Why integrate? P1.T2. Quantitative Analysis

[skoh]

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?)

Answer:
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?

Thanks!

 

[ShaktiRathore]

hi
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
thanks

 

[David Harper CFA FRM]

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,

 

[skoh]

Thanks! Oh yes..area under curve, how can I miss that?