Quant Analysis - Poisson Distrbn - Qn 9 - Random Set 1 !

[Chintan]

Hi David !

This is w.r.t. Qn 9 - Random Set 1 - Quant Analysis.

For calculating - P(X=0), going as per the formula - the denominator shud have been 0 factorial and hence the P(X=0) should be infinity.

However, in yr solution, you have not considered the denominator 0 factorial at all !

What do we understand from this ? I mean - wats the logic of ignoring 0 factorial ....

Kindly advise...

Regards,
Chintan

 

[Chintan]

Hi David !

I'm here again for a question on Poisson Distrn......

Well, in the random quiz Q No. 15 - Set 15 ....you computed P (x = 10) while ideally it should have been 5. And in the second set, you have computed P (x=20).

For me, ideally we shud take lambda for the first case as 10 (10% of 100) and x = 5 while for the second case, lambda shud be 20 (10% of 200) and x = 5. Solving with these nos., the ans will be 3.7833% and 0.005496%.

Pls tell me where am I going wrong in these calcls...

Regards,
Chintan

 

[David Harper CFA FRM]

Hi Chintan,

On the first problem:

0! = 1. So, Poisson[x=5] = [exp(-5)(5^0)]/0! = exp(-5). Although, the explanation could be better, I will fix the explanation to be more descriptive (but the answer looks good).

On the second problem, you are correct, it is an error.

Poisson[x=5 events, mean = 10] = 3.783%
Poisson[x=5 events, mean = 20] = 0.00549%

Apologies. David