How to treat x=0, y=0 in a linear model with no intercept? Very often in the research we want to establish a linear relationship without intercept: $y=\beta x + \epsilon$ and the sample has many double zero observations $x=0, y=0$. I am wondering how to deal with these many zeros. On one hand, they are the actual observations and we should not exclude them. On the other hand, they do not contribute estimating the real value of $\beta$, because with zero-zero observations, any $\beta$ value would be correct. 
The result of including all the zero-zero in the linear regression would end up with an estimated $\beta$ which is significant but relying on the only few non zero-zero observations.
Here is an extreme example of response $y$ and exploratory variable $x$. Both of them contains 100 zeros and only 2 non zeros.
x <- c(rep(0,100), 5, 10)
y <- c(rep(0,100), 10, 20)
fit <- lm(y~x-1)
summary(fit)

The model result shows that the estimated value of $\beta$ is 2 and it is significantly different from zero.
Coefficients:
Estimate Std. Error t value Pr(>|t|)    
 x        2          0     Inf   <2e-16 ***
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Residual standard error: 0 on 101 degrees of freedom
Multiple R-squared:     1,      Adjusted R-squared:     1 
F-statistic:   Inf on 1 and 101 DF,  p-value: < 2.2e-16 

The main purpose is to estimate what the value of β is. In this example, I could argue that with 100 zero-zero, β could be of any value, but the model output fixes it to 2, just based on the 2 non zero-zero values. In this case where many zero-zero observations are in the sample, is it the proper way to establish the linear relationship?
 A: The problem here seems to be not the occurence of $(0,0)$-values (since any other points $(x,y)$ will behave similarly), but rather a problem of model selection. If understand you right, you want to assume a linear model and then let the fitting procedure state doubts about either the fitted parameters or about the linear model itself.
Some points into three different directions:


*

*First, within the linear model, you could use Bayesian regression and then find that the slope has a much larger variance than the intercept (of course this depends also on the prior). Adding more (0,0) values will mainly narrow the distribtion of the intercept and not the distribution of the slope.

*Second, also assuming a linear model, you can use weighted least squares regression. If you have doubts about the (2,2) values, you can assume them to have a large variance and thus a small weight. Similar to the Bayesian setup, this will widen the error bars and draw higher doubt on the result for the intercept.

*Third, and more generally, if you want a procedure which doubts the model based on the training data, you can use model comparison schemes which will show that other models beside a straight line are possible and probably equally likely (e.g., a parabola will also fit the data well).
However, in the end you are stuck with the model you assume. Hence, you should choose it carefully and with regard to the data.
