P1.T2.316. Discrete distributions (Topic review)

[Nicole Seaman]

Questions:

316.1. Assume a 99.0% daily value at risk (VaR) model is perfectly accurate. Specifically, among 100 days, we expect a loss that exceeds VaR on exactly one day. The probability mass function of a binomial distribution is given by:

Over a series of 20 trading days, which is nearest to the probability that the daily loss will exceed VaR on exactly two (2) days?

a. 0.44%
b. 0.93%
c. 1.59%
d. 2.36%

316.2. The frequency of "Damage to Physical Assets," an operational risk event type, is characterized by a Poisson distribution. Over an average year, a company expects 36 of these particular loss events. During the next month, which is nearest to the probability the company will experience exactly zero (none) of these events?

a. 3.4%
b. 5.0%
c. 7.5%
d. 9.1%

316.3. The current price of an asset is S(0) and its future evolution is modeled with a binomial tree. At each node, there is a 62% probability of an (jump-up) increase. If the price increases, the next price is given by S(T+1)=S(T)*1.050; if the price decreases, the next price is given by S(T+1)=S(T)/1.050. Over the next year, there are twelve time steps, one for each month. Which is nearest to the probability that the final price of the asset is exactly the same as its current current price, S(0)?

a. 15.80%
b. 19.33%
c. 22.25%
d. 33.67%

Answers:

• Here in forum

 

[David Harper CFA FRM]

FYI, this begins a new sequence of fresh quantitative methods (topic 2) global topic review questions. We have pretty much surrounded the important P1 AIM-by-AIM, at this point, and we have already have P1.T3 and P1.T4 global topic reviews. So my plan is to build the P1.T2 global topic review (starting with this questions). By the time we finish, I imagine we will have received the 2014 draft AIMS, and we can then decide which sequence of P1 questions should come next (e.g., global T1 topic review, new key reading, etc). Thanks,

 

[MissJaguar]

To me it seems that the answer to question 316.3 provided in materials (15.824%) is totally wrong.

In a 12 - step-binomial tree there are 144 ( 12 power 2) states of nature at the end of the last step, with 13 different payoffs (12 +1). Moreover, there are 2 states of nature that can yield the same result as in the very beginning (USD 100), each with the probability (p^6) * ((1-p)^6) that is (0.62 ^ 6 )*(0.38^6). Thus 15.8 % is simply too large probability that cannot be the result.

Thanks for your feedback.

 

[David Harper CFA FRM]

Hi @MissJaguar (0.62^6)*(38^6) = 0.0171% is the probability for for one exact path to the terminal $100; e.g., 6 up then 6 down. The role of the binomial coefficient in the binomial distribution is, in this context, to produce the number of paths to the terminal value (conditional on a recombining tree, of course). Surprisingly, there are fully C(12,6) = 924 unique paths to the $100 at the end of 12 steps. That's why the probability = 15.80% = 0.0171% probability of an exact up/down path * 924 different paths (i.e., combinations of 6 up and 6 down).

Another way to look at this is: if we only summed the probabilities given by (0.62^x)*[(0.38)^(12-x)] for the final 13 outcomes, the sum is only 0.8320%, but it needs to be 100.0%!

Also, it's not 12^2 but rather 2^12 = 4,096 is the total number of unique paths which recombine to only 13 different final states, such fully 924 of the 4,096 finish at 100.

To verify the result, we can sum the probabilities, including the 15.80% to 100; and, the 13 different outcomes do sum to 100%. Also, the sum of C(12,x) with x running from 0 to 12, should sum to 4,096, which is does. So, this answer still looks fine to me. I hope that helps,

 

[MissJaguar]

Thanks, David. I trust I overlooked it. Thx + Best,