Hot answers tagged

31

This one is quite easy: Think of a man walking his dog. He will go along and his dog will stroll along running back and forth. Man and dog are mathematically "cointegrated". As an investor you bet that the dog is coming back to his master or that the leash has only a certain length.


28

The standard story (also told by @vonjd) is of "The Drunk and Her Dog". This is based on "A Drunk and Her Dog: An Illustration of Cointegration and Error Correction" (1994). The story is itself based on the standard illustration for a random walk known as the "drunkard's walk". The Dickey-Fuller test is used to check for a unit root. It can be used as ...


14

Two time series $X_1$ and $X_2$ are cointegrated if a linear combination $aX_1+bX_2$ is stationary i.e. it has constant mean, standard deviation and autocorrelation function for some $a$ and $b$. In other words, the two series never stray very far from one another. Cointegration might provide a more robust measure of the linkage between two financial ...


8

The somewhat tongue-in-cheek blog post http://www.portfolioprobe.com/2010/10/18/american-tv-does-cointegration/ includes the example of two classes of shares on the same company. In this case you have two assets that are essentially the same but with a few details different. The buying and selling of these assets will make the prices fluctuate from each ...


8

You should look at Paul Willmott's Frequently Asked Questions In Quantitative Finance. He offers 12 (I think) ways of deriving BS and I think you'll find what you look for there. The cool thing is that you really have many different approaches; one is the classic PDE, one is done using change of measure, one is done using binary trees, and so on.... Really ...


3

I have asked myself the very same question when I first read the book. As far as I can tell, the "scalability" condition is only imposed for technical reasons. It simplifies the subsequent proof of the Fundemental Theorem of Asset Pricing in constrained markets. There are several papers that have shown that the theorem is valid for conic constraints. ...


2

Let $u_t$ be the random walk $$ u_t = u_{t-i} + \varepsilon_t $$ where $\mathrm{E}[\varepsilon_t]=0$ and $\mathrm{var}[\varepsilon_t]=\sigma^2$ , i.e. $\varepsilon_t$ is stationary. Now let $$X_t = \alpha u_t +\nu_t$$ and $$Y_t = \beta u_t + \eta_t$$ where $\nu_t$ and $\eta_t$ are stationary processes similar to $\varepsilon_t$ Then both $X_t$ and $Y_t$...


2

I think to understand the martingale/local martingale distinction, it helps to bring in a third class of processes, the uniformly integrable martingale. I would argue that the local martingale and the non-uniformly integrable (true) martingale are actually fairly similar. The key property that a uniformly integrable martingale has is the so-called closure ...


2

Evans and Schmitz (2015) might give an answer to your question if the Fama-French factors are indeed working or not. Value, size and momentum have a long history as stock price predictors, and similar indicators have been applied to stock indices in order to predict the performance of one national index against another. Published back tests of trading ...


1

It is just partial answer to your question. The Fama and French three factor model can be written as: $$R_{it}=\beta_{im}R_{Mt}+ \beta_{iSMB}SMB_t+\beta_{iHML}HML_t + e_{it}$$ In this model the market index is supposed to capture systematic risk originating from macroeconomic factors. Whereas, SMB and HML are firm specific variables and are chosen ...


1

Your intuition is not exactly right. To start with often the facts that small minus big or high minus low explain the cross-section of returns is called a puzzle. It is called a puzzle precisely because there is no unifying explanation for them. It is fairly agreed among academics that the Size effect is most likely a January effect, or probably it even ...


1

You are asking an interesting question. Firstly, a Submartingale has increasing or equal expectation (not decreasing). Secondly, the process $dX_t=X_tdW_t$ is a true martingale (not strictly local), since its solution (by Ito): $$X_t=X_0e^{W_t-\frac{t}{2}}$$ has $E(X_t)=X_0$ constant expectation ($e^{-\frac{t}{2}}E(e^{W_t})=1, W_t\sim N(0,t)$). The ...


1

Here is an empirical strategy to test for cointegration. FIRST, check whether both $X_t$ and $Y_t$ contain an unit root. If they are both stationary then model $Y_t$ or $X_t$ in levels (and nothing is wrong). If one of the two is $I(1)$ (non-stationary for one level), then take differences to ensure stationarity. If they are both non-stationary, and hence ...



Only top voted, non community-wiki answers of a minimum length are eligible