Chapter 5 Probability of Loan Default

[yisuho97]

Hi, 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.

 

[ami44]

Hi,

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.

 

[yisuho97]

Thanks for the explanation!