Stock& Watson 207.2

[orit]

Hi David,
Can you please clarify a point regarding this question:
If the true probability of default is 1%, the entire sample default is n*p it means that since the bonds defaults are i.i.d the average default rate increases when the sample size n increases.
In this question the default rate is given by D/N, can you please explain?

Thanks,
Orit

 

[David Harper CFA FRM]

Hi Orit,

This is the Law of Large Numbers. The true probability is the population average (mu); i.e., each bond defaults with probability of 1%.
If there are 100 bonds, we expect 1 default (1%), but there might occur zero (0%) or 2 or even 3; if 3 defaults occur, D =3, N = 100, and default rate = 3%.
As N increases, the LLN says that conditional on i.i.d., the default rate (D/N) will approach (converge on) the "true" 1.0%. So, the number of defaults will increase (i.e., the expected value of n*p increases with n) but the default RATE tends toward 1.0%. Please note:

• Expected number of defaults = N*p; Expected default rate = (N*p/N) = p; this are ex ante as (p) is a population probability
• Actual realized (ex post) number defaults = D; LLN says D/N --> p, as N increases, if i.i.d.

 

[orit]

thanks a lot!! now its clear