Case study

A simulated share's daily and yearly returns

Model a share's returns with the normal: a day, a year by the square root of time, and the fat tails the model misses.

Draws on 7. The normal distribution, 10. Sampling distributions and the CLT

Step 1The question

An investor asks how risky a share is. The lab's data are 252 closing prices of a simulated share, one year of trading that includes a crash. The steps turn prices into returns, fit a normal model, put questions to it and then test it against the data.

Step 2From prices to log returns

A day's log return is \(r_t = \ln(P_t / P_{t-1})\), and log returns add up over days. The 251 daily log returns have a mean of 0.0012 and a standard deviation of 0.0279: on a typical day the price moves by a few per cent either way. The histogram is a tall, narrow bell, so the normal \(N(0.0012, 0.0279)\) is the first model to try.

251 values · mean 0.0012 · standard deviation 0.0279 · fitted N(μ = 0.0012, σ = 0.0279)

Open in the data-fit labDownload share-close.csv
The simulated share's 251 daily log returns with the fitted normal

Step 3A day: falls become log returns

The model describes log returns, so a question about a price change must be turned into one first. A fall of more than 5 % in a day leaves less than 0.95 of the price: a log return below \(\ln 0.95\), not below โˆ’0.05.

A 5 % fall in a day

With the daily log return modelled by the normal fitted to the 251 returns, \(N(0.00123, 0.0279)\), what is the probability that the price falls by more than 5 % in a day?

Your answer
A decimal or a fraction; within 0.0001 of the answer counts as right.

Step 4A year: the square root of time

Over 250 trading days the log returns add up. If the days are independent, the yearly log return has a mean of 250 times the daily one and a standard deviation of \(\sqrt{250}\) times the daily one (chapter 10): \(N(0.3072, 0.4412)\).

A fall of a quarter in a year

With the yearly log return modelled as \(N(0.3072, 0.4412)\), what is the probability that the price falls by more than 25 % over the year?

Your answer
A decimal or a fraction; within 0.0001 of the answer counts as right.

The median price after a year

Under the same yearly model, \(N(0.3072, 0.4412)\), the median yearly log return is 0.3072. By what factor does the median price after a year exceed today's?

Your answer
A decimal or a fraction; within 0.0001 of the answer counts as right.

Step 5The tails: where the model fails

Near the middle the model does well. It expects 7.5 of the 251 days to fall by more than 5 %, and 7 did. The tails are another matter: it expects 0.68 days beyond 3 standard deviations, and there were 4. The worst day, day 67, had a log return of โˆ’0.1595, 5.76 standard deviations below the mean: the model expects a day that bad once in about 236 million trading days.

Try this Open the returns in the data-fit lab: the Q-Q plot leaves the line at both ends, and the tail table counts the extreme days.

Step 6What the model can and cannot say

Read naively, the normal model makes large falls so rare that the share looks "safe to invest". The data say otherwise: the model understates exactly the extreme days that matter to an investor, and it was fitted to the very year that contained them. A year of data also says little about the average return: its standard error, 0.0018, is larger than the mean daily return itself (chapter 10). A careful report gives the model's figures together with the observed tails.

Remember: Turn every price change into a log return before using the model: a 5 % daily fall is \(\ln 0.95\) (0.0299), a 25 % yearly fall \(\ln 0.75\) (0.0888). Scale a day to a year with 250 for the mean and \(\sqrt{250}\) for the standard deviation. Then check the tails: this share's crash days are far beyond anything the normal allows.
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