Finance work
1.
(a) The formula of discount factor is . r is the interest rate, and it is constant at 0.03. t is the period number. First, getting the discount rate of each year. Second,based on the discount rate, getting the discount factor for each year.
(b)(i) Getting the PV through the discount factor, I use . The present value of the receiving the rent for 88 years is $30.86.
(ii) Using formula which from the class, I get $30.86. The result is the same as (i).
(c) Formula can be used to get the present value of receiving the rent forever .
(d) In (b), I have found the PV of 88-year lease. Following (b), I get the PV of 22 and 10-year lease. Using the PV of lease sell to divided the full market value, then getting the percentages.
(e) An 88-year lease would sell for 97 per cent of its full market price. We have formula , then . Canceling both side, 0.. To get r, . r=0.04065.
2.
(a) The earning for each year is $100K. I consume 30% of my earning, then I can saving $70K ever year. I put my saving to saving account at beginning of a year with 0.04 interest rate. Through , I get column 4 & 5. When i need to withdraw $70000 at beginning of T, I need to get the capital through .
b. As there is 5% saving rate, all previous years’ saving will increase by 5%. For example, in year 0, the total saving will be $5000 as the income is assumed to be saved at the end of year 0, thus no compounding. In year 1, the $5000 earned in year 0 will earn an interest rate of 5%. Total saving in year 1 will thus become I created a column to find the compounded saving each year,
c. Let saving rate be and earning be E. At t=T, the capital KT would be
As and E = $10000 in this question,
d. NPV of perpetuity of future consumption at t=T is . By equating KT and . we get . By solving for T, T=13.96 years.
Part 2:
e. As illustrated in part c, the amount of capital at the end of year T-1 would be , which would be equal to .
f. . Since E cancels out on both sides and the expression is independent of E, the minimum capital we need at the beginning of period T is E>0.
g. By equating and , we get
When s = 0.5, T = 13.96, which is the same as part d.
3
A
1.Use function =norminv(probability, mean, standard_dev) to generate 400 random rate of return with mean 10.7% and variance 3.2%.
2.To get arithmetic rate of return, the variance over 400 years, and the geometric rate of return, I use excel function =average(), =var.s(), =geomean(). For the geometric rate of return, I add 1 to each random rate to let them be positive.
3. Yes. I get the half of the variance, then get the difference of the arithmetic rate & geometric rate of return. I resample several times, and the half of the variance is close to the difference; so, the statement holds.
B
1. Following A1, I get other two column random rate of return. Using to get the holding rate over two periods.
2.
I used “varS” function to calculate variances of one-period holding rate of return and two-periods holding rate of return. The variance of two-periods holding rate of return is approximately 2 times of the variance of one-period holding rate of return. This implies that the standard deviation of two-periods rate of return is approximately times of the standard deviation of one-period rate of return. Thus, the conclusion that ‘Risk grows approximately with the square root of time’ is verified.
4
1. 1-day T-statistics = 2/50, N-day T-statistics = 2. I get the formula , so N= days.
2.Using the function IF (RAND()<0.5,-1,1), -1 & 1 have the same probability to show in the column.
3. Using the function 2+IF(RAND()<0.5,-1,1)*50 to get the signal.
4.(a) Using question 3 as the basis point change every year. Using the function to get 5000-days’ realized compounded rate of return.
1. I used the function “=2+IF(RAND()<0.5,-1,1)*70” to find the basis point change in each year. I then generate rate of return in column C and F with 5000 entries of daily rate of return. (a) Then I plot a scatter plot for column C as the time series plot of the realized rate of return. (b) I generated column D of year T by multiplying the drift in year T-1 by 2bp drift. I then plot a scatter plot for column D as the time series plot of the compounded rate of return from 2bp drift alone.
2. By drawing many samples of 5000 observations, I would expect to see that there are more positive rate of returns in columns C and F. I repeated the analysis using 4bp drift and 200 bp noise. The difference between two signals are sensible. It is harder to detect a drift in the case where drift is 4 bp and noise is 200 bp as there is less positive numbers in column I as compared to column C and F.
5
I chose AMZN as my company to compute beta.
1) I computed beta for AMZN using 2-year daily return, 2-year monthly return, and 4-year daily return data. I computed beta using , where Rs is the rate of return for stock and Rm is the rate of return for markets. After getting beta, I shrink it to by using the formula to allow beta to converge to 1 in the long run.
2) The result is as following: , , . The beta for AMZN provided by Yahoo Finance is 1.63. It matters to choose daily data rather than monthly data because stock returns are very noisy. By having daily data rather than monthly data, we can get more observations over the same period and thus we can have better estimates. Also, the longer the period of study, the less reliable the estimate as over a long horizon, the unobserved underlying true betas themselves tend to move around. Thus, we need to shrink the computed beta even more (50%-60%) when the horizon is longer (>5 years).
3) There is difference between the result obtained by calculation and the result provided by Yahoo Finance. This is probably because the beta provided by Yahoo Finance is not shrunk and it is computed using three-year data. The beta I computed using 2-year daily data is less than 1.63, which implies that the stock return of AMZN fluctuates less with the market return than what is predicted Yahoo Finance.