Why Regression Residuals May Not Be Reliable Pair Trading Signals
Summary
The document considers using a multiple linear regression of a stock’s price on market, commodity, competitor, and currency variables to estimate a model-implied price. The proposed trade goes long when the observed price is below that estimate, with a separate short leg as part of a broader pairs strategy. The response challenges the idea that a fitted price gap is automatically a signal: regression is designed to explain observed prices, and its residual captures what the fitted model did not explain.
The answer cautions that residuals centered near zero under standard regression conditions do not by themselves imply predictable returns, so trading their direction needs separate evidence. It also warns that correlated predictors can make coefficient estimates and significance difficult to interpret. These are reasons to test the signal carefully, not a complete evaluation of the strategy; the discussion does not specify an out-of-sample design, stationarity checks, or a full pairs-trading model.
Key ideas
- A gap between observed and regression-fitted price is a model residual, not automatically a forecast of future returns.
- A residual trading rule requires evidence that the gap predicts price movement beyond the fitting sample.
- Correlated explanatory variables can make regression coefficient estimates and significance harder to interpret.
- Residuals near zero on average do not establish that residual-based trades are profitable or random in every setting.
- The discussion does not provide a tested strategy or a complete pairs-trading framework.
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Full text
# Multiple (linear) regression
# Multiple (linear) regression
I am looking for some inputs on a pair trading strategy that I am trying to improve with some semi-fundamental input.
The basic idea is to use multiple linear regression to estimate the price of a stock ($Y$) based on fundamentals:
$$ Y(t) = \beta_0 + \beta_1*x_1(t) + \beta_2*x_2(t) + \cdots $$
Just to give you an idea of a case:
- $Y$ = Stock price, e.g. Starbucks
- $x_1$ = Market, e.g. S&P500
- $x_2$ = Coffe price
- $x_3$ = Competitor something..
- $x_4$ = Forex something..
My idea would be to only trade in the direction of the residual at the last date of the regression, meaning that if starbucks is at 55 and the regression gives a "fundamental" price at 60 I would only go long. And, since this is a part of a larger pair trading strategy I would short something else and include all the commonly used pair trading parameters.
After some extensive googling I have not been able to find any similar approaches. Anybody who has seen anything similar somewhere? Does it make any sense? Or is this just way off?
## Answer by SRKX (score 3)
https://quant.stackexchange.com/a/7233
I personally would not do that!
Your regression model has been fitted to approximate $Y(t)$ (the reality) as much as possible.
If I understand you well, you say:
- at the previous period $Y(t)=55$ (Starbucks traded at 55 USD)
- the last period's estimate from the regression is $\hat{Y}(t)=60$
- Since $\hat{Y}(t)-Y(t) > 0$, you want to invest.
This does not make sense to me because you are trusting more the model than the actual process $Y$ and the model has been optimized (fitted) on $Y$.
When you computed the regression parameters, you found the best linear relationship between $Y$ and your different dependent variables $X$. Yet, your model does not perfectly fit the data (it has residuals: 60-55=5). So in a sense, the model hasn't been able to completely understand Starbuck's price process. And yet, you are willing to trade in the direction of the residual, which is the error of the regression.
Besides, beware of the dependent variables $X$ you use. Multiple linear regression requires them to be uncorrelated. This would generate problem if you look at the statistical significance of the estimated parameters $\beta_1, ... \beta_k$
Finally, under the assumptions of the multiple regression, the residuals (which you trade on) are supposed to be normally distributed with mean 0! So this means that you are in fact trading on an indicator that is just random.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.