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Chapter 5 Probability of Loan Default


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Screenshot 2021-07-03 at 20.33.09.pngHi, I was wondering how E[PD^2] = p = E[PD] holds true, is it by definition? Thanks in advance!
I could not find a thread related to this when searching for one.


Well-Known Member

the original text is sloppy. PD is here a random variable that can be 1 or 0. The probability of PD being one is p.

Expected value of PD:
E[PD] = p * 1 + (1-p) * 0 = p
Expected value of PD^2:
E[PD^2] = p * 1^2 + (1-p) * 0^2 = p

it follows that E[PD] = E[PD^2] = p
Which is a property of the Bernoulli distribution. I’m not sure it’s unique to it, but for most distributions it is not true.