Hazard Function - Survival Analysis I just started taking survival analysis class and I'm stumped on this question.
Let $T_1,...,T_n$ random independent continuous variables, with hazard function of $h_1(t),...,h_n(t)$.
$T=min(T_1,...,T_n)$.
And we need to show that the hazard function of T is $\sum_j{h_j(t)} $
Any help or direction are welcome :)
 A: Before obtaining the hazard function of $T=\min\{T_1,...,T_n\}$, let's first derive its distribution and its density function, i.e. the CFD and PDF of the first-order statistic from a sample of independently but not identically distributed random variables.  
The distribution of the minimum of $n$ independent random variables is 
$$F_T(t) = 1-\prod_{i=1}^n[1-F_i(t)]$$
(see the reasoning in this CV post, if you don't know it already)
We differentiate to obtain its density function:
$$f_T(t) =\frac {\partial}{\partial t}F_T(t) = f_1(t)\prod_{i\neq 1}[1-F_i(t)]+...+f_n(t)\prod_{i\neq n}[1-F_i(t)]$$
Using $h_i(t) = \frac {f_i(t)}{(1-F_i(t)} \Rightarrow f_i(t) = h_i(t)(1-F_i(t)) $ and substituting in $f_T(t)$ we have
$$f_T(t) = h_1(t)(1-F_1(t))\prod_{i\neq 1}[1-F_i(t)]+...+h_n(t)(1-F_n(t))\prod_{i\neq n}[1-F_i(t)]$$
$$=\left(\prod_{i=1}^n[1-F_i(t)]\right)\sum_{i=1}^nh_i(t),\;\;\; h_i(t) = \frac {f_i(t)}{1-F_i(t)} \tag{1}$$
which is the density function of the minimum of $n$ independent but not identically distributed random variables.
Then the hazard rate of $T$ is 
$$h_T(t) = \frac {f_T(t)}{1-F_T(t)} = \frac {\left(\prod_{i=1}^n[1-F_i(t)]\right)\sum_{i=1}^nh_i(t)}{\prod_{i=1}^n[1-F_i(t)]} = \sum_{i=1}^nh_i(t) \tag{2}$$
A: Here is an informal way of looking at the matter.
Let $h(t)$ denote the hazard rate of a system. Then, $h(T)\Delta T$ is (approximately) the conditional probability that the system fails in the time interval $(T, T+\Delta T]$ given that the system is working at time $T$. Hence
$1-h(T)\Delta T$ is (approximately) the probability that a system working at
time $T$ is still functioning at time $T+\Delta T$. These approximations
improve in accuracy as $\Delta T \to 0$.
Now suppose that a system with hazard rate $h(t)$ is actually composed
of $n$ subsystems with hazard rates $h_i(t), 1 \leq i \leq n$, and the system
fails as soon as (at least) one subsystem fails. The subsystem failures are
independent. Consider the event that the
system is still working at time $T + \Delta T$ given the event
that the system is working at time $T$. But this means that all
$n$ subsystems were functional at time $T$ and continue to remain
functional at time $T+\Delta T$. Independence of the lack of failures
thus gives
$$\begin{align}
 1 - h(T)\Delta T &\approx \prod_{i=1}^n [1 - h_i(T)\Delta T]\\
&= 1 - \sum_{i=1}^n h_i(T)\Delta T 
+ \sum_{i, j: i\neq j} h_i(T)h_j(T)(\Delta T)^2 - \cdots \\
&\approx 1 - \sum_{i=1}^n h_i(T)\Delta T \quad \text{as}~ \Delta T \to 0
\end{align}$$
That is, $\displaystyle h(t) = \sum_{i=1}^n h_i(t)$.  If $A_i$ denotes 
the event that
the $i$-th subsystem fails in the interval $(T,T+\Delta T]$, the probability
that the system fails during $(T, T+\Delta T]$ is just $P(A_1\cup A_2\cup \cdots \cup A_n)$, the
probability that at least one subsystem fails. But this probability is
bounded above $\sum_i P(A_i)$ and the claim is, in effect, that this
union bound is tight and becomes an equality in the limit as $\Delta T \to 0$.
A: Since $T=\min(T_1,\ldots,T_n)$ and $T_1$,...,$T_n$ are independent, the survivor function $S(t)=P(T>t)$ of $T$ is
$$ \begin{align} S(t) &= P(min(T_1,\ldots,T_n)>t) \\
&=P(T_1>t,\ldots,T_n>t) \\
&=P(T_1>t)\cdots P(T_n>t) \\
&=S_1(t)\cdots S_n(t),
\end{align}$$
where $S_i(t)=P(T_i>t)$ is the survivor function of $T_i$.
Now, since $S_i(t)=\exp(-\int_0^t h_i(s)ds)$, we have that
$$ S(t) = \prod_i^n \exp\left(-\int_0^t h_i(s)ds\right)=\exp\left(-\int_0^t\sum_{i=1}^{n}h_i(s)ds\right).$$
Finally, since the hazard function of $T$ is linked to its survivor function by the relation $h(t)=-\frac{d\log S(t)}{dt}$, it follows that
$$h(t)=\sum_{i=1}^{n}h_i(t)$$ by the fundamental theorem of calculus.
