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.

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    $\begingroup$ The strategy is perfectly fine (I guess the third summand should have been $1*P(2<T\leq 3)$). I haven't checked the actual values, but it should be good. $\endgroup$ – A. Pongrácz Apr 20 at 17:36
  • $\begingroup$ The expected payout will be exactly that, yes. Just that you've taken 3 instead of 1 for payout in the 3rd year $\endgroup$ – Vizag Apr 20 at 17:36

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