Answer:
[tex]\bf h(x)=max(\lambda,\mu) e^{-max(\lambda,\mu) x}\;(x\geq0)[/tex]
Step-by-step explanation:
The PDF of X is
[tex]\bf f(x)=\lambda e^{-\lambda x}\;(x\geq0)[/tex]
The PDF of Y is
[tex]\bf g(x)=\mu e^{-\mu x}\;(x\geq0)[/tex]
The means of X and Y are respectively,
[tex]\bf \displaystyle\frac{1}{\lambda}\;,\displaystyle\frac{1}{\mu}[/tex]
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
[tex]\bf h(x)=max(\lambda,\mu) e^{-max(\lambda,\mu) x}\;(x\geq0)[/tex]
