Solution of investment Essay

Custom Student Mr. Teacher ENG 1001-04 15 April 2016

Solution of investment

1. The Fisher equation tells us that the real interest rate approximately equals the nominal rate minus the inflation rate. Suppose the (expected or realized) inflation rate increases from 3% to 5%. Does the Fisher equation imply that this increase will result in a fall in the real rate of interest? Explain.

The Fisher equation relates nominal rates required by investors to real rates required by investors and inflation. You can think about this from two perspectives:

i. Ex-ante (before) required Nominal Return as a function of required Real Return and Expected Inflation: (1 + rNominal) = (1 + rReal)(1 + E(i))

ii. Ex-post (afterward) realized Real Return as a function of Nominal Return and Realized Inflation: (1 + rReal) = (1 + rNominal)/(1 + i)

Assume the question asks this instead:

Suppose the expected inflation rate increases from 3% to 5%. Does the Fisher equation imply that this increase will result in a fall in the real rate of interest?

The theory is the ex-ante required real return is not a function of inflation. If expected inflation increase, required real return will remain unchanged and the required nominal return will increase (the security will have to pay more).

Now assume the question asks this:

Suppose realized inflation is 5% instead of the 3% expected inflation. Does the Fisher equation imply that this increase will result in a fall in the real rate of interest?

Since the nominal return was set when the asset was purchased, a realized inflation greater than expected inflation will decrease the realized real return.

2. You’ve just stumbled on a new dataset that enables you to compute historical rates of return on U.S. stocks all the way back to 1880. What are the advantages and disadvantages in using these data to help estimate the expected rate of return on U.S. stocks over the coming year?

If we assume that the distribution of returns remains reasonably stable (same expectation and standard deviation) over the entire history, then a longer sample period increases the precision of the sample statistic’s estimate of the actual expected return. The standard error of the estimate decreases as the sample size increases. (If you use the sample mean as an estimate of the “true” population mean, the standard error is the sample standard deviation divided by the square root of the sample size.)

However, if we believe the expected return of stocks is different now than it was in the late in the late 1800’s (due to changes in the structure of the economy for example), using data from that time period may not be appropriate for estimate modern stock price changes and only more recent data should be employed.

7. Suppose your expectations regarding the stock price are shown in the table below. Use equations 5.11 and 5.12 to compute the mean and standard deviation of the HPR on stocks.

State
Prob
End Price
HPR
Boom
0.35
$140
44.50%
Normal
0.30
$110
14.00%
Recession
0.35
$80
-16.50%

E(r) = [0.35 x 0.445] + [0.30 x 0.14] + [0.35 x (-0.165)] = 14% 2 = [0.35 x (0.445 – 0.14)2] + [0.30 x (0.14 – 0.14)2] + [0.35 x (-0.165 – 0.14)2] = 0.065127 = 25.52%

10. The continuously compounded annual return on a stock is normally distributed with a mean of 20% and standard deviation of 30%. With 95.44% confidence, we should expect its actual return in any particular year to be between which pair of values? Hint: Look again at Figure 5.4. With probability 0.9544, the value of a normally distributed variable will fall within two standard deviations of the mean; that is, between –40% and 80%.

11. Using historical risk premiums over the 1926–2009 period as your guide, what would be your estimate of the expected annual HPR on the S&P 500 stock portfolio if the current risk-free interest rate is 3%? From Table 5.3 and Figure 5.6, the average risk premium for large-capitalization U.S. stocks for the period 1926-2009 was: (11.63%  3.71%) = 7.92% per year Adding 7.92% to the 3% risk-free interest rate, the expected annual HPR for the S&P 500 stock portfolio is: 3.00% + 7.92% = 10.92%

13. During a period of severe inflation, a bond offered a nominal HPR of 80% per year. The inflation rate was 70% per year.

(a) What was the real HPR on the bond over the year?

(1 + R) = (1 + r)(1 + i)
r = (1 + R)/(1 + i) – 1 = (R – i)/(1 + i) = (0.80 – 0.70)/(1.70) = 5.88%

(b) Compare this real HPR to the approximation r ≈ R − i. r  R  i = 80%  70% = 10%
The greater inflation, the worse the estimate of real return.

15. An economy is making a rapid recovery from steep recession, and businesses foresee a need for large amounts of capital investment. Why would this development affect real interest rates? Real interest rates would be expected to rise as the demand for funds increased. See the shift in the demand curve in Figure 5.1).

Extra Questions:

1. A credit card charges 18% APR monthly. Calculate the EAR.
EAR = [1 + (APR/m)]m – 1 = (1 + 0.18/12)12 – 1 = 19.56%

2. The price of a stock on the last day of the year for the last 6 years in given in column 2 of the table below. (1)
(2)
(3)
(4)
Year
Price
Return
1 + Return
2009
$ 26.00
-7.14%
0.9286
2008
$ 28.00
16.67%
1.1667
2007
$ 24.00
-4.00%
0.9600
2006
$ 25.00
13.64%
1.1364
2005
$ 22.00
10.00%
1.1000
2004
$ 20.00

(a) Calculate the arithmetic average return. Interpret the value. E(r) = (-7.14% + 16.67% + -4.00% + 13.64% + 10.00%)/5 = 5.83%
The expected return next period is 5.83% assuming an equal weight on each of the last 5 years returns.

(b) Calculate standard deviation of the expectation.
σ = {[(-7.14% – 5.83%)2 + (16.67% – 5.83%)2 + (-4.00% – 5.83%)2 + (13.64% – 5.83%)2 + (10.00 – 5.83%)2]/(5 – 1)}1/2 = 10.73%

(c) Assume the returns for this stock are normally distributed. What is the likelihood of observing a return less than (5.83% – 10.73%) = -4.90%? The area under the Normal curve to the left of the mean less one standard deviation is approximately 16%, so there is a 16% chance of observing a return less that -4.90%

(d) Calculate expected return for the stock assuming an equal likelihood of each return over the last five years. E(r) = 0.2(-7.14%) + 0.2(16.67%) + 0.2(-4.00%) + 0.2(13.64%) + 0.2(10.00%) = 5.83%

(e) Now assume you believe that only the last two years are relevant. Calculate expected return for the stock assuming you assign a 50% chance to the 2009 return and a 50% chance to the 2008 return. E(r) = 0.5(-7.14%) + 0.5(16.67%) = 4.76%

(f) Calculate the standard deviation of the expectation assuming you assign a 50% chance to the 2009 return and a 50% chance to the 2008 return. σ = [0.5(-7.14% – 4.76%)2 + 0.5(16.67% – 4.76%)2]½ = 11.90%

(g) Calculate the value of $100 investment in the stock over the last 5 years. Wealth Index = (1 – 0.0714) (1 + 0.1667) (1 – 0.0400) (1 + 0.1364)(1 + 0.1000) = (0.9286)(1.1667)(0.9600)(1.1364)(1.100) = 1.3000 $100 dollars would be worth $100 x 1.3000 = $130

(h) Calculate the five-year holding period return for an investment in the stock over the last 5 years. HPR = [(1 – 0.0714)(1 + 0.1667)(1 – 0.0400)(1 + 0.1364)(1 + 0.1000)] – 1 = (0.9286)(1.1667)(0.9600)(1.1364)(1.100) = 1.3000 – 1 = 30.00%

(i) Calculate the ANUALIZED holding period return for five-year investment in the stock. Annualized HPR = [(1 – 0.0714)(1 + 0.1667)(1 – 0.0400)(1 + 0.1364)(1 + 0.1000)](1/5) – 1 = [1.3000]0.20 – 1 = 5.39%

(j) Calculate the Geometric Mean return for a five-year investment in the stock. Interpret the value. Geomean = [(1 – 0.0714) (1 + 0.1667) (1 – 0.0400) (1 + 0.1364)(1 + 0.1000)](1/5) – 1 = [1.3000]0.20 – 1 = 5.39%

= (1 + HPR) (1/5) – 1 = 1.3000 (1/5) – 1 = 5.39%
The return form this stock was equivalent to earning 5.39% each year. (1.0539)(1.0539)(1.0539)(1.0539)(1.0539) = 1.3000 = Wealth Index
[(1.0539)(1.0539)(1.0539)(1.0539)(1.0539)]– 1 = 30% = HPR

3. An investment has a 50% probability of an 80% return and a 50% probability of a -50% return.

(a) Calculate the expected one-period return of the investment and the standard deviation about that expectation. E(r) = 0.50(0.80) + 0.50(-0.50) = 15%
σ = [0.50(0.80 – 0.15)2 + 0.50(-0.50 – 0.15)2]½ = [0.4225]½ = 65.00%
The expected one-period return is positive.

(b) Estimate the expected periodic holding period return for the investment over a large number of periods. The geometric mean return is the expected periodic holding period return. Geometric Mean ≈ Arithmetic Mean – σ2/2 = 0.15 – (0.65)2/2 = 0.15 – 0.2113 = -6.13% The expected annual HPR is negative.

Now assume the investment has 25% probability of an 80% return, a 50% probability of a 15% return and a 25% probability of a -50% return.

(c) Calculate the expected one-period return of the investment and the standard deviation about that expectation.

E(r) = 0.25(0.80) + 0.50(0.15) + 0.25(-0.50) = 15%
σ = [0.25(0.80 – 0.15)2 + 0.50(0.15 – 0.15)2] + 0.25(-0.50 – 0.15)2]½ = [0.21125]½
σ = 45.96%

(d) Estimate the expected periodic holding period return for the investment over a large number of periods. The geometric mean return is the expected periodic holding period return. Geometric Mean ≈ Arithmetic Mean – σ2/2 = 0.10 – (0.4596)2/2 = 0.15 – 0.1056 = 4.44% The expected periodic holding period return is positive.

Note that the expected one period return did not change (15%) but the estimated holding period return increased from -6.13% to 4.44% just by lowering the deviation of returns (the risk) from 65% to 44.96%.

4. An investment has a 50% probability of a 40% return and a 50% probability of a -20% return.

(a) Calculate the expected one-period return of the investment and the standard deviation about that expectation.

E(r) = 0.50(0.40) + 0.50(-0.20) = 10%
σ = [0.50(0.40 – 0.10)2 + 0.50(-0.20 – 0.10)2]½ = [0.09]½ = 30.00%

(b) Estimate the expected periodic holding period return for the investment over a large number of periods. The geometric mean return is the expected periodic holding period return. Geometric Mean ≈ Arithmetic Mean – σ2/2 = 0.10 – (0.30)2/2 = 0.10 – 0.045 = 5.50%

Now assume the investment has 25% probability of returning 40%, a 50% probability of earning 10% and a 25% probability of returning -20%.

(c) Calculate the expected one-period return of the investment and the standard deviation about that expectation. E(r) = 0.25(0.40) + 0.50(0.10) + 0.25(-0.20) = 10%
σ = [0.25(0.40 – 0.10)2 + 0.50(0.10 – 0.10)2] + 0.25(-0.20 – 0.10)2]½ = [0.09]½ = 21.21%

(d) Estimate the expected periodic holding period return for the investment over a large number of periods. The geometric mean return is the expected periodic holding period return. Geometric Mean ≈ Arithmetic Mean – σ2/2 = 0.10 – (0.2121)2/2 = 0.10 – 0.0225 = 7.75%

Note that the expected one period return did not change (10%) but the estimated holding period return increased from 5.50% to 7.75% by lowering the deviation of returns (the risk).

5. You have calculated the following statistics for the historic returns for a portfolio of bonds and a portfolio of stocks.
Arithmetic Mean
Standard Deviation
Bonds
8.00%
10.67%
Stocks
15.00%
30.00%

a Calculate the probability of a realized loss next year for the bond portfolio if you assume the returns are normally distributed.

Calculate the P(r < 0%).
Calculate the number of Stdevs (10.67%) between the Mean (8%) and the target (0%): z =(x – Mean)/Stdev = [0 – 0.08)]/0.1067 = -0.75
P(r

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  • University/College: University of Chicago

  • Type of paper: Thesis/Dissertation Chapter

  • Date: 15 April 2016

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