# Distribution of minimum of hazard functions

Suppose I have two random variables, $X_1$ and $X_2$, that are independent (but not identically distributed) and assume both have hazard functions $\lambda_1(s)$ and $\lambda_2(s)$, for $s > 0$. Then how can I derive the hazard function of $\min(X_1, X_2)$?

EDIT: Let $S$ describe the lifetime of a person, then define $\theta_s(t) = P(S > s+t | S > s)$ as the survival function. Then define the hazard function as $\lambda(s) = \lim_{t \rightarrow 0} \frac{1}{t} (1-\theta_s(t)) = -\theta_s'(0)$

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Please define the term hazard function in the context that you are interested in. – Richard Apr 22 '14 at 7:26
Please see edit – siTTmo Apr 23 '14 at 15:40

Note that $$P(X_i >s)= \exp\Big(-\int_0^s \lambda_i(u) du \Big),$$ for $i=1, 2$. Then, $$P(\min(X_1, X_2) >s) = P((X_1>s)\cap (X_2>s)) = P(X_1>s)P(X_2>s) = \exp\Big(-\int_0^s (\lambda_1(u)+\lambda_2(u)) du \Big).$$ That is, the hazard function for $\min(X_1, X_2)$ is $\lambda_1(s)+\lambda_2(s)$.
Alternatively, note that $$\lambda_i(s) = \lim_{\Delta s \rightarrow 0} \frac{P(X_i \leq s+\Delta s | X_i > s)}{\Delta s} \\ = \lim_{\Delta s \rightarrow 0} \frac{P(X_i > s) - P(X_i > s+\Delta s)}{\Delta s P( X_i > s)},$$ for $i=1, 2$. Then the hazard function of $\min(X_1, X_2)$ is given by $$\lambda_{\min(X_1, X_2)}(s) = \lim_{\Delta s \rightarrow 0} \frac{P(\min(X_1, X_2) \leq s+\Delta s | \min(X_1, X_2) > s)}{\Delta s} \\ = \lim_{\Delta s \rightarrow 0} \frac{P(\min(X_1, X_2) > s) - P(\min(X_1, X_2) > s+\Delta s)}{\Delta s P( \min(X_1, X_2) > s)} \\ = \lim_{\Delta s \rightarrow 0} \frac{P(X_1 > s)P(X_2> s) - P(X_1 > s+\Delta s)P(X_2 > s+\Delta s)}{\Delta s P(X_1 > s)P(X_2> s)} \\ \small= \lim_{\Delta s \rightarrow 0} \frac{P(X_1 > s)P(X_2> s){\tiny-P(X_1 > s)P(X_2 > s+\Delta s) + P(X_1 > s)P(X_2 > s+\Delta s) }- P(X_1 > s+\Delta s)P(X_2 > s+\Delta s)}{\Delta s P(X_1 > s)P(X_2> s)} \\ \small=\lim_{\Delta s \rightarrow 0}\bigg(\frac{P(X_2> s)-P(X_2 > s+\Delta s)}{\Delta s P(X_2> s)} +\frac{P(X_2 > s+\Delta s)}{P(X_2> s)} \frac{P(X_1 > s)-P(X_1 > s+\Delta s)}{\Delta s P(X_1 > s)}\bigg) \\ = \lambda_1(s)+\lambda_2(s).$$