I have three factors a, b, and c. In the three univariate models

y ~ a
y ~ b
y ~ c

a and b are insignificant and c is significant. But in the multivariate model

y ~ a + b + c 

a and b are significant and c is insignificant.

I understand how adding a factor can make another factor significant (say by controlling for a bunch of variance that was obscuring its relationship). And I understand how adding a factor can make another factor insignificant (say by capturing all of its variance). But I don't understand how the factor that makes others significant can be made insignificant by them. What's the way to think or visualize to make this make obvious?


Are you sure you are using marginal sum of squares, and not sequential sum of squares? Because if you use sequential, and c is a linear combination of a and b, then of course putting it third would make it insignificant.

If you use marginal sum of squares, it is also possible. Say, $y=k_1a+k_2b$, and $c=k_3a+k_4b+noise$. If the level of this noise is high enough, you may still improve your explanation by adding $a$ as a factor on top of $b$ and $c$, or $b$ as a new factor on top of $a$ and $c$. But adding $c$ on top of $a$ and $b$ would not improve anything, as it would contain no new useful information compared to $a$ and $b$.

EDIT: user ahstat below provided a better response. I will only improve on it a bit: you don't seem to need fancy non-linear functions like min() and max() to make the linear combination of $a$ and $b$ much more powerful than any of them together. Here's R code:

    n = 1000
    x1 = rnorm(n) 
    x2 = rnorm(n)
    x3 = rnorm(n)*100
    a = x1+x3
    b = x1-x3


The scenario that I was trying to solve is a bit stronger: I wanted to find a combination where y~a+b would not be significant as well, but y~a+b+c would make $a$ and $b$ significant, but not $c$ (in case of sequential sum of squares). I'm still not sure whether it's only possible if $c$ tips $a$ just over the threshold for significance, by reducing the residual, or whether one could build some sort of a noisy 4-dimensional saddle with negative correlations where this pattern would always be the case.

  • $\begingroup$ Let's say it was sequential sum of squares: then of course being third would make c insignificant, but how could c added third make a and b significant? As for marginal SS, your example is great and concrete. I can see here how c added to a and b would become insignificant but, again, how did it simultaneously make a and b significant? $\endgroup$ – enfascination Sep 1 '17 at 4:11
  • $\begingroup$ The only idea I have is to make c a linear combination of a and b (so that it would be insignificant after a+b), but also make it carry a little of explanatory value, so that the residual after a+b+c would be smaller than for a alone. Then if a alone gives you something like p=0.06, reducing the residual may increase F-value just enough to make a significant once c is added. But overall I'm inclined to believe that your n is rather low and so the wild dance of numbers just happened to create this strange situation. $\endgroup$ – ampanmdagaba Sep 1 '17 at 5:32

You can have a linear combination of $a$ and $b$ significant to fit $y$ if there is some specific nonindependent noise while retrieving $a$ and $b$. For example, for some noise $\varepsilon$, if:

$ a = max(y,0) + \varepsilon,$

$ b = min(y,0) - \varepsilon,$

then, $a+b = y$ exactly.

Here is an example in R.

For individual regressions, you observe $\text{Pr}(>|t|) = 0.639$ for $a$; $\text{Pr}(>|t|) = 0.617$ for $b$; $\text{Pr}(>|t|) <2.10^{-16}$ for $c$ (because $c$ is constructed as $y + \text{noise}$).

But together, you have both $a$ and $b$ significant ($\text{Pr}(>|t|) <2.10^{-16}$), and there is no variance remaining to make $c$ useful ($\text{Pr}(>|t|) = 0.319$ for coefficient $c$).

N = 1000
y = rnorm(N)
eps = N*rnorm(N)
a = sapply(y, function(x){max(x,0)}) + eps
b = sapply(y, function(x){min(x,0)}) - eps
c = y + rnorm(N, 0, 4)

reg_a = lm(y ~ a)
plot(a, y)

reg_b = lm(y ~ b)
plot(b, y)

reg_c = lm(y ~ c)
plot(c, y)

reg = lm(y ~ a + b + c)

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