# Tag Info

6

Basically, prices usually have a unit root, while returns can be assumed to be stationary. This is also called order of integration, a unit root means integrated of order 1, I(1), while stationary is order 0, I(0). Time series that are stationary have a lot of convenient properties for analysis. When a time series is non-stationary, then that means the ...

5

The correct answer has some intuition though it doesn't generalize to continuous time very easily: Think about the paper below like this: $Var(X+Y) = Var(X) + Var(Y) + 2Cov(X,Y)$ The generalization is slightly hard because the dynamics of $\mu$ and $\sigma^2$ could be dependent for arbitrary returns. You can use a GMM estimator to derive the asymptotic ...

4

Perhaps overly simplistic and repeating the pt above, but when doing statistics, ideally we want to compare like with like. Returns can be comparable with each other. Prices on the other hand always depend on the previous price.

3

Just a bit of illustration added to @John's answer. Look at log prices $\log(P_t)$, assume that you know $P_0$ then $$\log(P_t) = \log(P_0) + r_1 + \cdots r_t$$ where $r_i = \log(P_i)-\log(P_{i-1})$ are the log returns. By modelling the log-returns (which as already said take values on the whole real line which is a nice property for modelling) we model ...

2

Actually, neither of your two results are quite correct. As explained in the Details for the Return.calculate function, most of the PerformanceAnalytics functions use discrete returns, not log returns. To get the correct results, you will have to convert your data from log returns to simple returns. Compare the charts from the following: ...

2

The answer is that it depends. In addition to the Lo paper above, there are a number of excellent references that go into depth about annualizing or time scaling non-i.i.d. returns, one of which is Roger Kauffman, "Long-Term Risk Management", 2005 which can be found at http://www.rogerkaufmann.ch/all-Budapest.pdf. There are some well known cases where the ...

2

If you wanted to see the following (price $S_t$, log return $r$, simple return $R$) then $$r = \log(S_{t+1}) - \log(S_t) = \log(S_{t+1}/S_{t}),$$ and $$R = S_{t+1}/S_{t}-1,$$ thus $$R = \exp(r)-1$$ and $$r = \log(1+R).$$ Was this the question?

1

The high serial correlation you are getting in the first case is a spurious correlation. The correct way to do it is with returns. The price series has a unit root. You need to take diff(log(prices))) in order to have a stationary time series, on which you can then estimate autocorrelations, auto regressive coefficients, etc. properly. This was shown by ...

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