# Future value of an appliance who's time of life is modeled by exponential distrib?

1. Oct 14, 2011

### Hodgey8806

1. The problem statement, all variables and given/known data
The initial value of an appliance is $700 and it's value in the future is given by: v(t)=100(2^(3-t)-1), 0<=t<=3 where t is time in years. Thus, after the first 3 years the appliance is worth nothing as far as the warranty is concerned. If it fails in the first three years, the warranty pays v(t). Compute the expected value of the payment on the warranty if T has an exponential distribution with mean 5. 2. Relevant equations I would think to use E(x) = int(x*v(t), 0<=t<=3) which is approximately 574.14977 I would also use the probability of it failing in the first three years which would be: P(T<=3) = 1- e^-(3/5) which is approximation .4512 . 3. The attempt at a solution Thus, my solution would be .4512*574.14977 which is approximately$259.05.

Is this correct? Thanks!

2. Oct 15, 2011

### Ray Vickson

Re: Future value of an appliance who's time of life is modeled by exponential distrib

I get \$121.73, and my calculation is nothing like yours. I am computing the expected value of v(T); I have no idea what you are computing.

RGV

3. Oct 15, 2011

### Hodgey8806

Re: Future value of an appliance who's time of life is modeled by exponential distrib

Would you care to share how you calculated the Expected value of v(t)?

4. Oct 15, 2011

### Ray Vickson

Sorry, no. I am allowed to give hints only, and I have already said exactly what I did: I computed the expectation of a function of a random variable.

RGV