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: The following is how I define the hazard function (instantaneous rate of mortality):
Let $S$ denote the lifetime of an object, then define $\theta_s(t) = P(S>s+t | S>s)$. Then the hazard function, $\lambda(s)$ is defined as:
$\displaystyle \lambda(s) = \lim_{t \rightarrow 0} -\frac{\theta_s(t) – \theta_s(0)}{t} = -\theta_s'(0)$
EDIT 2: My solution:
Let $Y = \min\left(X_1, X_2\right)$. We need to find the hazard function for $Y$, which we denote as $\lambda_Y(s)$. We have,
\begin{align*}
\lambda_Y(s) = \frac{f_Y(s)}{1-F_Y(s)}
\end{align*}
where $f_Y(s)$ and $F_Y(s)$ denote the probability density function and cumulative distribution function of $Y$, respectively. Note that,
\begin{align*}
F_Y(s) & = P\left(Y \le s\right) \\
& = 1-P(Y>s) \\
& = 1-P\left(\min\left(X_1, X_2\right) > s\right) \\
& = 1 – P\left(X_1 > s, X_2 > s\right) \\
& = 1-P(X_1>s)P(X_2>s) \ \ \ \ \text{Due to the independence between} \ X_1 \ \text{and} \ X_2 \\
& = 1-\left[1-P(X_1 \le s) \right]\left[1-P(X_2 \le s) \right] \\
& = 1-\left[1-F_{X_1}(s) \right]\left[1-F_{X_2}(s) \right]
\end{align*}
where $F_{X_1}(s)$ and $F_{X_2}(s)$ denote the cumulative distribution functions of $X_1$ and $X_2$, respectively. Hence,
\begin{align*}
1-F_Y(s) & = 1-\left(1-\left[1-F_{X_1}(s) \right]\left[1-F_{X_2}(s) \right] \right) \\
& = \left[1-F_{X_1}(s) \right]\left[1-F_{X_2}(s) \right]
\end{align*}
We also have,
\begin{align*}
f_Y(s) & = \frac{d}{dy} F_Y(s) \\
& = F_{X_1}'(s)\left[1-F_{X_2}(s) \right]+F_{X_2}'(s)\left[1-F_{X_1}(s)\right] \\
& = f_{X_1}(s) \left[1-F_{X_2}(s) \right] + f_{X_2}(s)\left[1-F_{X_1}(s)\right]
\end{align*}
where $f_{X_1}(s)$ and $f_{X_2}(s)$ denote the probability density function $X_1$ and $X_2$, respectively. Putting everything together yields,
\begin{align*}
\lambda_Y(s) & = \frac{f_{X_1}(s) \left[1-F_{X_2}(s) \right] + f_{X_2}(s)\left[1-F_{X_1}(s)\right]}{ \left[1-F_{X_1}(s) \right]\left[1-F_{X_2}(s) \right]} \\
& = \frac{f_{X_1}(s)}{1-F_{X_1}(s)} + \frac{f_{X_2}(s)}{1-F_{X_2}(s)} \\
& = \lambda_1(s) + \lambda_2(s)
\end{align*}
Best Answer
Answer:
I have given the answer for n random variables which can be simplified to two.
$$F_{min}(x) = P(X_{min} \leq x)$$ $$= 1- P(X_{min} > x)$$$$= 1-P(X_1 > x; X_2 > x;\cdots X_n > x)$$ $$= 1 - [1 - F_1(x)][1 - F_2(x)]\cdots[1 - F_n(x)]$$ For x > 0,$$ F_{i}(x) = 1 - e^{-\lambda x}$$ $$ 1- F_{i}(x) = e^{-\lambda x}$$
$$F_{max}(x) = F_{1}x.F_{2}x.F_{3}x.\cdots F_{n}x$$
Hazard function $= \frac{P(x)}{1-F(x)}$.(probability function/survival fuction) For this to be minimum, we need to have $F_{min}(x)$. Do the calculation appropriately.
Thanks
Satish