Fight Finance

Question 718  arithmetic and geometric averages

The symbol ##\text{GDR}_{0\rightarrow 1}## represents a stock's gross discrete return per annum over the first year. ##\text{GDR}_{0\rightarrow 1} = P_1/P_0##. The subscript indicates the time period that the return is mentioned over. So for example, ##\text{AAGDR}_{1 \rightarrow 3}## is the arithmetic average GDR measured over the two year period from years 1 to 3, but it is expressed as a per annum rate.

Which of the below statements about the arithmetic and geometric average GDR is NOT correct?

(a) Arithmetic average gross discrete return is ##\text{AAGDR}_{0\rightarrow T} = \dfrac{\text{GDR}_{0 \rightarrow 1}+\text{GDR}_{1\rightarrow 2}+...+\text{GDR}_{T-1 \rightarrow T}}{T}##

(b) Geometric average gross discrete return is ##\text{GAGDR}_{0\rightarrow T} = \dfrac{\text{GDR}_{0 \rightarrow 1} . \text{GDR}_{1\rightarrow 2} ... \text{GDR}_{T-1 \rightarrow T}}{T}##

(c) The geometric average gross discrete return can be quickly found using the first ##(P_0)## and last ##(P_T)## prices. ##\text{GAGDR}_{0\rightarrow T} = \left( \dfrac{P_T}{P_0} \right)^{1/T}##

(d) The arithmetic average is always bigger than or equal to the geometric average, ##\text{AAGDR} \ge \text{GAGDR}##.

(e) The arithmetic and geometric averages of returns will be equal if the variance of the stock's returns is zero.

Question 719  mean and median returns, return distribution, arithmetic and geometric averages, continuously compounding rate

A stock has an arithmetic average continuously compounded return (AALGDR) of 10% pa, a standard deviation of continuously compounded returns (SDLGDR) of 80% pa and current stock price of $1. Assume that stock prices are log-normally distributed. The graph below summarises this information and provides some helpful formulas.

In one year, what do you expect the median and mean prices to be? The answer options are given in the same order.

(a) P1median = $1.105171,   P1mean = $1.521962

(b) P1median = $1.1,             P1mean = $1.42

(c) P1median = $1.521962,   P1mean = $1.105171

(d) P1median = $0.802519,   P1mean = $1.105171

(e) P1median = $1.42,           P1mean = $1.1

Question 720  mean and median returns, return distribution, arithmetic and geometric averages, continuously compounding rate

A stock has an arithmetic average continuously compounded return (AALGDR) of 10% pa, a standard deviation of continuously compounded returns (SDLGDR) of 80% pa and current stock price of $1. Assume that stock prices are log-normally distributed.

In 5 years, what do you expect the median and mean prices to be? The answer options are given in the same order.

(a) $0.332871, $1.648721

(b) $8.16617, $1.648721

(c) $1.61051, $1.648721

(d) $1.61051, $5.773534

(e) $1.648721, $8.16617

Question 721  mean and median returns, return distribution, arithmetic and geometric averages, continuously compounding rate

Fred owns some Commonwealth Bank (CBA) shares. He has calculated CBA’s monthly returns for each month in the past 20 years using this formula:

###r_\text{t monthly}=\ln⁡ \left( \dfrac{P_t}{P_{t-1}} \right)###

He then took the arithmetic average and found it to be 1% per month using this formula:

###\bar{r}_\text{monthly}= \dfrac{ \displaystyle\sum\limits_{t=1}^T{\left( r_\text{t monthly} \right)} }{T} =0.01=1\% \text{ per month}###

He also found the standard deviation of these monthly returns which was 5% per month:

###\sigma_\text{monthly} = \dfrac{ \displaystyle\sum\limits_{t=1}^T{\left( \left( r_\text{t monthly} - \bar{r}_\text{monthly} \right)^2 \right)} }{T} =0.05=5\%\text{ per month}###

Which of the below statements about Fred’s CBA shares is NOT correct? Assume that the past historical average return is the true population average of future expected returns.

(a) The returns ##r_\text{t monthly}## are continuously compounded monthly returns and are likely to be normally distributed, so the mean, median and mode of these continuously compounded returns are all equal to 1% per month.

(b) The annualised average continuously compounded return is ##\bar{r}_\text{cc annual}=12×0.01=0.1=12\%\text{ pa}##. The annualised standard deviation is ##\sigma_\text{annual} = \sqrt{12}×0.05=0.173205081=17.32\%\text{ pa}##.

(c) Over the next 10 years the mean, median and mode continuously compounded 10 year returns will all equal ##0.01 \times 12 \times 10 = 120\%##.

(d) Over the next 10 years the expected mean gross discrete 10 year return is ##e^{(0.01 - 0.05^2/2)×12×10} = 2.857651118## and the median gross discrete 10 year return is ##e^{(0.01 + 0.05^2/2)×12×10}=3.857425531##.

(e) If the current price of the CBA shares is $75, then in 10 years they’re expected to have a mean value of ##75×e^{(0.01 + 0.05^2/2)×12×10} = 289.3069148## and a median value of ##75×e^{0.01×12×10}=249.0087692##.

Question 722  mean and median returns, return distribution, arithmetic and geometric averages, continuously compounding rate

Here is a table of stock prices and returns. Which of the statements below the table is NOT correct?

Price and Return Population Statistics
Time	Prices	LGDR	GDR	NDR
0	100
1	50	-0.6931	0.5	-0.5
2	100	0.6931	2	1
Arithmetic average		0	1.25	0.25
Arithmetic standard deviation		0.9802	1.0607	1.0607

 

(a) The geometric average of the gross discrete returns (GAGDR) equals 1 which is 100%.

(b) ##\text{GAGDR} = \exp \left( \text{AALGDR} \right)##. The GAGDR is equal to the natural exponent of the arithmetic average of the logarithms of the gross discrete returns (AALGDR).

(c) ##\text{GAGDR} = \left( P_T/P_0 \right)^{1/T}##. The GAGDR equals the ratio of the last and first prices raised to the power of the inverse of the number of time periods between them.

(d) ##\text{LGAGDR} = \text{AALGDR}##. This is always true, regardless of the distribution of the prices or returns and the number of return observations. The logarithm of the geometric average of the gross discrete returns (LGAGDR) is always equal to the arithmetic average of the logarithms of the gross discrete returns (AALGDR).

(e) ##\text{LAAGDR} = \text{AALGDR} + \text{SDLGDR}^2/2##. This is always true, regardless of the distribution of the prices or returns and the number of return observations. The logarithm of the arithmetic average of the gross discrete returns (LAAGDR) equals the arithmetic average of the logarithms of the gross discrete returns (AALGDR) plus half the variance of the LGDR's.

Question 723  mean and median returns, return distribution, arithmetic and geometric averages, continuously compounding rate

Here is a table of stock prices and returns. Which of the statements below the table is NOT correct?

Price and Return Population Statistics
Time	Prices	LGDR	GDR	NDR
0	100
1	99	-0.010050	0.990000	-0.010000
2	180.40	0.600057	1.822222	0.822222
3	112.73	0.470181	0.624889	0.375111
Arithmetic average		0.0399	1.1457	0.1457
Arithmetic standard deviation		0.4384	0.5011	0.5011

 

(a) The geometric average of the gross discrete returns (GAGDR) equals 1.04075 which is 104.075%.

(b) ##\exp \left( \text{GAGDR} \right) = \text{AALGDR}##. The natural exponent of the GAGDR equals the arithmetic average of the logarithms of the gross discrete returns (AALGDR).

(c) ##\text{GAGDR}^T = \left( P_T/P_0 \right)##. The GAGDR raised to the power of the number of time period that it's measured over equals the ratio of the first and last prices.

(d) ##\text{GAGDR} = \exp \left( \text{AALGDR} \right)##. This is always true, regardless of the distribution of the prices or returns. The logarithm of the geometric average of the gross discrete returns (LGAGDR) is always equal to the arithmetic average of the logarithm of the gross discrete returns (AALGDR).

(e) ##\text{AAGDR} \approx \exp \left( \text{AALGDR} + \text{SDLGDR}^2/2 \right)##. This is only approximately true and depends on the distribution of the prices and returns. It's asymptotically true as the time period that the returns are measured over reaches infinity.

Question 779  mean and median returns, return distribution, arithmetic and geometric averages, continuously compounding rate

Fred owns some BHP shares. He has calculated BHP’s monthly returns for each month in the past 30 years using this formula:

###r_\text{t monthly}=\ln⁡ \left( \dfrac{P_t}{P_{t-1}} \right)###

He then took the arithmetic average and found it to be 0.8% per month using this formula:

###\bar{r}_\text{monthly}= \dfrac{ \displaystyle\sum\limits_{t=1}^T{\left( r_\text{t monthly} \right)} }{T} =0.008=0.8\% \text{ per month}###

He also found the standard deviation of these monthly returns which was 15% per month:

###\sigma_\text{monthly} = \dfrac{ \displaystyle\sum\limits_{t=1}^T{\left( \left( r_\text{t monthly} - \bar{r}_\text{monthly} \right)^2 \right)} }{T} =0.15=15\%\text{ per month}###

Assume that the past historical average return is the true population average of future expected returns and the stock's returns calculated above ##(r_\text{t monthly})## are normally distributed. Which of the below statements about Fred’s BHP shares is NOT correct?

(a) The returns ##r_\text{t monthly}## are continuously compounded monthly returns.

(b) The mean, median and mode of the continuously compounded annual return is expected to equal 0.8% per month.

(c) The annualised average continuously compounded return is ##\bar{r}_\text{cc annual}=12×0.008=0.096=9.6\%\text{ pa}##. The annualised standard deviation is ##\sigma_\text{annual} = \sqrt{12}×0.15=0.519615242 = 52\%\text{ pa}##.

(d) If the current price of the BHP shares is $20, they're expected to have a median value of $52.2339 ##(= 20×e^{0.008×12×10})## in 10 years.

(e) If the current price of the BHP shares is $20, they're expected to have a mean value of $13.5411 ##(= 20×e^{(0.008 - 0.15^2/2)×12×10})## in 10 years.

Question 790  mean and median returns, return distribution, arithmetic and geometric averages, continuously compounding rate, log-normal distribution, VaR, confidence interval

A risk manager has identified that their hedge fund’s continuously compounded portfolio returns are normally distributed with a mean of 10% pa and a standard deviation of 30% pa. The hedge fund’s portfolio is currently valued at $100 million. Assume that there is no estimation error in these figures and that the normal cumulative density function at 1.644853627 is 95%.

Which of the following statements is NOT correct? All answers are rounded to the nearest dollar.

(a) Future portfolio values have a log-normal distribution. The mean (expected or arithmetic average) portfolio value in one year is $115,603,957.

(b) The median (50th percentile) portfolio value in one year is $115,603,957.

(c) The 90% confidence interval of continuously compounded portfolio returns over the next year ##(r_{0 \rightarrow 1})## is ##-39.3456088\% < r_{0 \rightarrow 1} < 59.3456088\%##.

(d) The 90% confidence interval of portfolio values in one year ##(V_1)## is ##\$67,472,095 < V_1 < \$181,023,395##.

(e) The annual 95% relative VaR is $48,131,863 and the absolute VaR is $32,527,905.

Question 791  mean and median returns, return distribution, arithmetic and geometric averages, continuously compounding rate, log-normal distribution, VaR, confidence interval

A risk manager has identified that their pension fund’s continuously compounded portfolio returns are normally distributed with a mean of 5% pa and a standard deviation of 20% pa. The fund’s portfolio is currently valued at $1 million. Assume that there is no estimation error in the above figures. To simplify your calculations, all answers below use 2.33 as an approximation for the normal inverse cumulative density function at 99%. All answers are rounded to the nearest dollar. Which of the following statements is NOT correct?

(a) The mean (expected or arithmetic average) portfolio value in one year is $1,072,508. The median (50th percentile) portfolio value in one year is $1,051,271.

(b) The annual 99% relative VaR is $391,591.

(c) The annual 99% absolute VaR is $340,320.

(d) The 98% confidence interval of portfolio values in one year ##(V_1)## is ##\$659,680 < V_1 < \$1,675,312,977##.

(e) The 98% confidence interval of continuously compounded portfolio returns over the next year ##(r_{0 \rightarrow 1})## is ##-41.6\% < r_{0 \rightarrow 1} < 51.6\%##.

Question 792  mean and median returns, return distribution, arithmetic and geometric averages, continuously compounding rate, log-normal distribution, confidence interval

A risk manager has identified that their investment fund’s continuously compounded portfolio returns are normally distributed with a mean of 10% pa and a standard deviation of 40% pa. The fund’s portfolio is currently valued at $1 million. Assume that there is no estimation error in the above figures. To simplify your calculations, all answers below use 2.33 as an approximation for the normal inverse cumulative density function at 99%. All answers are rounded to the nearest dollar. Assume one month is 1/12 of a year. Which of the following statements is NOT correct?

(a) The median portfolio value in one year is $1,105,171. In one month it's $1,008,368.

(b) The mean portfolio value in one year is $1,197,217. In one month it's $1,015,113.

(c) The worst one percentile of portfolio values in one year is $435,178.

(d) The worst one percentile of portfolio values in one month is $933,016.

(e) The 98% confidence interval of portfolio values in two years is between $326,918 and $4,563,305.

Question 811  log-normal distribution, mean and median returns, return distribution, arithmetic and geometric averages

Which of the following statements about probability distributions is NOT correct?

(a) A normally distributed variable’s mean, median and mode are all expected to be equal.

(b) A log-normally distributed variable’s mean, median and mode are all expected to be different. Mean > median > mode.

(c) Future stock prices ##(P_t )## are often assumed to be log-normally distributed.

(d) Stocks’ annual net discrete returns ##(P_t/P_0-1)## must be normally distributed when stock prices ##(P_t)## are log-normally distributed.

(e) The total area under a probability density function (pdf) is always one.

Question 874  utility, return distribution, log-normal distribution, arithmetic and geometric averages

Who was the first theorist to endorse the maximisiation of the geometric average gross discrete return for investors (not gamblers) since it gave a "...portfolio that has a greater probability of being as valuable or more valuable than any other significantly different portfolio at the end of n years, n being large"?

(a) Daniel Bernoulli.

(b) John Larry Kelly Jr.

(c) Henry Allen Latane.

(d) Ole Peters.

(e) Paul Anthony Samuelson.

Question 877  arithmetic and geometric averages, utility, utility function

Gross discrete returns in different states of the world are presented in the table below. A gross discrete return is defined as ##P_1/P_0##, where ##P_0## is the price now and ##P_1## is the expected price in the future. An investor can purchase only a single asset, A, B, C or D. Assume that a portfolio of assets is not possible.

Gross Discrete Returns
In Different States of the World
Investment	World states (probability)
asset	Good (50%)	Bad (50%)
A	2	0.5
B	1.1	0.9
C	1.1	0.95
D	1.01	1.01

 

Which of the following statements about the different assets is NOT correct? Asset:

(a) A's Arithmetic Average Gross Discrete Return (AAGDR) is the highest.

(b) B's Geometric Average Gross Discrete Return (GAGDR) is the lowest.

(c) C's GAGDR is higher than its AAGDR.

(d) D's AAGDR and GAGDR are equal since it's volatility is zero.

(e) C's expected utility of wealth would be the highest for an investor with a log utility function.

Question 921  utility, return distribution, log-normal distribution, arithmetic and geometric averages, no explanation

Who was the first theorist to propose the idea of ‘expected utility’?

(a) Daniel Bernoulli.

(b) John Larry Kelly Jr.

(c) Henry Allen Latane.

(d) Ole Peters.

(e) Paul Anthony Samuelson.

Question 925  mean and median returns, return distribution, arithmetic and geometric averages, continuously compounding rate, no explanation

The arithmetic average and standard deviation of returns on the ASX200 accumulation index over the 24 years from 31 Dec 1992 to 31 Dec 2016 were calculated as follows:

###\bar{r}_\text{yearly} = \dfrac{ \displaystyle\sum\limits_{t=1992}^{24}{\left( \ln⁡ \left( \dfrac{P_{t+1}}{P_t} \right) \right)} }{T} = \text{AALGDR} =0.0949=9.49\% \text{ pa}###

###\sigma_\text{yearly} = \dfrac{ \displaystyle\sum\limits_{t=1992}^{24}{\left( \left( \ln⁡ \left( \dfrac{P_{t+1}}{P_t} \right) - \bar{r}_\text{yearly} \right)^2 \right)} }{T} = \text{SDLGDR} = 0.1692=16.92\text{ pp pa}###

Assume that the log gross discrete returns are normally distributed and that the above estimates are true population statistics, not sample statistics, so there is no standard error in the sample mean or standard deviation estimates. Also assume that the standardised normal Z-statistic corresponding to a one-tail probability of 2.5% is exactly -1.96.

Which of the following statements is NOT correct? If you invested $1m today in the ASX200, then over the next 4 years:

(a) Continuously compounded mean and median returns are both 37.96% over the 4 year period.

(b) 95% confidence interval of the continuously compounded returns is between -28.3664% and 104.2864% over the 4 year period.

(c) Mean dollar value is $1.547835m while its median dollar value is $1.4617m in 4 years.

(d) 95% confidence interval of the dollar value is between $0.75302m and $2.17038m in 4 years.

(e) 95% confidence interval of the effective annual return is between -24.698% and 183.7332% over the 4 year period.

Question 926  mean and median returns, return distribution, arithmetic and geometric averages, continuously compounding rate

The arithmetic average continuously compounded or log gross discrete return (AALGDR) on the ASX200 accumulation index over the 24 years from 31 Dec 1992 to 31 Dec 2016 is 9.49% pa.

The arithmetic standard deviation (SDLGDR) is 16.92 percentage points pa.

Assume that the log gross discrete returns are normally distributed and that the above estimates are true population statistics, not sample statistics, so there is no standard error in the sample mean or standard deviation estimates. Also assume that the standardised normal Z-statistic corresponding to a one-tail probability of 2.5% is exactly -1.96.

If you had a $1 million fund that replicated the ASX200 accumulation index, in how many years would the median dollar value of your fund first be expected to lie outside the 95% confidence interval forecast?

(a) 23.17 years.

(b) 41.31 years.

(c) 134.19 years.

(d) 536.75 years.

(e) Never.

Question 927  mean and median returns, mode return, return distribution, arithmetic and geometric averages, continuously compounding rate

The arithmetic average continuously compounded or log gross discrete return (AALGDR) on the ASX200 accumulation index over the 24 years from 31 Dec 1992 to 31 Dec 2016 is 9.49% pa.

The arithmetic standard deviation (SDLGDR) is 16.92 percentage points pa.

Assume that the log gross discrete returns are normally distributed and that the above estimates are true population statistics, not sample statistics, so there is no standard error in the sample mean or standard deviation estimates. Also assume that the standardised normal Z-statistic corresponding to a one-tail probability of 2.5% is exactly -1.96.

If you had a $1 million fund that replicated the ASX200 accumulation index, in how many years would the mean dollar value of your fund first be expected to lie outside the 95% confidence interval forecast?

(a) 23.17 years.

(b) 41.31 years.

(c) 134.19 years.

(d) 536.75 years.

(e) Never.

Question 928  mean and median returns, mode return, return distribution, arithmetic and geometric averages, continuously compounding rate, no explanation

The arithmetic average continuously compounded or log gross discrete return (AALGDR) on the ASX200 accumulation index over the 24 years from 31 Dec 1992 to 31 Dec 2016 is 9.49% pa.

The arithmetic standard deviation (SDLGDR) is 16.92 percentage points pa.

Assume that the log gross discrete returns are normally distributed and that the above estimates are true population statistics, not sample statistics, so there is no standard error in the sample mean or standard deviation estimates. Also assume that the standardised normal Z-statistic corresponding to a one-tail probability of 2.5% is exactly -1.96.

If you had a $1 million fund that replicated the ASX200 accumulation index, in how many years would the mode dollar value of your fund first be expected to lie outside the 95% confidence interval forecast?

Note that the mode of a log-normally distributed future price is: ##P_{T \text{ mode}} = P_0.e^{(\text{AALGDR} - \text{SDLGDR}^2 ).T} ##

(a) 23.17 years.

(b) 41.31 years.

(c) 134.19 years.

(d) 536.75 years.

(e) Never.

Question 929  standard error, mean and median returns, mode return, return distribution, arithmetic and geometric averages, continuously compounding rate

The arithmetic average continuously compounded or log gross discrete return (AALGDR) on the ASX200 accumulation index over the 24 years from 31 Dec 1992 to 31 Dec 2016 is 9.49% pa.

The arithmetic standard deviation (SDLGDR) is 16.92 percentage points pa.

Assume that the data are sample statistics, not population statistics. Assume that the log gross discrete returns are normally distributed.

What is the standard error of your estimate of the sample ASX200 accumulation index arithmetic average log gross discrete return (AALGDR) over the 24 years from 1992 to 2016?

(a) 43.9125 percentage points per annum.

(b) 3.4538 percentage points per annum.

(c) 1.9371 percentage points per annum.

(d) 0.705 percentage points per annum.

(e) 0.3954 percentage points per annum.

Question 1085  inflation, consumer price index, price gains and returns over time, arithmetic and geometric averages

Calculate the annual inflation rate between 30th September 1948 when the consumer price index (CPI) was 3.7 and 31st December 2023 when the CPI was 136.1, as published by the ABS. Note that you should find the annual CPI inflation rate between these dates.
The annual inflation rate over these 75.25 (=2024 - 1948.75) years was:

(a) 36.783784% pa

(b) 6.541001% pa

(c) 4.907386% pa

(d) 2.539572% pa

(e) 1.782742% pa