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Why integrate? P1.T2. Quantitative Analysis


Hi David,

The question:
A credit asset has a principal value of $6.0 with probability of default (PD) of 3.0% and a loss given default (LGD) characterized by the following probability density function (pdf): f(x) = x/18 such that 0 ≤ x ≤ $6. Let expected loss (EL) = E[PD*LGD]. If PD and LGD are independent, what is the asset's expected loss? (note: why does independence matter?)

If PD and LGD are not independent, then E[PD*LGD] <> E(PD) * E(LGD); for example, if they are positively correlated, then E[PD*LGD] > E(PD) * E(LGD).
For the E[LGD], we integrate the pdf: if f(x) = x/18 s.t. 0 < x < $6, then F'(x) = (1/18)*(1/2)*x^2 = x^2/36
(note this satisfied the definition of a probability over the domain (0,6) as 6^2/36 = 1.0).
The mean of f(x) integrates xf(x) where xf(x) = x*x/18 = x^2/18, which integrates to 1/18*(x^3/3) = x^3/54, so E[LGD] = 6^3/54 = $4.0.
E[PD * LGD] = 3.0%*$4.0 = $0.120.

I'm confused as to why we need to integrate. Isn't f(x) the probability by itself?



Well-Known Member
let x be the principal value than for this value of principal,
EL(x)= PD*Principal*LGD= 3%*x*x/18 (LGD and PD need to be independent since any two events independent implies P(A*B)=P(A)*P(B), so PD does not depends on the LGD whcih depends on the the value of x so that PD is in itself independent and constant if it were not independent then it would also depend on x and on LGS in some way..i think this makes sense)
Also for each and every $ we find need to find the expected loss as the LGD is the function of this principal x so we need to integrate since the LGD is not constant that is same for each dollar of the principal for which we need to find the expected loss.
total EL(0 ≤ x ≤ $6)= for (0 ≤ x ≤ $6) integral of(3%*x*x/18)
total EL(0 ≤ x ≤ $6)=3%*(x^3/54) =3%*216/54=3%*4=$.12

hope its clear

David Harper CFA FRM

David Harper CFA FRM
Staff member
Hi skoh, just to insert an idea which is even more basic than Shakti's good detail explanation, which is really easy to miss:
  • For discrete distributions, f(x) = Prob [X = x], as you suggest. However:
  • For a continuous distribution, probability [a < X < b] = [f(x)dx] over the interval a,b; i.e., the area under the curve from a to b.
A way to think about this is, all probability distributions must sum to 1.0. In a discrete distribution, the f(x) has width, so it has "area under the curve" as a fraction of the overall 1.o. But in a continuous distribution f(x) has no width and no area. Thanks,