Distribution function of an exponential random variable

[endlessend25]

Let Y ⇠ Exp(2) be the size of an insurance loss. If the insurance company pays min(Y, 4), what is the variance of that payment?

I know that lambda =2

If I put that in the pdf I get: 2*e^(-2x). I don't understand how to use the information given by min(Y, 4)

 

[Mean Joe]

Instead of integrating your pdf from 0 to infinity, split into two integrals (0 to 4, and 4 to infinity). You figure out what to put for your integrand.

 

[endlessend25]

If i get you:
\(\int_{0}^{4}2e^{-2x}dx + \int_{4}^{\propto } 4 dx \)

 

[Mean Joe]

OK clearly your 2nd integral is divergent, because you forget the cdf for x>4.

I would use \(Var(X) = E[X^2] - E^2[X]\), but that's because I forgot my cool equations for insurance payments.