Bernoullis distribution Vs Poisson's distribution (Miller Ch4 EOC Q&A)

[Biju George]

Hi All, Preparing for the Part-1 2015 Nov. Got a question on distributions from the below two questions given in the Miller's question at the end of Distributions (page 89, Q1 & Q2)

My Confusion: This is solved by Bernoullis distribution. Why can't we use poisons distribution to solve these?

Question 1:

XYZ Corporation announces its earnings four times per year. Based on historical data, you estimate that in any given quarter the probability that XYZ Corporation’s earnings will exceed consensus estimates is 30%. Also, the probability any other quarter. What is the probability that XYZ Corporation will exceed estimates three times in a given year?

Answer:

The number of times XYZ Corporation exceeds consensus estimates follows a binomial distribution; therefore:




Question 2:

The market risk group at your firm has developed a value at risk (VaR) model. In Chapter 7 we examine VaR models more closely. In the meantime, assume the probability of an exceedance event on any given day is 5%, and the probability of an exceedance event occurring on any given day is independent of an exceedance event having occurred on any previous day. What is the probability that there are two exceedances over 20 days?

Answer: The number of exceedance events follows a binomial distribution; therefore:




Thanks in Advance

 

[Arka Bose]

Hi,

In case of a Bernoulli distribution, we are given an exact probability of the happening of an event. So we find out the probability of the event over n periods of time. The same thing applies to your problems here, we are given an exact probability that earnings will exceed consensus is 0.3. now our n=4 so we find the probability by using Binomial distribution.

However, in case of a poisson distribution, we EXPECT some events to occur with RELATIVE TO TIME. For example, if in this question it was given that on an average we expect 3 times out of 4 quarters (or 75% of time) that earning will exceed consensus, then find probability of having consensus met 2 times, we have to use poisson distribution.

 

[Biju George]

Thanks a lot for the clarification

 

[Sixcarbs]

It's a binomial for P[X=3]. p=.30

In solution .3 is used for (1-p) but .03 is used for p^3. Can't be both. Also, then there is an extra ".07" right before "=" sign. The three terms turn into four. I don't know where that comes into play.


Please confirm.

Thank you

 

[Nicole Seaman]

It's a binomial for P[X=3]. p=.30

In solution .3 is used for (1-p) but .03 is used for p^3. Can't be both. Also, then there is an extra ".07" right before "=" sign. The three terms turn into four. I don't know where that comes into play.


Please confirm.

Thank you

Hello @Sixcarbs

Please note that I moved your question here where this specific end of chapter practice question from Miller has already been discussed. I found this in the forum by using the search function and searching for the first sentence of the question. We ask that everyone attempts to search the forum before posting new threads to make sure that their question has not already been discussed. If the above discussion does not answer your question, I'm sure someone can help further.

Thank you,

Nicole

 

[David Harper CFA FRM]

Hi @Sixcarbs Apologies but the answer to question #1 contains two typos (in the interim calculation, which is sloppy I admit), it should read:

p[X = 3] = C(4,3) * 0.30^3 * 0.70^1 = 4 * 0.30^3 * 0.70 = 7.56% (ie., same result ... but as you suggest, there are only three multiplicands in the solution to a binomial. Thank you!