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Variance of EDF

Thread starter #1
Hi all

I may have missed something in the forum, and I am happy for someone to point me to an existing thread rather that opening a new one if this question already exists.

I cannot figure out why the Var(EDF) is equal to EDF*(1-EDF). This may be a simple question, which may also indicate that I need way more studying until I'm ready.


Many thanks.
 

David Harper CFA FRM

David Harper CFA FRM
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#2
Hi @nikogeorgiev If you search "Bernoulli" you'll find discussions, but I can help quickly. EDF is a Bernoulli because the obligor either defaults (X = 1) with probability (p) or survives (X = 0) with probability (1-p). Variance(EDF; aka, PD) is a formula you'll want to memorize; https://en.wikipedia.org/wiki/Bernoulli_distribution. But just as importantly, we want to be able to apply the variance formula: Variance(X) = E[X^2] - E[X]^2; actually, best is to recognize this variance as a special case of Cov(X,Y) = E[X*Y] - E[X]*E[Y] where X = Y <--this is very useful relationship!

For a Bernoulli, X = {0, 1} and X^2 ={0^2, 1^2} = {0, 1}. So E[X^2] = p*1 + (1-p)*0 = p and E[X] = p such that E[X]^2 = p^2, and therefore the variance(EDF) = E[X^2] - E[X]^2 = p - p^2 = p*(1-p). I hope that's helpful!
 
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