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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 :)

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    $\begingroup$ Start with the DF of the order statistic, then write out the identities relating it to its survivor function and, subsequently, its hazard. $\endgroup$ – AdamO Nov 5 '13 at 23:47
  • $\begingroup$ Besides the proof: Does this result make sense intuitively? $\endgroup$ – Michael M Nov 6 '13 at 7:54
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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

$$h_T(t) = \frac {f_T(t)}{1-F_T(t)} = \frac {\left(\prod_{i=1}^n[1-F_i(t)]\right)\sum_{i=1}^nh_i(t)}{\prod_{i=1}^n[1-F_i(t)]} = \sum_{i=1}^nh_i(t) \tag{2}$$

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    $\begingroup$ This is a good answer, although not my favored approach. However, when a user uses the tag "self-study", you'll want to give hints rather than complete answers. $\endgroup$ – AdamO Nov 5 '13 at 23:46
  • $\begingroup$ Although I do not know what @AdamO's "favored approach" might be, perhaps he was alluding to the fact that once readers recognize the hazard function is the derivative of $\log(1-F)$, then the result is an immediate and obvious consequence of the first equation, allowing them to skip all the intervening algebraic manipulation. $\endgroup$ – whuber Sep 6 '14 at 18:59
  • $\begingroup$ @whuber That's a good guess, I had wondered also myself at the time about the "favored approach". $\endgroup$ – Alecos Papadopoulos Sep 6 '14 at 20:02
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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 $\Delta T \to 0$.

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    $\begingroup$ It's nice to see such an argument based on the concept of hazard rate (+1). But since you're not giving a mathematically rigorous proof, wouldn't it be simpler--and more in keeping with the spirit of this approach--just to notice that (conditional on the system still working at time $T$) the chance of failure by time $T+\Delta T$ for sufficiently short increments $\Delta T$ is essentially the chance of a single failure, which will be the sum of the $h_i$? That makes the result intuitive and immediate. $\endgroup$ – whuber Sep 7 '14 at 15:01
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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.

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