Distribution function of an exponential random variable

endlessend25

New Member
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)

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Mean Joe

TS Contributor
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

New Member
If i get you:
$$\int_{0}^{4}2e^{-2x}dx + \int_{4}^{\propto } 4 dx$$

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Mean Joe

TS Contributor
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.