When A and B are positively related variables, can they have opposite effect on their outcome variable C?

Question

A is positively related to B.

C is the outcome of A and B, but the effect of A on C is negative and the effect of B on C is positive.

Can this happen?

Answer 1

The other answers are truly marvelous - they give real life examples.

I want to explain why this can happen despite our intuition to the contrary.

See this geometrically!

Correlation is the cosine of the angle between the (centered) vectors. Essentially, you are asking whether it is possible that

• $A$ makes an acute angle with $B$ (positive correlation)
• $B$ makes an acute angle with $C$ (positive correlation)
• $A$ makes an obtuse angle with $C$ (negative correlation)

Yes, of course:

In this example ($\rho$ denotes correlation):

• $A=(0.6,0.8)$
• $B=(1,0)$
• $C=(0.6,-0.8)$
• $\rho(A,B)=0.6>0$
• $\rho(B,C)=0.6>0$
• $\rho(A,C)=-0.28<0$

Your Intuition is Right!

However, your surprise is not misplaced.

The angle between vectors is a distance metric on the unit sphere, so it satisfies the triangle inequality:

$$\measuredangle AB \le \measuredangle AC + \measuredangle BC$$

thus, since $\cos \measuredangle AB = \rho(A,B)$,

$$\arccos\rho(A,B) \le \arccos\rho(A,C) + \arccos\rho(B,C) $$

therefore (since $\cos$ is decreasing on $[0,\pi]$)

$$\rho(A,B)\ge\rho(A,C)\times\rho(B,C) - \sqrt{(1-\rho^2(A,C))\times(1-\rho^2(B,C))} $$

So,

• if $\rho(A,C)=\rho(B,C)=0.9$, then $\rho(A,B)\ge 0.62$
• if $\rho(A,C)=\rho(B,C)=0.95$, then $\rho(A,B)\ge 0.805$
• if $\rho(A,C)=\rho(B,C)=0.99$, then $\rho(A,B)\ge 0.9602$

Answer 2

Yes, two co-occuring conditions can have opposite effects.

For example:

• Making outrageous statements (A) is positively related to being entertaining (B).
• Making outrageous statements (A) has a negative effect on winning elections (C).
• Being entertaining (B) has a positive effect on winning elections (C).

Answer 3

I've heard this car analogy which applies well to the question:

• Driving uphill (A) is positively related to the driver stepping on the gas (B)
• Driving uphill (A) has a negative effect on vehicle speed (C)
• Stepping on the gas (B) has a positive effect on vehicle speed (C)

The key here is the driver's intention to maintain a constant speed (C), therefore the positive correlation between A and B naturally follows from that intention. You can construct endless examples of A, B, C with this relationship thus.

The analogy comes from an interpretation of Milton Friedman's Thermostat and comes from an interesting analysis of monetary policy and econometrics, but that's irrelevant to the question.

Answer 4

Yes, this is trivial to demonstrate with a simulation:

Simulate 2 variables, A and B that are positively correlated:

> require(MASS)
> set.seed(1)
> Sigma <- matrix(c(10,3,3,2),2,2)
> dt <- data.frame(mvrnorm(n = 1000, rep(0, 2), Sigma))
> names(dt) <- c("A","B")
> cor(dt)

          A         B
A 1.0000000 0.6707593
B 0.6707593 1.0000000

Create variable C:

> dt$C <- dt$A - dt$B + rnorm(1000,0,5)

Behold:

> (lm(C~A+B,data=dt))

Coefficients:
(Intercept)            A            B  
    0.03248      0.98587     -1.05113  

Edit: Alternatively (as suggested by Kodiologist), just simulating from a multivariate normal such that $\operatorname{cor}(A,B) > 0$, $\operatorname{cor}(A,C) > 0$ and $\operatorname{cor}(B,C) < 0$

> set.seed(1)
> Sigma <- matrix(c(1,0.5,0.5,0.5,1,-0.5,0.5,-0.5,1),3,3)
> dt <- data.frame(mvrnorm(n = 1000, rep(0,3), Sigma, empirical=TRUE))
> names(dt) <- c("A","B","C")
> cor(dt)
    A    B    C
A 1.0  0.5  0.5
B 0.5  1.0 -0.5
C 0.5 -0.5  1.0

Answer 5

$$ C = mB + n (A-proj_B(A)) $$

then $$ \left<C,A\right> = m\left<B,A\right> + n\left<A,A\right> -n \left<B,A\right> $$

Then covariance between C and A could be negative in two conditions:

1. $n>m,\ \left<A,A\right> < \left<B,A\right> (n-m)/n $
2. $n<-m,\ \left<A,A\right> > \left<B,A\right> (n-m)/n$