10.Paul Graham, FRM is analyzing the sales growth of a baby product launched threeyears ago by a regional company. He assesses that three factors contributeheavily towards the growth and comes up with the following results:
Y = b +1.5 X1 + 1.2X2 + 3X3
Sum ofSquared Regression [SSR] = 869.76
Sum ofSquared Errors [SEE] = 22.12
Determinewhat proportion of sales growth is explained by the regression results.
a. 0.36
b. 0.98
c. 0.64
d. 0.55
Coefficient of Determination i.e. R2 explains proportion of variation explained by the
regression. R2 = SSR/SST, SEE = (SSE/(n-2) )^1/2 , SST = SSR + SSE. Therefore, SSE = 489.29, SST =
1359.05, R2 = 0.64
愣是没看懂什么意思
33.Imagine a stack-and-roll hedge of monthly commodity deliveries that youcontinue for the next five years.
Assumethe hedge ratio is adjusted to take into effect the mistiming of cash flows butis not adjusted for the basis risk of the hedge. In which of the following situationsis your calendar basis risk likely to be greatest?
a.Stack and roll in the front month in oil futures.
b.Stack and roll in the 12-month contract in natural gas futures.
c.Stack and roll in the 3-year contract in gold futures.
d. Allfour situations will have the same basis risk.
答案是a. 没看懂他的解释：The oil term structure is highly volatile at the short end, making a front-month stack-androll
hedge heavily exposed to basis fluctuations. In natural gas, much of the movement occurs at the front
end, as well, so the 12-month contract won’t move as much. In gold, the term structure rarely moves
much at all and won’t begin to compare with oil and gas.
17. A junior bond with a face value of 200 matures in 5 years. A senior bond on the same firms also matures
in 5 years, and has a face value of 100. Assume -A = .5 and the riskfree rate=.04. Firm value is equal to
400. Using the Merton model, what is the value of the junior bond? (The following table includes figures
that will reduce the time required to answer this but it also includes figures that are irrelevant to the
problem and some that are strictly wrong).
d1 d2 N(d1) N(d2)
1.978 0.860 .9761 .8051
1.978 0.860 .8051 .9761
0.9952 -.1228 .8413 .5478
0.9952 -1.505 .8413 .0661
0.9952 -.1228 .8413 .4522
a. 99.07
b. 174.55
c. 75.48
d. 16.63
答案是a 官方解释这么说：
To get this answer use the first and last row of the table to make the calculation of the
Black-Scholes formula go faster. Plug in the values given into the B-S formula and use the first row to
get the value of the equity if the leverage were just 100 (the face value of the senior bond). Subtract
this off 400 to get the value of the senior bond. Plug in the values of the B-S formula assuming a
strike of 300 (the total debt) to get the value of the equity. Use the last row of the table to make this
go faster. Subtracting the value of the equity and the senior debt from 400 gives us the answer, a.
但是官方解释没有说为什么用第一行和第四行的数据。还有那个"-A" 是什么意思？
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