16

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 ...


8

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.


8

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 ...


7

I think most people agree that aggregate (index) stock returns have negative skewness. However, this does not appear to be the case for individual stock returns. These two papers find that average skewness of individual stock returns is positive though time-varying: Paper 1 R. Albuquerque, Skewness in Stock Returns: Reconciling the Evidence on Firm Versus ...


6

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 ...


6

The Black-Scholes-Merton (1973) model implies that the prices of the underlying asset at maturity $S_T$ are log-normally distributed $$ln(S_T)\sim N\big[ln(S_0)+(\mu-\frac{\sigma^2}{2})T,\;\sigma^2T\big]$$ so that the logarithmic returns to maturity $ln(\frac{S_T}{S_0})$ are normally distributed $$ln(\frac{S_T}{S_0})\sim N\big[(\mu-\frac{\sigma^2}{2})T, \;\...


6

Let $R$ denote the arithmetic return and $r$ the log returns. $$R=\frac{V_f-V_i}{V_i} \textrm { and } r=\ln\left(\frac{V_f}{V_i}\right)$$ Arithmetic and log returns are connected as: $$R=\frac{V_f-V_i}{V_i} =\frac{V_f}{V_i}-1$$ Hence, $R+1=\frac{V_f}{V_i}$. Taking log on both sides. $$\ln\left(\frac{V_f}{V_i}\right)=\ln(R+1) \textrm{ and } r=\ln(R+1)$$


5

An example of non-overlapping one month returns: the return in January, the return in February, the return in March, etc. An example of overlapping 30 day returns: the return from January 1 to January 30, the return from January 2 to January 31, the return from January 3 to February 1, the return from January 4 to February 2, and so on. There are far fewer ...


5

In Python, simple geometric returns: import numpy as np import pandas as pd sp500 = pd.io.data.DataReader('^GSPC', 'yahoo')['Close'] simple_ret = sp500.pct_change() (1+simple_ret).cumprod()[-1] -1 0.74751768460019963 Log-returns: log_ret = np.log(1+simple_ret) np.exp(log_ret.cumsum()[-1]) -1 0.74751768460020074 In ...


5

Transmuting one to the other is pretty straightforward without the underlying sequence of prices. To go from log to simple: $R = exp(r) - 1$ To go from simple to log: $r = log(R+1)$


5

Stock prices cannot be negative which means that they are not normally distributed due to the fact they cannot be negative as result of this stock prices behave similarly to exponential functions. To transform this exponential values back to a normally distributed variable, you need to take the natural logarithm, and therefore can take a lognormal value and ...


5

In reality, neither are stock prices log-normally distributed nor are returns normally distributed. More sophisticated models drop this assumption. For instance, returns are more peaked and have fatter tails than a normal distribution would suggest. In simple models, such as the Black and Scholes (1973) model, it is however assumed that the stock price ...


5

You're right but a GBM doesn't assume that percentage returns are normally distributed. It's about log-returns. If the log-return $r_t=\ln\left(\frac{S_{t+dt}}{S_t}\right)$ is normally distributed (GBM assumption), then $r_t$ can indeed be any arbitrarily large (positive or negative) number with positive probability. This also implies that stock prices are ...


4

Large? ? The relationship between normal and log returns is $$(normal return) = exp(log return)-1$$ Therefore log-returns can be from $-\infty$ to $+\infty$ while normal ones can only be between $-1$ and $+\infty$.


4

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?


4

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 ...


4

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: charts....


4

The above relation really only approximately. If you consider arithmetic retunrs then it is exact. For the approximation you just need to look at the Taylor series of the exponential: $$ e^x = 1 + x + \text{ terms of higher order}. $$ These terms of higher order ($x^2$ and $x^3$) become small if $x$ is much smaller than one - which holds true for returns. ...


4

There is no possibility to convert any two of your mentioned variables into the remaining one. For the compound and arithmetic return you can derive an inequality, but that's the best you can do. The definitions for your statements are: $$r_{\mathrm{compound}}= \prod_{t=0}^{n}{\left( 1+r_t \right)}$$ $$r_{\mathrm{arithmetic}}=\frac{1}{n} \sum_{t=0}^n{r_t}$...


4

In the colloquial sense of the word "justified," it is not justified. I will describe why it is justified mathematically and under what circumstances and in what case it is not justified. Let me begin with the simplest of equations $$\tilde{w}=R\bar{w}+\epsilon,\epsilon\sim\mathcal{N}(0,\sigma^2).$$ Let us assume that this equation is an element ...


4

OK, this need have nothing to do with any single sample of data. It's an inherent difference between the behaviour of linear vs logarithmic numbers. Which is what make up your respective simple and log returns, and associated averages. Imagine I offered you a bet in which you put a pound on the table, I put two down, we flipped a fair coin, and winner ...


4

You can simply start with the definition of gross returns \begin{align*} R_{t+1}&=\frac{D_{t+1}+P_{t+1}}{P_t} \\ &=\frac{1+P_{t+1}/D_{t+1}}{P_t/D_t}\frac{D_{t+1}}{D_t}, \end{align*} where the first fraction contains now your price dividend ratio. Going to log-returns, \begin{align*} r_{t+1} &= \ln\left(1+\frac{P_{t+1}}{D_{t+1}}\right) - \ln\left(\...


3

The geometric mean of quantities $\{a_1, \dots, a_n\}$ is $$ \bar{a}_g = \left( \prod_{i=1}^n a_i \right)^{1/n} $$ Taking the logarithm of both sides gives $$ \log \bar{a}_g = \frac{1}{n} \sum_{i=1}^n \log a_i $$ so the log of the geometric mean is equal to the arithmetic mean of the logs. In your case, the relevant quantities $a_i$ are the growth rates over ...


3

To summarise what "John" just explained above: Say that you have stock portfolio for several years: $t_0, t_1, ..., t_m$. Say that you have $n$ stocks, so that stock $i$ has a vector of prices $X_i$. The length of each price vector is $m$ because there are $m$ years. Then, for the first year $t_1$: Calculate the $n$ different arithmetic returns for ...


3

Well, it wasn't easy because you didn't mentioned how your data is formatted. I create my own data.frame() basing on data you provided. You can skip this part if your data.frame is ready. Here's code I used to create a dataframe: > #given dates > dates=c("2000-1-3","2000-1-4","2000-1-5","2000-1-6","2000-1-7","2000-1-10","2000-1-11") > #formating ...


3

Create a new price series that has a value for every minute, e.g. by carrying the last observation forward. Then compute returns from this new price series. (There are simpler approaches for this particular case, but I'd prefer the one outlined above as it is conceptually clear.) A sketch in R. (Disclosure: I am the maintainer of packages PMwR, from which ...


3

As you pointed out, not necessarily: I know that Fama-French have developed their 3 factor model using only simple returns That's because of a very common misconception: and, after all, the dependent variable in a regression needs to be iid. In fact, the dependent variable does not have to be normally distributed or i.i.d. for that matter. That ...


3

BS assumes prices NOT returns are log-normally distributed. Why making that assumption? 1.log-normal is not perfect but OK to fit potential prices distribution. 2.The nature of log-normal distribution will force the left tail to be above zero. 3. There are definitely distributions work better than log-normal in terms of fitting stock price data, but that ...


3

Because $\mathbb{E}\left(e^{\sigma W_t}\right) = e^{\frac{1}{2}\sigma^2T} > 1$, you need that correction to ensure that your asset grows on average at rate $\mu$ (or $r$ in the risk-neutral measure). This is pretty well explained in the chapter on BS model from Hull’s book Options, futures and other derivatives!


3

The return $R_i$ as expressed in $$R_{i+1,i}=\frac{S_{i+1}-S_i}{S_i}=\mu \Delta t + \sigma \Delta W(t_{i+1},t_i)$$ is not possible. To see this, let's get the returns over two small time steps of $\Delta t$ each. Then $$R_{i+2,i+1}=\frac{S_{i+2}-S_{i+1}}{S_{i+1}}= \mu \Delta t + \sigma \Delta W(t_{i+2},t_{i+1})$$ but $$R_{i+2,i}=\frac{S_{i+2}-S_{i}}{S_{i}}= ...


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