Showing $λ_V(x) \leq min\{λ_1(x),\cdots,λ_n(x)\}$ Hazard function

Suppose $X_1, \cdots, X_n$ are independent, nonnegative continuous functions, each $X_i$ has hazard function $\lambda_i(x)$.

If $V=\max\{X_1, \cdots, X_n\}$, I need to show that $\lambda_V(x)\leq \min\{\lambda_1(x),\cdots, \lambda_n(x)\}$.

I know that $\displaystyle\lambda_V(x)=\frac{f_V(x)}{1-F_V(x)}=\frac{\sum_j f_j(x)\prod_{i\neq j}F_i(x)}{1-\prod_{i=1}^n F_i(x)}$.

However, I'm stuck here.

Any help would be appreciated.

1 Answer

The claim that you need to show seems to be false. For a counter example, let $X_1 \sim \textrm{Exp}(1)$ so that $\lambda_1(x) = 1$. Let us define $X_2$ by a piecewise hazard function: $$\lambda_2(x) = \left\{ \begin{array}{ll} a & \quad x \leq 1 \\ b & \quad x > 1 \end{array} \right.,$$ where $a,b>0$. The CDF of $X_2$ is $$F_2(x) = \left\{ \begin{array}{ll} 1-e^{-ax} & \quad x \leq 1 \\ 1-e^{-a}e^{-b(x-1)} & \quad x > 1 \end{array} \right.$$ Then, intermediate steps omitted, we work out the hazard rate of $V = \max(X_1,X_2)$ for $x>1$ to be $$\lambda_V(x) = \frac{e^{-x}(1-e^{-a}e^{-b(x-1)}) +be^{-a}e^{-b(x-1)}(1-e^{-x})}{1 - (1-e^{-a}e^{-b(x-1)})(1-e^{-x})}.$$ Now, let us consider the limit of this expression as $b\rightarrow 0$ (for some fixed $x>1$). $$\lim_{b\rightarrow 0} \lambda_V(x) = \frac{e^{-x}(1-e^{-a})}{1-(1-e^{-a})(1-e^{-x})}>0.$$ As the limit is some positive constant, selecting a suitably small $b>0$ will lead to $\lambda_V(x)>b$ which is then a contradiction with the claim.

Some intuition: say we have some system that breaks down when both components $1,2$ break. The claim is saying that the hazard rate of the system is smaller than either hazard rate of the components. However, our counterexample has one component that is likely to break during the first day of use, but after surviving the first day is extremely reliable. The system is still not extremely reliable during the second day: it is well plausible that the second component has broken during the first day and that the first component breaks now.