Question

Let X and Y be independent continuous random variables with respective hazard rate functions λX(t) and λY(t), and set W = min(X, Y).

(a) Determine the distribution function of W in terms of those of X and Y.

(b) Show that λW(t), the hazard rate function of W, is given by λW(t) = λX(t) + λY(t)

Short answer

a. $F_w\left(t\right)=1-(1-F_x(t\left)\right)(1-F_y(t\left)\right)$

b.${\lambda }_w\left(t\right)={\lambda }_x\left(t\right)+{\lambda }_y\left(t\right)$

Step-by-step solution

Content Introduction

X and Y are independent continuous random variable with respect hazard rate function${\lambda }_x\left(t\right)$and${\lambda }_y\left(t\right)$

Explanation (Part a)

Let us determine the cumulative distribution function of W using the fact that W is larger than t when both X and Y are larger than t.

$$\begin{gathered} F_w\left(t\right)=P(W\le t) \\ =1-P(W>t) \\ =1-P(X>t,Y>t) \end{gathered}$$

X and Y are both independent

$$\begin{gathered} =1-P(X>t)(Y>t) \\ =1-(1-P(X\le t\left)\right)(1-P(Y\le t\left)\right) \\ =1-(1-F_x(t\left)\right)(1-F_y(t\left)\right) \end{gathered}$$

Explanation (Part b)

We use the definition of the hazard rate function $\lambda \left(t\right)=\frac{f\left(t\right)}{1-F\left(t\right)}$

$$\begin{gathered} {\lambda }_w\left(t\right)=\frac{f_w\left(t\right)}{1-F_w\left(t\right)} \\ =\frac{(1-F_x(t\left)\right)f_y\left(t\right)+(1-F_y(t\left)\right)f_x\left(t\right)}{1-(1-(1-F_x\left(t\right))(1-F_y\left(t\right)} \\ =\frac{(1-F_x(t\left)\right)f_y\left(t\right)}{1-(1-(1-F_x\left(t\right))(1-F_y\left(t\right)}+\frac{(1-F_y(t\left)\right)f_x(t}{1-(1-(1-F_x\left(t\right))(1-F_y\left(t\right)} \\ =\frac{f_y\left(t\right)}{1-F_y\left(t\right)}+\frac{f_x\left(t\right)}{1-F_x\left(t\right)} \\ ={\lambda }_x\left(t\right)+{\lambda }_y\left(t\right) \end{gathered}$$