Answers _ Mathematics and Statistics for Financial Risk Management
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Answers
CHAPTER 1
1.
a. y
= 5
a.
y
= ln(1) – ln(
e
) = 0 – 1 = –1
b.
y
= ln(10) + ln(
e
) = ln(10) + 1 = 3.3026
2.
Annual rate = 5.12%; semiannual rate = 5.05%; continuous rate = 4.99%.
3.
4.
5.
6.
7.
8.
ln(ln(10)) = 0.8340.
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9.
10.
11.
The bond will pay 10 coupons of $2, starting in a year's time. In addi-
tion, the notional value of the bond will be returned with the final coupon
payment in 10 years. The present value, V, is then:
We start by evaluating the summation, using a discount factor of δ
=
1/1.05 ≈
0.95:
Inserting this result into the initial equation we obtain our final result:
Note that the present value of the bond, $78.83, is less than the notional
value of the bond, $100. This is what we would expect, given that there is
no risk of default, and the coupon rate is less than the discount rate.
CHAPTER 2
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1.
Probability that both generate positive returns = 60% × 70% = 42%.
Probability that both funds lose money = (1 – 60%) × (1 – 70%) = 40% ×
30% = 12%.
2.
88%. The sum of all three events—upgrade, downgrade, and no change
—must sum to one. There is no other possible outcome. 88% + 8% + 4% =
100%.
3.
50%. The outcomes are mutually exclusive; therefore, 20% + 30% =
50%.
4.
5.
6.
32.14%. By applying Bayes’ theorem, we can calculate the result:
Even though the model is 90% accurate, 95% of the bonds don't default
and it predicts that 10% of them will. Within the bond portfolio, the mod-
el identifies 9.5% of the bonds as likely to default, even though they won’t.
Of the 5% of bonds that actually default, the model correctly identifies
90%, or 4.5% of the portfolio. This 4.5% correctly identified is over-
whelmed by the 9.5% incorrectly identified.
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7.
Given the density function, we can find c
by noting that the sum of
probabilities must be equal to one:
8.
First we check that this is a valid CDF, by calculating the value of the
CDF for the minimum and maximum values of x
:
Next we calculate the PDF by taking the first derivative of the CDF:
9.
We first calculate the CDF by integrating the PDF:
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We first try to find c
using the fact that the CDF is zero at the minimum
value of x
, x
= 0.
As it turns out, any value of c
will satisfy this constraint, and we cannot
use this to determine c
.
If we use the fact that the CDF is 1 for the maximum value of x
, x
= e
, we
find that c
= 1:
The CDF can then be expressed simply as:
10.
P
(both bonds default) = 9%.
P
(one defaults) = 42%.
P
(neither defaults) = 49%.
11.
We can start by summing across the first row to get W
:
In a similar fashion, we can find X
by summing across the second row:
To calculate Y
, we can sum down the first column, using our previously
calculated value for W
:
Using this result, we can sum across the third row to get Z
:
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The completed probability matrix is:
The last part of the question asks us to find the conditional probability,
which we can express as:
We can solve this by taking values from the completed probability matrix.
The equity underperforms in 40% of scenarios. The equity underper-
forms and the bonds are downgraded in 15% of scenarios. Dividing, we
get our final answer, 37.5%.
12.
The probability that a B-rated bond defaults over one year is 2%. This
can be read directly from the last column of the second row of the ratings
transition matrix.
The probability of default over two years is 4.8%. During the first year, a
B-rated bond can either be upgraded to an A rating, stay at B, be down-
graded to C, or default. From the transition matrix, we know that the
probability of these events is 10%, 80%, 8%, and 2%, respectively. If the
bond is upgraded to A, then there is zero probability of default in the sec-
ond year (the last column of the first row of the matrix is 0%). If it re-
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mains at B there is a 2% probability of default in the second year, the
same as in the first year. If it is downgraded to C, there is a 15% probabili-
ty of default in the second year. Finally if a bond defaulted in the first
year it stays defaulted (the last column of the last row is 100%). Putting
this all together we have:
CHAPTER 3
1.
Mean = 6.43%; median = 5%.
2.
Mean = 3.00%; standard deviation = 6.84%.
3.
4.
Using the results of Question 3, we first calculate the variance of the es-
timator of the mean:
where σ
is the standard deviation of r
. For the last step we rely on the fact
that, because the data points are i.i.d., the covariance between different
data points is zero. We obtain the final answer by taking the square root
of the variance of the estimator:
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5.
Covariance = 4.87%; correlation = 82.40%.
6.
Series #1: Mean = 0, standard deviation = 39, skewness = 0.
Series #2: Mean = 0, standard deviation = 39, skewness = –0.626.
7.
Series #1: Mean = 0, standard deviation = 17, kurtosis = 1.690.
Series #2: Mean = 0, standard deviation = 17, kurtosis = 1.
8.
The variance is approximately 5.56. From a previous example we know
the mean to be 20/3; thus the variance can be found as:
9.
We start by expanding the mean:
By carefully rearranging terms, we are left with:
Assuming that all the different values of X are uncorrelated with each
other, we can use the following two relationships:
Then:
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10.
First we note that the expected value of X
A
plus X
B
is just the sum of
the means:
Substituting into our equation for variance, and rearranging, we get:
Expanding the squared term and solving:
Using our definition of covariance we arrive at our final answer:
11.
If the bond does not default, you will receive $100. If the bond does
default, you will receive 40% × $100 = $40. The future value, the expected
value of the bond at the end of the year, is then $94:
The present value of the bond is approximately $89.42:
CHAPTER 4
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1.
The number of times XYZ Corporation exceeds consensus estimates fol-
lows a binomial distribution; therefore:
2.
The cumulative number of exceedance events follows a binomial distri-
bution; therefore:
3.
Because the annual returns of both funds are normally distributed, the
difference in their returns is also normally distributed:
The mean of this distribution is 10%, and the standard deviation is 50%.
At the end of the year, the difference in the expected returns is 92%. This
is 82% above the mean, or 1.64 standard deviations. Using Excel or con-
sulting the table of confidence levels in the chapter, we see that this is a
rare event. The probability of more than a 1.64 standard deviation event
is only 5%.
4.
The average number of defaults over five months is 10; therefore:
5.
If the returns of the fund are normally distributed with a mean of 10%
and a standard deviation of 15%, then the returns of $200 million invest-
ed in the fund are also normally distributed, but with an expected return
of $20 and a standard deviation of $30. A loss of $18.4 million represents
a –1.28 standard deviation move:
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This is a one-tailed problem. By consulting the table of confidence inter-
vals or using a spreadsheet, we determine that just 10% of the normal dis-
tribution lies below –1.28 standard deviations.
6.
The return of –30% is approximately a –1.64 standard deviation event:
According to the table of confidence intervals, 5% of the normal distribu-
tion lies below –1.64 standard deviations. The probability of a return less
than –30% is then 5%.
7.
For the mean:
From a previous example, we know that c
= 1/(
x
2
– x
1
); therefore:
For the variance:
Substituting in for c
and μ
from above:
For the final step, we need to know that:
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Substituting in and solving, we have:
8.
Using integration by substitution, define a new variable y
and solve:
9.
Using the same substitution as in the previous question:
10.
Using the same substitution as before:
For the final step we need to know that:
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Using this result, we achieve the desired result:
11.
First we note that the mean of X
A
is zero:
Similarly, the mean of X
B
is zero.
Next, we want to calculate the variance. In order to do that, it will be use-
ful to know two relationships. First we rearrange the equation for vari-
ance, Equation 3.20
, to get:
Similarly, we can rearrange our equation for covariance, Equation 3.26
,
to get:
With these results in hand, we now show that the variance of X
A
is one:
Next we calculate the covariance of X
A
and X
B
:
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Putting the last two results together completes the proof:
12.
For the portfolio consisting of 50% A
and 50% B
, we can proceed two
ways. The PDF of the portfolio is a triangle, from –0.5 to +0.5, with height
of 2.0 at 0. We can argue that the mean is zero based on geometric argu-
ments. Also, because both distributions are just standard uniform vari-
ables shifted by a constant, they must have variance of 1/12; 50% of each
asset would have a variance of ¼ this amount, and—only because the
variables are independent—we can add the variance of the variable, giv-
ing us:
Alternatively, we could calculate the mean and variance by integration:
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This confirms our earlier answer.
For the 50/50 mixture distribution, the PDF is bimodal and symmetrical
around zero, giving a mean of zero:
For the variance we have:
Notice that, while the mean is the same, the variance for the mixture dis-
tribution is significantly higher.
CHAPTER 5
1.
Mean = 45.0; standard deviation = 29.3; standard deviation of mean =
9.3. For the hypothesis that the mean is greater than 40, the appropriate t
-
statistic has a value of 0.54. For a one-sided t
-test with 9 degrees of free-
dom, the associated probability is 70%. There is a 30% chance that the
true mean is found below 40, and a 70% chance that it is greater than 40.
2.
The mean is 6.9%, and the standard deviation of the returns is 23.5%,
giving a standard deviation of the mean of 7.4%. The t
-statistic is 0.93.
With 9 degrees of freedom, a one-sided t
-test produces a probability of
81%. In other words, even though the sample mean is positive, there is a
19% chance that the true mean is negative.
3.
A negative return would be greater than 2 standard deviations. For a
normal distribution, the probability (one-tailed) is approximately 2.28%.
If we do not know the distribution, then, by Chebyshev's inequality, the
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probability of a negative return could be as high as 12.5% = ½ × 1/(2
2
).
There could be a 25% probability of a +/–2 standard deviation event, but
we're interested only in the negative tail, so we multiply by ½. We can
only perform this last step because we were told the distribution was
symmetrical.
4.
The expected return is +10%. The 95% VaR is 35% (i.e., 5% of the re-
turns are expected to be worse than –35%). The expected shortfall is
37.5% (again the negative is implied).
5.
For a normal distribution, 5% of the weight is less than –1.64 standard
deviations from the mean. The 95% VaR can be found as: 0.40% – 1.64 ·
2.30% = –3.38%. Because our quoting convention for VaR, the final answer
is VaR = 3.38%.
6.
We can use Equation 5.4
to calculate the expected variance of the sam-
ple variances. Because we are told the underlying distribution is normal,
the excess kurtosis can be assumed to equal zero and n
= 33; therefore:
The standard deviation of the sample variances is then 4.0%.
7.
An appropriate null hypothesis would be: H
0
: σ
= 40%. The appropriate
test statistic is:
Using a spreadsheet, or other program, we calculate the corresponding
probability for a chi-squared distribution with 32 degrees of freedom.
Only 2.23% of the distribution is greater than 50. At a 95% confidence lev-
el, we would reject the null hypothesis.
8.
Answer: 12.5%. This is a –2 standard deviation event. According to
Chebyshev's inequality, the probability of being more than 2 standard de-
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viations from the mean is less than or equal to 25%.
Because the distribution of returns is symmetrical, we assume that half of
these extreme events are greater than +2 standard deviations, and half
are less than –2 standard deviations. This leads to the final result, of
12.5%.
9.
The standard deviation of the mean is 2%:
This makes the difference between the average fund return and the
benchmark, 18% – 14% = 4%, a +2 standard deviation event. For a t
distri-
bution with 35 degrees of freedom, the probability of being more than +2
standard deviations is just 2.67%. We can reject the null hypothesis, H
0
:
μ
= 14% at the 95% confidence level. The difference between the average
return and the benchmark return is statistically significant.
10.
To find the 95% VaR, we need to find v
, such that:
Solving, we have:
The VaR is a loss of 90. Alternatively, we could have used geometric argu-
ments to arrive at the same conclusion. In this problem, the PDF de-
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scribes a rectangle whose base is 200 units and whose height is 1/200. As
required, the total area under the PDF, base multiplied by height, is equal
to one. The leftmost fraction of the rectangle, from –100 to –90, is also a
rectangle, with a base of 10 units and the same height, giving an area of
1/20, or 5% of the total area. The edge of this area is our VaR, as found by
integration before.
11.
In the previous question we found that the VaR, v
, was equal to –90. To
find the expected shortfall, we need to solve the following equation:
Solving, we find:
The final answer, a loss of 95 for the expected shortfall, makes sense. The
PDF in this problem is a uniform distribution, with a minimum at –100.
Because it is a uniform distribution, all losses between the (negative) VaR,
–90, and the minimum, –100, are equally likely; therefore, the average
loss, given a VaR exceedance, is halfway between –90 and –100.
12.
To find the 95% VaR, we need to find v
, such that:
By inspection, half the distribution is below 5, so we need only bother
with the first half of the function:
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We can use the solution to the quadratic equation:
Because the distribution is not defined for π
< –15, we can ignore the neg-
ative, giving us the final answer:
The one-day 95% VaR for Pyramid Asset Management is approximately
8.68.
CHAPTER 6
1.
2.
For the first part of the question, because matrix addition is commuta-
tive and associative, the order in which we perform the operations does
not matter:
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3.
4.
5.
A matrix multiplied by an appropriately sized identity matrix is itself.
This is true when the identity matrix is multiplied by itself, too.
For U
2
:
For AU
:
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6.
To prove that J
is the inverse of K
, we need to show that the two matri-
ces multiplied together produce an identity matrix.
7.
To solve this problem we could multiply M
by itself five times.
Alternatively, we can reexpress M
as the product of a constant and an
identity matrix:
8.
At the end of the year, it is expected that 61% of the bonds will have an
A rating, 36.4% B, 2.2% C, and 0.4% D. To get the answer, we can proceed
one rating at a time. Of the 60% of bonds that are rated A at the start of
the year, we expect 95% will still be rated A at the end of the year. Of the
40% of bonds that are rated B at the start of the year, we expect 10% to
have been upgraded to A by the end of the year. Putting the two together,
we have:
We can calculate the other three ratings similarly:
We can check our answer by noting that the sum of the answers is 100%.
At the end of the year each bond must be either A, B, C, or D; therefore,
the sum of the expected values must be 100%.
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9.
To calculate the two-year transition matrix, we simply square the one-
year matrix. Using T
1
and T
2
to denote our one-year and two-year matri-
ces, respectively, we have:
Though not necessary, we can reformat this to match the original one-
year matrix:
10.
We can use our Cholesky algorithm to find the elements of the matrix:
We can express the full lower triangular matrix as:
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We can verify this answer by noting that LL
′ is indeed equal to our origi-
nal covariance matrix, Σ
.
CHAPTER 7
1.
Vectors a
and b
are not orthogonal, but b
and c
are orthogonal. We
know this from their inner products, which we can calculate as follows:
2.
In order for A
to be an orthonormal basis, we require that the column
vectors are orthogonal and have a magnitude of one. For the two column
vectors to be orthogonal, we require that their inner product is zero:
We next check that the column vectors have a magnitude of one:
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Both vectors are normal; therefore, the solution:
makes A
an orthonormal basis.
3.
In order for B
to be an orthonormal basis, we require that the column
vectors are orthogonal and have a magnitude of one. For the two column
vectors to be orthogonal, we require that their inner product is zero:
Using the fact that the magnitude of the first column vector must be one:
Substituting in our previous result:
Both the positive and negative root are legitimate solutions. There are ac-
tually two possible final answers.
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4.
Because B
is an orthonormal basis, we can find the coordinate vector
for x
:
5.
A coordinate vector, c
, for x
should satisfy the following equation:
Working through produces three simultaneous equations:
By solving and substituting in, we arrive at the final answer: c
1
= 3, c
2
= 1,
c
3
= 4.
CHAPTER 8
1.
The expected return of XYZ is 6.01%:
2.
The expected value of r
XYZ
is 0.07%:
The variance of r
XYZ
is:
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To get to the last line, we use the fact that the covariance between the re-
gressor and the disturbance term is zero in a linear regression, which
implies:
Taking the square root of the variance, we get a standard deviation of
2.06%.
3.
We start by calculating the covariance:
The correlation is then:
4.
The R
2
is 20%:
5.
The corresponding F
-statistic is 12:
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Using a spreadsheet or other program, we see that the probability associ-
ated with this F
-statistic is 0.11%; that is, there is only a 0.11% chance that
an F
-statistic of this magnitude (or greater) would have happened by
chance. The F
-statistic is significant at the 95% confidence level.
6.
We compute the adjusted R
2
for each model. The univariate model has
two regressors, including the constant. The second has four:
On this basis, the original univariate model is slightly better.
7.
30.5%:
8.
One possible solution is to drop X
3
from the model:
Another possibility, if the spread between X
2
and X
3
is of interest, is:
where β
5
is taken from the regression of X
2
on X
3
. Based on the assump-
tion of the OLS model, the term in parentheses will be uncorrelated with
X
2
.
9.
We start by writing the equation for the covariance of X
and Y
:
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Using our linear regression equation and making use of the OLS assump-
tions, we see that the second term can be expressed in terms of X
, β
, and :
Substituting this into our covariance equation:
All that remains is to divide both sides by the variance of X
, and to ex-
pand the correlation term:
10.
First, we find the optimal value of α
, α
*:
Next, we solve for β
:
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At this point we substitute in our optimal value of α
, α
*:
CHAPTER 9
1.
The models are:
a.
AR(2)
b.
AR(1)
c.
ARMA(2,1)
d.
Drift-diffusion
2.
3.
The expected value of r
t
is 0.80% the next period, and 0.84% the follow-
ing period:
To get the two-period-ahead forecast, we can use the previous result:
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Alternatively, we can substitute the original equation into itself to get r
t
in
terms of r
t
–2
and r
t
–3
:
4.
The expected log return over one year is 0.0%. The standard deviation
of the annual log return is 24%.
We can get this result by recognizing the annual return as a collection of
i.i.d. variables, and using our square root rule to calculate the standard
deviation. More formally, we can construct the annual return series (re-
member, log returns are additive):
where r
256,
t
is our 256-day annual return. We can find the expected value
as follows:
We can then calculate the variance as follows:
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For each term in the final summation, we can determine the value by not-
ing the following:
Now we have:
The variance of the annual returns is 256 times as great as the variance of
the daily returns. To get the standard deviation, we just take the square
root of both sides:
5.
The expected log return over one year is 25.6%. The standard deviation
of the annual log return is 24%.
As before, we can get this result by recognizing the annual return as a col-
lection of i.i.d. variables, and using our square root rule to calculate the
standard deviation. More formally, we can construct the annual return
series (remember, log returns are additive):
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where r
256,
t
is our 256-day annual return. We can find the expected value
as follows:
Using this result, we can then calculate the variance as follows:
This is exactly the same as what we had in the previous question. The ad-
dition of the drift term does not impact the variance or standard devia-
tion. The final result is the same as before:
6.
The expected log return over two days is 0.40%. The standard deviation
of the two returns is 3.0%. For the case where λ
equals –0.50, the mean of
the two-day return would be approximately 0.13%, and the standard de-
viation would be approximately 1.73%.
We start by expressing the original AR(1) equation as an infinite sum of
lags of the disturbance term:
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Constructing the two-period return is fairly straightforward. Paying care-
ful attention to the time subscripts, we can group the disturbance terms
into one summation:
where r
2,
t
is our two-day return. We can find the expected value as
follows:
We then proceed to find the variance:
The standard deviation of the two-day return is then:
7.
The unconditional mean of the model is equal to θ
, 4%. If interest rates
start out at 6%, then we would expect interest rates to be 5.00%, then
4.50%, and then 4.25% over the next three periods. This result is obtained
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by noting that the conditional expectation for the next period's interest
rate is, in this case, simply the average of the previous period's rate and
the long-term mean of 4%:
8.
By iteratively substituting the equation into itself, we see that this
process can be written as an infinite moving average:
The unconditional mean is 0:
Similarly, we can find the unconditional variance. First we note that, be-
cause the covariance between different disturbance terms is zero and the
expected value of any individual disturbance terms is zero, we have the
following:
Using this and the fact that the unconditional mean is also zero:
9.
Using the results from the previous question, we first derive an expres-
sion for the covariance:
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For the last line we were able to eliminate the first term in the expecta-
tions by noting that all the products contained different disturbance
terms. From the preceding problem we know that the expected value of
these cross products is zero. Because the unconditional variance is the
same for both r
t
and r
t
–
n
, finding the correlation is just a matter of divid-
ing Var[
r
t
]:
10.
We start by expressing both r
t
, and r
t
–1
as infinite series:
Next we find the mean of both series:
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Because the error terms are uncorrelated, we know that:
Using this and the previous results, we calculate the variances and
covariance:
Finally, the correlation is:
11.
To annualize the log return, we simply multiply by the number of
months in a year, 12. To get the annualized standard deviation, we multi-
ply by the square root of the number of periods. For skewness and kurto-
sis, we divide by the square root of 12 and 12, respectively. This gives:
mean = 24%; standard deviation = 5.20%; skewness = –0.29; kurtosis =
0.20.
CHAPTER 10
1.
We need to find h
, such that:
Solving, we find:
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Alternatively, the formula for the half-life can be expressed as:
2.
We start by computing decay factors and values for x
2
:
For the mean, using Equation 10.2
, we have:
For the variance, using Equation 10.12
, we have:
Finally, we can take the square root of our answer for the variance, to get
the standard deviation, 22.48.
3.
We start by calculating the following values:
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We then use Equation 10.2
to calculate our estimates of the mean: mean
(no decay) = 0.5236; mean (decay = 0.99) = 0.5263; mean (decay = 0.90) =
0.5486.
4.
We start by expanding the table from our answer to question #3:
In the last line, we have used our estimate of the mean (no decay) from
the previous problem.
For the first estimator with no decay factor we can use Equation 3.19
to
calculate the variance:
For the second and third estimators, we use Equation 10.12
, and our esti-
mates of the mean from the previous question:
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Taking the square root of the variances, we arrive at our final answers:
standard deviation (no decay) = 0.2688; standard deviation (decay =
0.99) = 0.2676; standard deviation (decay = 0.90) = 0.2610.
5.
The new estimates are 10.10%, 9.82%, and finally 9.78%. These can be
found as follows:
6.
The half-lives are:
7.
The half-life of the EWMA estimator is approximately 11.11 days. A rec-
tangular window with 22 days would have the most similar half-life, 11
days.
8.
9.
Approximately 19.72%. We first update our estimate of the variance,
and then take the square root:
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10.
Approximately 10.25%. We can use our updating rule:
to calculate successive estimates of the variance. The estimate of the stan-
dard deviation is just the square root of the variance estimator:
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Five years of realized returns for CCC are given in the following table. Remember:
1. While CCC was started 40 years ago, its common stock has been publicly traded for the past 25
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Office
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2012
2014
2015
2016
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Quantitative Problem 1: You plan to deposit $2,200 per year for 6 years into a money market account with an annual return of 2%. You plan
to make your first deposit one year from today.
a. What amount will be in your account at the end of 6 years? Do not round intermediate calculations. Round your answer to the nearest cent.
4
b. Assume that your deposits will begin today. What amount will be in your account after 6 years? Do not round intermediate calculations.
Round your answer to the nearest cent.
6
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Review the FVAN definition and its equation.
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Understand the difference between an ordinary annuity and an annuity due.
Be careful about the order of mathematical operations if using the equation.
If using a financial calculator, be…
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E Chapter 10 Homework - FINANC X
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E10.1 (LO 1) (Acquisition Costs of Realty) The following expenditures and receipts are related to land, land improvements, and buildings acquired for use in a business enterprise. The receipts are
enclosed in parentheses.
a. Money borrowed to pay building contractor (signed a note)
$(275,000)
b. Payment for construction from note proceeds
275,000
c. Cost of land fill and clearing
8,000
d. Delinquent real estate taxes on property assumed by purchaser
7,000
e. Premium on 6-month insurance policy during construction
6,000
f. Refund of 1-month insurance premium because construction completed early
(1,000)
g. Architect's fee on building
22,000
h. Cost of real estate purchased as a plant site (land $200,000 and building $50,000)
250,000
i. Commission…
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8-19
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1 of 2
On December 21 2017. Wildhorse Co. purchased $730,000 of commercial paper from Voight, Inc. The commerc
paper matures January 14, 2018.
Currency and coin on hand amounted to $10,446,
A money market fund account held at Rogers, Inc. (a mutual fund organization) permits Wildhorse Co. to write chee
on the full account balance, which was $930,000 at December 31, 2017.
Wildhorse Co. has agreed to maintain a cash balance of $200,000 at all times at First National Bank of Alexander to
ensure future credit availability.
The balance in Wildhorse Co's petty cash fund is $400.
During December 2017, Wildhorse Co. made travel advances to employees totaling $30,000 for executive travel that
9.
10.
will occur during the first quarter of 2018. The employees will not reimburse the company for the advance, but are
required to spend the funds on company-related travel and to submit receipts.
The company received an LO.U. trom Stiller LLP, a company customer, on December 12. 2017. in the amount of…
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Pears
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Chapter: Recording Business TraNSACTIONS
Sep. 1
hed:course:7710589/products/79c3fa4c-a84f-42ba-b87a-e36a400bca00/pages/urn:pe
6
7
Proble X
P-F:2-30A. Journalizing transactions, posting journal entries to T-
accounts, and preparing a trial balance (Learning Objectives 3, 4)
15
G
New
Ann Simpson started her practice as a design consultant on September 1,
2024. During the first month of operations, the business completed the
following transactions:
3. Prepare the trial balance of Vince York, M.D., as of July 31, 2024.
30
Received $48,000 cash and issued common stock to Simpson.
Purchased office supplies, $1,200, and furniture, $1,300, on account.
Performed services for a law firm and received $1,900 cash.
X
Paid $18,000 cash to acquire land to be used in operations.
Performed services for a hotel and received its promise to pay the $1,200
within one week.
Paid for the furniture purchased on September 4 on account.
Paid assistant's semimonthly salary, $1,500.
Received cash on…
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Question 4 of 5
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Current Attempt in Progress
Van Occupanther is the bookkeeper for Roscoe Company. Van has been trying to get the balance sheet of Roscoe Company to balance.
Roscoe's balance sheet is as follows.
ROSCOE COMPANY
Balance Sheet
December 31, 2022
Assets
Liabilities
Cash
$ 9,400
Accounts payable
$25,000
Supplies
7,100
Accounts receivable
(19,500)
Equipment
45,000
Common stock
40,000
Dividends
9,200
Retained earnings
25,200
Total assets
$70,700
Total liabilities and stockholders' equity
$70,700
Prepare a correct balance sheet. (List Assets in order of liquidity.)
ROSCOE COMPANY
Balance Sheet
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Quick ratio
Adieu Company reported the following current assets and current liabilities for two recent years:
Dec. 31, 20Y4 Dec. 31, 20Y3
Cash
$830
$1,090
Temporary investments
1,200
1,500
Accounts receivable
820
910
Inventory
2,100
2,500
Accounts payable
1,900
2,500
a. Compute the quick ratio on December 31 for each year. Round to one decimal place.
20Υ4
20Υ3
Quick Ratio
b. Is the quick ratio improving or declining?
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Question 9 of 10 <
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At August 31, Coffman Company has this bank information: cash balance per bank $6,450; outstanding checks $2,762; deposits in
transit $1,700; and a bank service charge $20.
Determine the adjusted cash balance per bank at August 31, 2021.
Adjusted cash balance per bank
2$
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Harper Industries has $400 million of common equity on its balance sheet; its stock price is $70 per share; and its market value added (MVA) is $90 million. How many common shares are
currently outstanding? Write out your answer completely. For example, 5 million shares should be entered as 5,000,000. Round your answer to the nearest whole number, if necessary.
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In its most recent financial statements, Nessler Inc. reported $35 million of net income and $420 million of retained earnings. The previous retained earnings were $391 million. How much in
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during the year? Assume
dividends declared were actually paid. Write out your answer completely. For example, 25 mill
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2022 Spring
PROBLEM 1
lome
Use the Excel Template File to answer Problem 1 and Problem 2
Exam II Excel Template File. Problem1 and Problem2.xlsx
nnouncements
The accounts listed below for the year ended 2021, in random order, are for Polanyi Knowledge, Inc.
Construct a classified balance sheet for the business in good form.
yllabus
Modules
Cash
32,900
Discount on bonds payable
103.000
rades
Accumulated depreciation-Buildings 630,600
Wages payable
92,000
eople
Accounts Receivable
90,200
Note payable due 2023
300,000
uJa
Taxes payable
27,000
Retained earnings
980,000
Inventory
180,000
Bonds payable
500,000
Additional paid in capital common
Common stock
264,400
150,000
stock
Long-term investments
77,000
Preferred stock
110,000
Buildings
1,500,000
Accounts payable
167.000
Additional paid in capital preferred
Accumulated depreciation:
13,400
stock
210,000
equipment
Equipment
760,000
Land
310,600
Prepaid…
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6.4 Mortgage Financing
Jessica's mortgage payment, including principal, interest, taxes, and insurance, is $750 per month. She also has other
monthly debt payments of $440. If her monthly gross income is $3,950. How would a bank assess Jessica's capacity to
take on an additional credit card?
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X Bb Microsoft Word - Cash-to- X Bb Homework 5
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Cash-to-accrual assignment
1. Slater Company uses the cash basis of accounting. Slater Company collected $950,000
from its customers during 2019. Customers owed Slater $150,000 of accounts receivable
at the beginning of 2019, and $190,000 of accounts receivable at the end of 2019. What is
Slater's sales revenue for 2019 under the accrual basis of accounting?
2. Merry Company uses the cash basis of accounting. Merry Company made $600,000 in
payments to its suppliers during the year. Merry's beginning inventory was $20,000, and
its ending inventory was $35,000. In addition, Merry had a beginning accounts payable of
$40,000 and an ending accounts payable of $70,000. What is Merry's cost of goods sold
under the accrual basis of accounting?
(((
3. Little Company uses the cash basis of accounting.…
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The adjustment of overapplied manufacturing overhead cost results in all the following, EXCEPT:
O a. decrease in net operating income.
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O b. increase in net operating income.
out of
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decrease in cost of goods sold
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XYZ Company had the following information for the year:
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OMR 110,000
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OMR 150,000
Actual manufacturing overhead incurred
OMR 166,000
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The Company used a predetermined overhead rate of OMR 30 per direct labor hour for the year. Assume the only inventc
an ending Work in Process Inventory balance of OMR 17,000. What was cost…
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Foundations Of Finance
Finance
ISBN:9780134897264
Author:KEOWN, Arthur J., Martin, John D., PETTY, J. William
Publisher:Pearson,
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Fundamentals of Financial Management (MindTap Cou...
Finance
ISBN:9781337395250
Author:Eugene F. Brigham, Joel F. Houston
Publisher:Cengage Learning
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Corporate Finance (The Mcgraw-hill/Irwin Series i...
Finance
ISBN:9780077861759
Author:Stephen A. Ross Franco Modigliani Professor of Financial Economics Professor, Randolph W Westerfield Robert R. Dockson Deans Chair in Bus. Admin., Jeffrey Jaffe, Bradford D Jordan Professor
Publisher:McGraw-Hill Education