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?
Solved – Difference between hazard function and intensity function
hazardpoisson distribution
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Actually, it sounds like the hazard ratio you obtained was the adjusted ratio, whereas by definition the crude ratio has no adjustment for covariates.
When you did your multivariable Cox regression, you included some covariates in the model, and the crude ratio would be the hazard ratio without these covariates included in the model.
To get this, you can re-run your Cox regression without including the covariates.
I would recommend reading the following link: http://faculty.chass.ncsu.edu/garson/PA765/cox.htm
In a slide presentation, Karlis and Ntzoufras define a bivariate Poisson as the distribution of $(X,Y)=(X_1+X_0,X_2+X_0)$ where the $X_i$ independently have Poisson $\theta_i$ distributions. Recall that having such a distribution means
$$\Pr(X_i=k) = e^{-\theta_i}\frac{\theta_i^k}{k!}$$
for $k=0, 1, 2, \ldots.$
The event $(X,Y)=(x,y)$ is the disjoint union of the events
$$(X_0,X_1,X_2) = (i,x-i,y-i)$$
for all $i$ that make all three components non-negative integers, from which we may deduce that $0 \le i \le \min(x,y)$. Because the $X_i$ are independent their probabilities multiply, whence
$$F_{(\theta_0,\theta_1,\theta_2)}(x,y)=\Pr((X,Y)=(x,y)) \\= \sum_{i=0}^{\min(x,y)} \Pr(X_0=i)\Pr(X_1=x-i)\Pr(X_2=y-i).$$
This is a formula; we are done. But to see that it is equivalent to the formula in the question, use the definition of the Poisson distribution to write these probabilities in terms of the parameters $\theta_i$ and (assuming neither of $\theta_1,\theta_2$ is zero) re-work it algebraically to look as much as possible like the product $\Pr(X_1=x)\Pr(X_2=y)$:
$$\eqalign{ F_{(\theta_0,\theta_1,\theta_2)}(x,y)&= \sum_{i=0}^{\min(x,y)} \left( e^{-\theta_0} \frac{\theta_0^i}{i!}\right) \left( e^{-\theta_1} \frac{\theta_1^{x-i}}{(x-i)!}\right) \left( e^{-\theta_2} \frac{\theta_2^{y-i}}{(y-i)!}\right) \\ &=e^{-(\theta_1+\theta_2)}\frac{\theta_1^x}{x!}\frac{\theta_2^y}{y!}\left(e^{-\theta_0}\sum_{i=0}^{\min(x,y)} \frac{\theta_0^i}{i!}\frac{x!\theta_1^{-i}}{(x-i)!}\frac{y!\theta_2^{-i}}{(y-i)!}\right). }$$
If you really want to--it is somewhat suggestive--you can re-express the terms in the sum using the binomial coefficients $\binom{x}{i}=x!/((x-i)!i!)$ and $\binom{y}{i}$, yielding
$$F_{(\theta_0,\theta_1,\theta_2)}(x,y) = e^{-(\theta_0+\theta_1+\theta_2)}\frac{\theta_1^x}{x!}\frac{\theta_2^y}{y!}\sum_{i=0}^{\min(x,y)}i!\binom{x}{i}\binom{y}{i}\left(\frac{\theta_0}{\theta_1\theta_2}\right)^i,$$
exactly as in the question.
Generalization to multivariate scenarios could proceed in several ways, depending on the flexibility needed. The simplest would contemplate the distribution of
$$(X_1+X_0, X_2+X_0, \ldots, X_d+X_0)$$
for independent Poisson distributed variates $X_0, X_1, \ldots,X_d$. For more flexibility additional variables could be introduced. For instance, use independent Poisson $\eta_i$ variables $Y_1, \ldots, Y_d$ and consider the multivariate distribution of the $X_i + (Y_i + Y_{i+1} + \cdots + Y_d)$, $i=1, 2, \ldots, d.$
Best Answer
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.
is correct.