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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:

Copyright © 2014 Keith Woodward