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I have two continuous variables $x_1$ and $x_2$, such that their sum is a constant: $x_1+x_2=c$. Clearly, I cannot run the following OLS model due to perfect multicollinearity:

Model (1): $y = \alpha + \beta_1 x_1 + \beta_2 x_2 + \epsilon$

If I run the Model (2) below, $\beta_1$ is significantly negative:

Model (2): $y = \alpha + \beta_1 x_1 + \epsilon$

I have reason to believe that $x_2$ is also a determinant of $y$ and must be in the regression model alongside $x_1$. In Model (3), which constrains the intercept to zero, $\beta_1$ is significantly positive:

Model (3): $y = \beta_1 x_1 + \beta_2 x_2 + \epsilon$

Models (2) and (3) reach opposite conclusions regarding $\beta_1$. Model (3) yields the theoretically predicted result ($\beta_1>0$). My question is whether I can rely on Model (3), since it excludes the intercept in order to include $x_2$ alongside $x_1$.

To address the comments, think of $c$ as the size of a pie that is equal for everyone, and each individual slices the pie into two pieces $x_1$ and $x_2$, such that $x_1+x_2=c$. What I'm investigating is whether one type of slice $x_2$ matters more than the other $x_1$. In other words, whether $\beta_2>\beta_1$ in Model (3).

To be more specific, $c$ is the average lottery return, which is decomposed into two parts for each individual as follows:

$c = o_i r_{oi} + p_i r_{pi}$ where $o_i$ ($p_i$) is the proportion of lotteries individual $i$ observes (plays), such that $o_i+p_i=1$, and $r_{oi}$ ($r_{pi}$) is the average return of the lotteries that are observed (played) by individual $i$. Finally, $x_1\equiv o_i r_{oi}$ and $x_2\equiv p_i r_{pi}$, such that $x_1+x_2=c$; and $y$ is the future participation rate in the lotteries.

Rational learning theories predict that individuals will give equal importance to the returns that they observe ($x_1$) versus those that they experience ($x_2$): $\beta_1=\beta_2>0$. On the other hand, reinforcement learning theories predict personally experienced outcomes matter more: $\beta_2>\beta_1>0$. That is why I'm trying to find out whether Model (3) is an appropriate way of testing these two theories.

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    $\begingroup$ As you've written this, you don't need to include both $x_1$ & $x_2$ in the model - either one will do as it determines the other. Are you sure that's what you meant? $\endgroup$ – Scortchi Jul 30 '13 at 17:16
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    $\begingroup$ Models (2) and (3) are identical, because $\beta_1x_1+\beta_2x_2$ = $\beta_1x_1+\beta_2(c-x_1)$ = $\beta_2 c + (\beta_1-\beta_2)x_1$. Thus $\alpha=\beta_2 c$ and the $\beta_1$ of model (2) equals $\beta_1-\beta_2$ from model (3). Whence they cannot reach "opposite conclusions": the tests of coefficient "significance" (i.e., whether the coefficients might be zero) and the signs of the estimated coefficients are merely making conclusions about different things. $\endgroup$ – whuber Jul 30 '13 at 17:36
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    $\begingroup$ You imply by a pie example that your $x_1$ and $x_2$ are (equivalent to) proportions of a total, which wasn't implied merely by their sum being a constant. Suppose the variables were proportion male and proportion female and their sum is 1. It is natural to think that the proportions necessarily depend on how many males and females there are. But if you feed a model one variable that is proportion male, a second variable proportion female adds precisely no extra information, and any statistical model (and any statistical program) will necessarily reflect that. So, the question is ill-posed. $\endgroup$ – Nick Cox Jul 31 '13 at 10:52
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    $\begingroup$ This doesn't help much, partly because the precise rules for this lottery aren't clear. In particular, if you can lose money, then your pie analogy is invalid. But I think my answer is the same. All the comments you are getting seem to point to the same thing: regardless of your substantive interest in comparing two things, the constraint means that statistically it doesn't work out that way. $\endgroup$ – Nick Cox Jul 31 '13 at 14:10
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    $\begingroup$ The model in the edit does not appear to capture what you want to know. If future participation rates ($y$) among $n$ future lotteries depend on a combination of average observed ($r_o$) and played ($r_p$) returns, then the model ought to be $\mathbb{E}[y]=h(\alpha+\beta_o r_o + \beta_p r_p)$, with $y$ having a Binomial$(n,\cdot)$ distribution and $h$ being a "link" function. (This is a GLM.) Weighting $r_o$ and $r_p$ by rates of participation and observation seems unwarranted. If you like, you could include one of these (interdependent) rates, say $p$, as a third explanatory variable. $\endgroup$ – whuber Jul 31 '13 at 14:26

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