Comparing (hidden) regression coefficients in simple linear regression? I am doing text modeling, where each document can belong to multiple topics. I specify the desired topic number to be 3 and the three topics to be weather, sports, and education. So the model will output a triplet representing the topic mixture of each document.
Let $p_1,p_2,p_3$ denote Pr(weather), Pr(sports), Pr(education), respectively. So a news article that talks about a football match being canceled due to the bad weather may have a triplet $(p_1,p_2,p_3)=(0.4,0.5,0.1)$. The three elements in the tuple sum to $1$.
Now I wish to explore the relationship between topics and the popularity of the document. I start with multiple linear regression. Since $p_1+p_2+p_3=1$, I should only include two of $p_1,p_2,p_3$ like
$$y=\beta_0+\beta_1p_1+\beta_2p_2,$$
where $y$ is the reader-rated popularity of the documents.
Denote $p_3$'s coefficient by $\beta_3$ (not in the current model above). Is it possible to test all of the following three hypotheses with the model above?


*

*$\beta_1=\beta_2$;

*$\beta_2=\beta_3$;

*$\beta_1=\beta_3$.


Is my existing answer below correct? If not, how can I do that?
 A: EDIT to respond to the altered question:
Again you phrase your hypothesis based on parameters outside this model, which makes it a little uncertain what exactly you are going at. But interpreting your hypotheses to be referring to the marginal effect of each of the three topics being covered in one text, I think what you are trying to do can be done. Basically then the answer is  given by following the suggestion in D. Stroet's answer, which is (sort of) equivalent to your own answer  and the edited part in @EdM's answer:

*

*$\beta_1=0$ indicates that increasing the content of weather [holding sports constant and consequently reducing education]  has no bearing on reader rating. This is equivalent to saying that weather and education are contributing equally to reader popularity.


*$\beta_2=0$ the same reasoning applies. This is equivalent to saying that sports and education contribute equally to reader popularity.


*$\beta_1=\beta_2$ can be tested just as you intended in you question. Finally, this implies that sports and weather contribute equally to reader popularity
Original Anwer:

I can not comment, hence I'm putting this in an answer. Whoever can, please convert this into a comment.
@Sibbs, can you give a reasonable example, where testing equality of these partial effects (I assume that's what you mean, since you never define $\beta_3$) makes sense, yet such a restriction ($x_1+x_2+x_3=1$) holds? As @EdM points out your "solution" already points to the conceptual flaw of your model/question. Maybe if you elaborate on what brought you there, someone could help you.
It would help to know more about the actual data underlying your model, as presented in the first question. You provided a 3-variable case "for simplicity" but that leads to this problem with only 2 independent coefficients when all $x_i$ must add up to 1. If these are data on fruit, however, there may be additional "components" (like water content, fiber) that have no bearing on perceived "sweetness" other than their influences on the effective concentrations of the sweetness-associated $x_i$. Knowing more about all the original data, as opposed to how some variables have already been transformed in a way that requires them to add to 1, may help resolve your underlying issue without trying to perform a set of comparisons that can't really be done with the restrictions you have imposed.

A: To test if $\beta_1=\beta_3$, we need to "reframe" the regression a bit.
$$Y=\beta_0+\beta_1x_1+\beta_2x_2$$
$$Y=\beta_0+\beta_1x_1+\beta_2(1-x_1-x_3)$$
$$Y=\beta_0+\beta_2+(\beta_1-\beta_2)x_1-\beta_2x_3$$
Now $\beta_1=\beta_3$ is equivalent to $\beta_1-\beta_2=-\beta_2$, becoming $\beta_1=0$. So just test if $\beta_1=0$ for $\beta_1=\beta_3$.
A: It's not possible to do all 3 pairwise comparisons of regression coefficients that you wish, because you only really have two regression coefficients.
The value of $\beta_3$ in a linear regression would be the effect of a change in $x_3$ on $Y$ with $x_1$ and $x_2$ held constant. That is not possible in your situation where the $x_i$ must all add up to 1.
All of the information about your regression is included in any choice of 2 of your 3 $x_i$. This is why your algebra came up with the result that the only way for $\beta_1=\beta_3$ is for both to be zero.
Edit based on further reflection:
What you can do in this case comes from the answer by @Glen_b to your previous question. The $\beta_0$ intercept in the model you present on this page represents the value of the Sweetness outcome variable ($Y$) when $x_3$ is the only fruit component present ($x_1$, $x_2$ both 0).
In that context, $\beta_1$ represents the change in Sweetness per change in $x_1$ when $x_2$ is held constant. Since all the $x_i$ must add to 1 and all are (presumably) non-negative, the only way for this to occur is for $x_3$ to go down as $x_1$ goes up. Thus $\beta_1$ represents the excess Sweetness of $x_1$ as it replaces $x_3$ in the mix; the same argument holds for $\beta_2$ as the excess Sweetness provided by $x_2$ as it replaces $x_3$.
So to compare the relative contributions of the $x_i$ to Sweetness, significantly non-zero values of $\beta_1$ and $\beta_2$ mean that $x_1$ resp. $x_2$ contribute differently to sweetness than does $x_3$. The comparison between $\beta_1$ and $\beta_2$ distinguishes $x_1$ from $x_2$.
Although this answers your fundamental question (not the question you posed for the way you originally wanted to proceed), I urge you to consider the original data underlying this work and whether you would be better served by analyzing data closer to their original forms rather than in a form forced to add up to 1, as the answer by @sheß suggests. When data take on necessarily restricted ranges I worry that linear regression might not model the data adequately; I trust that you have extensively examined diagnostics to make sure that your model actually works here.
A: My suggestion is to interpret the t-statistics of $x_1$ and $x_2$ in relation to $x_3$, because the latter is the reference group. 
That will answer 2. and 3. Test the equality of $x_{1}$ and $x_2$ separately with an appropriate test which will answer 1.
