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Same here. Certified at 2pm EST. Submitted on 28th of June at 8am EST

Anyone else certified already ?

Anyone certified ?

Passed with 4-4-4-3-3

I saw the your enrollment will expire on may 2024 but cant see he results yet..

And it doen not matter if it was year 2 given year 1 OR year 3 given year 2 OR year 4 given year 3 because the hazard rate is constant : (exp[-0.12] - exp[-0.24]) /exp[-0.12] = (exp[-0.24] - exp[-0.36]) /exp[-0.24] = (exp[-0.36] - exp[-0.48]) /exp[-0.36] = (exp[-0.48] - exp[-0.60]) /exp[-0.48]...

In the mock exam, you are asked for the joint probability (survival in 1st year AND THEN default in 2nd year => two events). In the real exam, you were asked for the conditional probability (default in 2nd year GIVEN survival in 1st year => one event). In the mock exam, the perspective was...

This is mock exam. Real exam was : Assuming a constant hazard rate model, what is the probability that the company will default in the second year given it has survived in the first year ? This is not the same...

Not exactly. As I said above, there may be several exam versions ... But I am 100% sure that the question was about PofD 2y given SURVIVAL 1y.. So it is not the same question as it was in the mock exam.

Yeah this one

yes.. to me, it was a big trap. The formula of lognormal VAR i.e P*[1 - exp(mu - z*sigma)] gave 387 The formula of normal VAR i.e P*[-mu + z*sigma] gave 416 The option were 414 and 416 and the question was "what is the closest to the lognormal VAR ?" Thus, I put 414........ Am I the only one ?

What the calculation of lognormal VAR with options of 414 and 416 ? The question was "which one is the closest to the Var level ?" The normal Var gave 416 and lognormal gave 387... so I pick 414... I am 99% sure I am wrong though

No it is not 10. The wording is different. In the real exam we week asked the proba of A given B. In the mock it is the proba of A and then B

The question was : what is the probability of default in the second year given survival in first year

Which was the answer proposed

Yes I remember the question and the frequency of loss plays a role in the reliability of the chosen distribution. But the distribution itself mesures the size of losses.

PD = 1 - exp(-s/(1-RR))

it is size of loss.

BCVA = CVA - DVA so it was 40K to me