# Thread: Distribution function of an exponential random variable

1. ## Distribution function of an exponential random variable

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)

2. ## Re: Distribution function of an exponential random variable

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.

3. ## Re: Distribution function of an exponential random variable

If i get you:

4. ## Re: Distribution function of an exponential random variable

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

I would use , but that's because I forgot my cool equations for insurance payments.

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