Question

The lifetime of two light bulbs are modeled as independent and exponential random variables X and Y, with parameters lambda and mu, respectively. The time at which one of those two light bulb first burns out is Z = min (X, Y), what is the PDF of Z?

Answer 1

Answer:

$\bf h(x)=max(\lambda,\mu) e^{-max(\lambda,\mu) x}\;(x\geq0)$

Step-by-step explanation:

The PDF of X is

$\bf f(x)=\lambda e^{-\lambda x}\;(x\geq0)$

The PDF of Y is

$\bf g(x)=\mu e^{-\mu x}\;(x\geq0)$

The means of X and Y are respectively,

$\bf \displaystyle\frac{1}{\lambda}\;,\displaystyle\frac{1}{\mu}$

so we can see that the larger the parameter, the smaller the mean. Hence the PDF of Z = min(X, Y) is an exponential with the largest parameter of the two.

Therefore, the PDF of Z is

$\bf h(x)=max(\lambda,\mu) e^{-max(\lambda,\mu) x}\;(x\geq0)$