The origin of Ong's Unexpected Loss (UL)

[Liming]

Dear David,

Can I check with you if Ong's unexpected loss (Unexpected loss (UL) = SQRT[(EDF)(variance of LGD) + (LGD^2)(variance of EDF)]*(Adjusted exposure) ) can be traced back to the formula for calculating the variance of the product of two independent variables (V(x*y) = E(x)2V(y) + E(y)2V(x) + V(x)V(y) ). However, the last item "V(x)V(y)" seems to be missing in the UL formula.

Thanks!
Liming
4/11/09

 

[David Harper CFA FRM]

Hi Liming,

It is a tempting comparison, but unfortunately I don't think it applies. Ong gives the derivation in Appendix A to Chapter 5, but it is decidely non-trivial and, to the extent it invokes the familiar, it's more dependent on variance (X) = E(X^2) - (E(X))^2 ... since Ong was introduced to the FRM, I've seen many attempts to directly intuit his UL but i don't know if it is possible...thanks, David

As @MarekH shows, clearly I was wrong!

 

[MarekH]

Hi Liming,

It is a tempting comparison, but unfortunately I don't think it applies. Ong gives the derivation in Appendix A to Chapter 5, but it is decidely non-trivial and, to the extent it invokes the familiar, it's more dependent on variance (X) = E(X^2) - (E(X))^2 ... since Ong was introduced to the FRM, I've seen many attempts to directly intuit his UL but i don't know if it is possible...thanks, David

There is a clear link between the UL formula and the formula for variance of a product of two variables.

The Ong's UL formula can be written in a simplified form as follows:

UL = EAD*\sqrt(E(D)*var(LGD) + E(LGD)^2 * var(D)) where D is a binomial random variable with mean equal to PD and variance equal to PD(1-PD). Hence,

UL = EAD*sqrt( PD * var(LGD) + var(D)E(LGD)^2)=EAD*sqrt( (PD-PD^2+PD^2) * var(LGD) + var(D)E(LGD)^2)=
EAD*sqrt( (PD(1-PD)var(LGD)+PD^2 * var(LGD) + var(D)E(LGD)^2)=
EAD*sqrt(var(D)var(LGD)+E(D)^2 var(LGD) + var(D)E(LGD)^2)

which indeed equals EAD * STD(LGD * D) as LGD and D are assumed to be independent.