The initial cost of a machine is 3\$. The life-time, T, of that machine has an exponential distribution with an expected value equal to 3 years. The maker wants to offer a warranty that pays 3\$ if the machine gets broken during the first year, 2\$ if the machine gets broken during the second year, and 1\$ if the machine gets broken during the third year.
I know that T ~ Exp(1/3) and that the warranty payment function looks like this:
X = 3 (for T <= 1), 2 (for 1 < T <= 2), 1 (for 2 < T <= 3)
To find E[X] I tired the following: 3*P(T <= 1) + 2*P(1 < T <= 2) + 1*P(2 < T <= 3), which gives me approximately 1.402. However, I don’t know if that is the right way to do it.