GARP Practice Q12: t-Distribution

[ami44]

Hallo,

I hope this hasn't been asked elsewhere here, but I'm stuck on Question 12 of the 2014 practice Exam. The question is:

A risk manager is examining a Hong Kong trader's profit and loss record for the last week, as shown in the table below:

Trading Day | Profit/Loss (HKD million)
Monday..... | 10
Tuesday.....| 80
Wednesday...| 90
Thursday....| -60
Friday......| 30

The profits and losses are normally distributed with a mean of 4.5 million HKD and assume that transaction costs can be ignored. Part of the t-table is provided below:

DF| alpha=0.3 | alpha=0.2 | alpha=0.15
4 | 0.569 ....| 0.941 ....| 1.19
5 | 0.559 ....| 0.920 ....| 1.156

According to the information provided above, what is the probability that the trader will record a profit of at least 30 million HKD on the first trading day next week.

a. About 15%
b. About 20%
c. About 80%
d. About 85%

The official Answer goes like this:
Calculate the mean xbar = 30 and std. dev. of the mean s_xbar = 27.0185
Calculate a t-statistics = (30 - 4.5) / 27.0185 = 0.9438
With this value go into the t-table (4 degrees of freedom) and come up with the probability 0.2

My Problem with this is:
We know that (xbar - 4.5) / s_xbar has a t-distribution with 4 degrees of freedom. But the question asks about the distribution of the profits and losses of one trading day (next Monday), lets call that value x6. If my understanding is correct, the question asks about the value of P(x6 > 30).

The answer above seems to suggest that (x6 - 4.5) / s_xbar is also t-distributed with 4 degrees of freedom and I doubt that (because xbar und x6 have different standard deviation).
Instead I believe (x6 - 4.5) / s might be t-distributed which would result in an t-value of 0.422 and a probability of 0.34.

Maybe I'm missing something. I'm learning from the Schweser books, so something might have been left out there. Can anybody tell me where my error might be or confirm, that there is a problem with the question?

 

[David Harper CFA FRM]

Hi @ami44 I absolutely agree with you, this Q&A reads like a misunderstanding to me. I can't say it much better than you have already. It is only reasonable to assume the next trading day's profit (your x6) is independent of the prior five given. We are already told the P/L are normally distributed, therefore the probability that this single day's profit will exceed 30 is Pr[ Z > (X6 - 30)/s(population P&L)]. To my knowledge, however, we cannot infer (X6 - 30)/S(sample P&L) is t-distributed given the same reason you cite: it's not a distribution of the sample mean. If for some reason we *assumed* the sample standard deviation estimates the population standard deviation, then I would also get Pr [ Z > (30-4.5)/60.415] = 1 - NORM.S.DIST(0.422) ~= 34%

It's interesting the next day's threshold equals the current sample mean, which leads me to wonder if the question is thinking to simply query the probability that, given a population mean, the sample mean will exceed 30. This is the question explored by the answer, actually. The given explanation to the answer, in so many words, is a test of the null hypothesis that (given the) population mean of 4.5, and observed sample mean (with 4 df) in excess of 30 is 20% probable. So, it seems likely the question's intent is "given an additional trading day, which if 30 would not change the sample mean of 30, what is the probability the sample mean will exceed 30?" The answer to that depends on whether you want to realistically account for the prior fact that the five trading days have already occured ...
... if we add another day, and make an unrealistic leap back to an unconditional state (neglecting the 5 trading days have already occurred), then we could ask, given six trading days, what is the prob the sample mean exceeds 30? By my calculation that reduces the variances and produces a critical t of 1.156 which happens to be displayed in the exhibit (?!); just as Pr[X-bar > 30 | 5 df] ~= 20%, the Pr[X-bar > 30 | 6 df] ~= 15%. You can't really do this, either, just interesting .... the text of the answer suggests they confused a single observation with the properties of a sample mean. So I do agree with you!

 

[ami44]

David, thank you for confirming my suspicion.
I agree, that the sample mean and the threshold are both 30 seems somehow suspicious. It feels as if it should mean something, but than it doesn't seem to.