcal{F}_{t^-}\right)=\mathbb{E}\left(N(t+dt)-N(t)\vert \mathcal{F}_{t^-}\right)=\lambda(t) dt$$
The number at risk is the link between the intensity process and the hazard rate
If we define the process $Y$ as $Y(t)=\# \{ i:X_i\geq t\}$  which represents the number at risk at time $t$ (i.e. the number of events $X_i$ that have not yet been observed at $t$), we have that $$\lambda(t)=Y(t)\alpha(t)$$
Which gives a nice relation between the intensity function $\lambda$ and the hazard rate $\alpha$.
More details, including the consideration of censored random times, can be found in [1, section II.1].

References
[1] P. K. Andersen, Ø. Borgan, R. D. Gill and N. Keiding. (1993) "Statistical Models Based on Counting Processes". Springer-Verlag: New York.
[2] Intensity of counting processes, https://en.wikipedia.org/w/index.php?title=Intensity_of_counting_processes&oldid=960041135 (last visited Sept. 30, 2020).
A: The hazard rate and intensity function can take the same functional form but conceptually they are different, hazard rates are based on continuous values taken from a population, intensity measures the rate at which events in time occur. 
A: A Poisson process is a model for a stream of "random" arrivals and has the properties that 


*

*there can be at most one arrival at any instant $t$

*the number of arrivals in any interval $(t_1,t_2]$ is a Poisson random variable which is here denoted as $\mathbb N(t_1,t_2]$

*For $t_1 < t_2 \leq t_3 < t_4 \leq t_5 < t_6 < \cdots \leq t_{2n-1} < t_{2n}$, the Poisson random variables 
$\mathbb N(t_1,t_2], \mathbb N(t_3,t_4], \mathbb N(t_5,t_6], \cdots , \mathbb N(t_{2n-1},t_{2n}]$, which count the numbers of arrivals in the 
$n$ disjoint or
non-overlapping time intervals 
$(t_1,t_2],(t_3,t_4], (t_5, t_6], \cdots , (t_{2n-1},t_{2n}]$, are
independent random variables

*The probability of exactly one arrival in a small time
interval $(t, t+\Delta t]$ is proportional to the length 
$\Delta t$ of the time interval; the probability of two or more arrivals
in this small interval is $o(\Delta t)$ and can be neglected in the
limit as $\Delta t \to 0$.
The constant of proportionality in this last item is assumed to be a
constant $\lambda > 0$ for homogeneous Poisson processes but is assumed to be
varying with time for nonhomogeneous processes. That is, the probability
of one arrival in the vanishingly small interval
$(t, t+\Delta t]$ is $\lambda(t)\Delta t$ while the probability of no
arrivals during this interval is $1 - \lambda(t)\Delta t$. Here, of course,
we assume that $\lambda(t) > 0$ for all $t$.  $\lambda(t)$ is called the
intensity of the process at time $t$.
Let $P_0(t)$ denote the probability that there are no arrivals in
the interval $(0,t]$. If no arrivals occurred in $(0,t+\Delta t]$,
then it must be that there are no arrivals in $(0,t]$ and that
there are no arrivals in $(t,t+\Delta t)$. The numbers of arrivals in these
two disjoint time intervals are independent random variables and so we see
that
$$\begin{align}
P_0(t+\Delta t) &= P_0(t)(1-\lambda(t)\Delta t)\\
P_0(t+\Delta t) - P_0(t) &= - \lambda(t)P_0(t)\Delta t\\
\frac{P_0(t+\Delta t) - P_0(t)}{\Delta t} &= -P_0(t)\lambda(t)\\
\frac{\mathrm dP_0(t)}{\mathrm dt} &= -P_0(t)\lambda(t)\\
P_0(t) &= \exp\left(-\int_0^t \lambda(\tau)\,\mathrm d\tau\right)\tag{1}\\
&= \exp\left(-t\cdot\bar{\lambda}(0,t]\right)\tag{2}
\end{align}$$
where $\bar{\lambda}(t_1,t_2]$ denotes the average value 
$\displaystyle\frac{1}{t_2-t_1}\int_{t_1}^{t_2}\lambda(t)\,\mathrm dt$
over the time interval $(t_1,t_2]$
Skipping additional details, I will assert that the parameter of
the Poisson random variable
$\mathbb N(t_1,t_2]$ is $\displaystyle \int_{t_1}^{t_2}\lambda(t)\,\mathrm dt$.
Thus, the _average number of arrivals in $(t_1,t_2]$ is
$$E\left[\mathbb N(t_1,t_2]\right] = \int_{t_1}^{t_2}\lambda(t)\,\mathrm dt
= (t_2-t_1)\bar{\lambda}(t_1,t_2].\tag{3}$$
Note that the average number of arrivals in $(t_1,t_2]$ per unit time is
$\bar{\lambda}(t_1,t_2]$ and is called the average intensity over this time
interval while $\lambda(t)$ is called the (instantaneous) intensity at time $t$.

Poisson processes deal with a stream of arrivals whereas hazard rates
and survival analysis deal with only one arrival -- 
the arrival of the Angel of Death!  Consider a system
that is put into operation at time $0$ and fails at some random time
$X > 0$. The hazard rate
function $h(t)$ tells us the conditional probability of the system
failing in the interval $(t,t+\Delta t]$ conditioned on the system
being in working condition at time $t$.  Thus,
$$\begin{align}
h(t)\Delta t &= P\{X \in (t,t+\Delta t]\mid X > t\}\\
&= \frac{P\left(\{X \in (t,t+\Delta t]\}\cap P\{X > t\}\right)}{P\{X > t\}}\\
&= \frac{P\{X \in (t,t+\Delta t]\}}{P\{X > t\}}\\
&= \frac{f_X(t)\Delta t}{1 - F_X(t)}.
\end{align}$$
Consequently,
$$\begin{align}
\int_0^t h(\tau)\,\mathrm d\tau 
&= \int_0^t \frac{f_X(\tau)}{1 - F_X(\tau)}\,\mathrm d\tau\\
&= - \ln (1-F_X(\tau))\big|_0^t\\
&= -\ln (1-F_X(t))\\
1-F_X(t) = P\{X > t\} 
&= \exp \left(- \int_0^t h(\tau)\,\mathrm d\tau\right)\tag{4}
\end{align}$$
which of course looks a lot like $(1)$, and both integrals
are telling us the probability that there are no arrivals
in $(0,t]$. However, if the first arrival after $0$ occurs
at time $T$, then analysis of the time of the next arrival
is based on $\lambda(t)$ for $t \geq T$ whereas there are no
new arrivals in survival analysis: the system is dead and that's
all there is to it. Now, we can extend the paradigm to say
that the failed system is instantaneously replaced by a brand-new
system that begins operating at time $T$, but the analysis 
now begins anew and the hazard rate $\hat{h}(t)$ for the replacement
is $h(t-T)$ etc. In other words, the probability that the
replacement is struck dead in $(T, T+\Delta t]$ is $h(0)\Delta t$,
not $h(T)\Delta t$.
What the OP conjectured, viz.

To me, it seems like the intensity function deals with reoccurring failure, while the hazard function deals only with the time to first failure?

is correct. 
A: The hazard function is typically encountered in event history models/survival analysis models. The hazard is the probability of experiencing the event in a given time period/by a given point in time (depending on how time is operationalized in the model), conditional on not having experienced the event up to that period/before that point in time.
This is in contrast to the survival function, which, also indexed by time, is the probability of not experiencing the event by that point/period in time across all of study time.
