question 123.3 (Rachev Chapter 2 &3)

[sarita]

Dear David,

How are you? i have a general question with respect to exponential and Poisson.

There are two formulas:

exponential: 1-exp^(-lambda*x)
Possion: (lambda^x * exp^-lambda)/ X!

I don't quite understand when to use which one.

For instance for question 123.1 the question asks for the probability of default; and the solution uses 1-exp^-0.03*3 to get the answer.

for question 123.3, the question refer to possion distribution. Why don't we use (lambda^x*exp^-lambda)/x!?

why is the probability of next call within 3 min is 1-exo^-lambda*x?

Kindly explain when you have time. and how should we know at the exam which formula to use.

many thanks,
SY

 

[David Harper CFA FRM]

Hi SY,

It's a good question; but the FRM is unlikely to force you into a false choice between these two, it so far has not.

Please note that, in this case, you could use either:

If c is the average number of calls per hours, then c/60 = average number of calls per minute.
And, for example, if your question concerns whether a call will occur in the next (n) minutes, then:
Poisson lambda = n*c/60 (i.e., the average number of calls in n minutes)
and Exponential beta = c/60 average calls per minute

Then:
Probability of no calls in next (n) minutes = Poisson P[X = 0] = lambda^0/0!*EXP(-lambda) = EXP(-lambda)
Probability of at least one call = 1- EXP(-lambda) = 1 - EXP(n*c/60) = 1 - EXP(-n*beta) = exponential CDF!
i.e., the same result!

Then, for example, assume we are interested in whether there is a call in the next 10 minutes (the "waiting time"), n = 10: c = average of 20 calls per hour, so that c/60 = 1/3 average call per minute

Poisson with lambda = 10*1/3 = 3.333 is the average number of calls in the 10 minute period.
P[at least one call in next 10 minutes] = 1 - P[X=0] = 1 - 1/1*EXP(-0.33333) ~= 3.6%

Compare to the exponential CDF
= 1 - EXP(-10 minutes * 0.33333 per minute average/harzard) ~= 3.6%

See how the exponential, as a "waiting time" until next event, is the solution to a Poisson where we are concerned with the P[X=0] or 1 - P[X=0]?

In case it is helpful, i attached a PDF from my evernote database, I don't know the source, it is what i found to help myself understand
Also, this is a more direct demonstration of the relationship:
http://stats.stackexchange.com/questions/2092/relationship-between-poisson-and-exponential-distribution

Thanks, David

 

[sarita]

Dear David,
please see this question:

A call center receives an average of two phone calls per hour. The probability that they will receive 20 calls in an 8 hour day is closest to:

1)5.59%
2) 16.56%
3)3.66%
4)6.40%

The answer is A using formual : (lambda'-x * e'(-lambda))/X! .. x=20 and lambda=2*8=16.

My quesion is that why can't i use 1-e'(-lambda*x) formula? sorry, i still don't understand when to use which one. Shouldn't they both give the same answer? kindly help.

Best,
S

 

[David Harper CFA FRM]

Hi saray,

This problem calls for the Poisson because we want the probability of a number of calls in a time interval; i.e., 20 calls in 8 hours.

The exponential refers to the probability that the next call will happen in the next X minutes/hours. For example, we would want the exponential if the question were, "What is the probability that the next call will occur in the next hour?" Answer: since beta = 2 calls/hour, exponential CDF = 1 - exp(-1 hour * 2 calls per hour) = 86.5%

The equivalence (as illustrated above) is a special case: the probability that the NEXT call will occur within the next X hours/minutes (i.e., exponential) is equal to 1 - Prob [zero calls in the next X hours/minutes] (i.e., Poisson).

Hope that helps, David

 

[[email protected]]

Saray,

Just to second what David said, you consider exponential if you need to MEASURE (not technically) TIME till an event ... you use Poisson to measure the number of events in a given time ......... In exponential you measure TIME... in Poisson you measure the number of events...

In your question above you want to measure a likelihood for the number of events (i.e. 20 calls in 8 hours)... use POISSON...

Alan

 

[Jiew Kwang]

Hi,

I don't myself remember seeing any exponential distribution problem so far. But regarding your question, this was how i did it:

There is an average of 2 calls / hr.
So lambda would be 16 calls in 8 hrs.
P(x=20) = [16^20*exp(-16)] / 20! = 0.05591950595450295675
Answer is a.

 

[David Harper CFA FRM]

@Alan: that is a helpful summary re Poisson/exponential

There is a direct connection, that took me a while to grasp, between the two. At least i think! It is to view the exponential as a sort of special case of the Poisson. What i mean is:

1. Poisson: Poisson tells us the probability an (X) number of events happening in a time interval (given the average events/interval)
2. Exponential: If we specifically define the time interval as FROM now (0) to time (T), the exponential is telling us the probability of 1 - P[no events happening in the time interval], where p[0 events during the interval] is a Poisson pmf; i.e., exponential P() = 1 - P[n=0]

to summarize (and this is not my insight, i got this from http://stats.stackexchange.com/questions/2092/relationship-between-poisson-and-exponential-distribution): the exponential CDF ("what is 1 - probability of no events during time interval T; meaning what is the probability of at least one event during T?") is equal to 1 - poisson pmf = 1 - P[0 events during T].

Hopefully that is agreeable, thanks, David