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Say you have an independent variable, $x$, and two dependent variables $y_1$ and $y_2$. I want to calculate whether these two variables have a significantly different slope. I can do it by calculating separate regressions, and then using he method mentioned at Test a significant difference between two slope values.

I was wondering, would be equivalent to testing for the significance of the slope of the series $(y_1-y_2)$ [e.g. null hypothesis slope=0]? I tried it with a couple of hundred independent series, and the results look very similar (similar distribution of p-values), but I'd think if it were that easy, more people would use it, as the computation is simpler. So, why is it not the same?

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naught101
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1 Answers1

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If the x's are the same for both, you're effectively making the values paired (on $x$).

More generally - as in the case where you have two separate fitted lines - you don't have that.

When you do have the two sets of $y$ paired on $x$, then you could test for a difference in slope by taking differences $y_{2i}-y_{1i}$ and regressing that on $x$.

If the y-values for a given $x$ are independent (conditional independence), it should make no difference which you do.

$\hat{\beta_2}-\hat{\beta_1}=(X'X)^{-1}X'y_2-(X'X)^{-1}X'y_1=(X'X)^{-1}X'(y_2-y_1)$

showing the estimates are the same, and (assuming conditional independence):

$\text{Var}((X'X)^{-1}X'(y_2-y_1))=(X'X)^{-1}X'\text{Var}(y_2-y_1)X'(X'X)^{-1}$

$=(X'X)^{-1}X'((\sigma_1^2+\sigma_2^2)I)X'(X'X)^{-1}$

$=(\sigma_1^2+\sigma_2^2)(X'X)^{-1}X'X'(X'X)^{-1}$

$=(\sigma_1^2+\sigma_2^2)(X'X)^{-1}$

$=\text{Var}(\hat{\beta_2})+\text{Var}(\hat{\beta_1})$

(though the sample values won't be perfectly uncorrelated). So I think the two are at least asymptotically equal, and should be equally efficient at smaller samples. I don't think it matters which you use.

However, if the y-values are positively correlated, it would be important to work with the differences.

Glen_b
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  • Cool, thanks, I had a vague idea that it might require error independence, but I wasn't sure. – naught101 Feb 28 '15 at 06:19
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    It was a fun question actually, I'd never been through that particular calculation before. (I already had come to the answer from general reasoning, but it's always best to follow through a couple of lines of algebra to make sure.) – Glen_b Feb 28 '15 at 06:55