Bounding a Linear Causal Effect Using Relative Correlation Restrictions

A.1 Proposition 1

Proof: To establish result 1, note that:

λ(bx; m)=corrm(x, y−bxx−(y−bxx)p)corrm(x, (y−bxx)p)=covm(x, y−bxx−(y−bxx)p)varm(x)varm(y−bxx−(y−bxx)p)covm(x, (y−bxx)p)varm(x)varm((y−bxx)p)=(covm(x, y)−bxvarm(x)covm(x, yp)−bxcovm(x, xp)−1)varm(y−bxx)varm((y−bxx)p)−1

We can apply several properties of the best linear predictor, specifically that cov(x, y^p)=cov(x^p, y^p), cov(x, x^p)=var(x^p) and var(y–y^p)=var(y)–var(y^p), to further derive:

where p₁, p₂, p₃, and p₄ are all polynomials (and thus differentiable) in b_x. They are also differentiable in m. Application of the quotient and product rules implies that λ(b_x; m) is differentiable provided that (a) p₂≠0, (b) p₄≠0, and (c) p3p4>1. Condition (a) fails if and only if:

p2=covm(xp, yp)−bxvarm(xp)=0

Since var_m(x^p)>0 by equation (11), we can solve to get

bx=covm(xp, yp)varm(xp)=βx∞(m)

Condition (b) fails if and only if:

p4=varm(yp−bxxp)=0

which implies that y^p–b_xx^p is constant. Since the covariance of any random variable with a constant is zero, this in turn implies that cov(x^p, y^p–b_xx^p)=cov(x^p, y^p)–b_xvar(x^p)=0. Again we can solve for b_x to get:

bx=covm(xp, yp)varm(xp)=βx∞(m)

Condition (c) fails if and only if p₃≤p₄, or equivalently:

varm(y−bxx)≤varm(yp−bxxp)

Note that y^p–b_xx^p is the best linear predictor of y–b_xx, so:

varm(y−bxx)=varm(yp−bxxp)+varm(y−bxx−(yp−bxxp))

This implies that var_m(y–b_xx–(y^p–b_xx^p))=0, which also implies that:

y−bxx−(yp−bxxp)=0

Rearranging, we get:

y=yp−bxxp+bxx

which implies that y is an exact linear function of (x, c) and equation (9) is violated. Therefore, condition (c) must hold. Since conditions (a), (b), and (c) hold for all bx≠βx∞(m), λ(bx; m), is differentiable at all bx≠βx∞(m).

To establish result 2, note that var_m(x) is strictly positive by (9) and var_m(x^p) is strictly positive by (11). Therefore:

limbx→∞(covm(x, y)−bxvarm(x))=−∞limbx→∞(covm(xp, yp)−bxvarm(xp))=−∞

So by L’Hospital’s rule:

limbx→∞covm(x, y)−bxvarm(x)covm(xp, yp)−bxvarm(xp)=varm(x)varm(xp)

By the same reasoning:

limbx→∞(varm(x)−2bxcovm(x, y)+bx2varm(x))=∞limbx→∞(varm(yp)−2bxcovm(xp, yp)+bx2varm(xp))=∞limbx→∞(−2covm(x, y)+2bxvarm(x))=∞limbx→∞(−2covm(xp, yp)+2bxvarm(xp))=∞

So by two applications of L’Hospital’s rule:

limbx→∞varm(y)−2bxcovm(x, y)+bx2varm(x)varm(yp)−2bxcovm(xp, yp)+bx2varm(xp)=varm(x)varm(xp)

Result 2 can then be derived by substitution, and the argument repeated for limbx→−∞.

To prove result 3 I first show how the behavior of λ(b_x; m) near βx∞(m) depends on some special cases:

1. Suppose that m implies an exact linear relationship between y^p and x^p, i.e.

   (22)Em((yp−am−bmxp)2)=0 (22)

   for some a_m and b_m. Then equation (14) is satisfied for all λ when bx=βx∞(m)=bm.

   Proof: To show that βx∞(m)=bm:

   βx∞(m)=covm(xp, yp)varm(xp)=covm(xp, am+bmxp)+covm(xp, yp−am−bmxp)varm(xp)=bm varm(xp)+0varm(xp)=bm

   To show that equation (14) is satisfied at βx∞(m) for all λ, note that cβ_c (b_x; m)=y^p–b_xx^p. This implies that:

   varm(cβc(βx∞(m);m))=varm(yp−βx∞(m)xp)=varm(yp−bmxp)=varm(yp−bmxp)−2covm(yp−bmxp, am)+varm(am)=varm(yp−am−bmxp)=0

   and by the same argument covm(x, cβc(βx∞(m); m))=0. Equation (14) thus reduces to 0=λ0, a condition that is satisfied by any λ.

2. Suppose that m implies:

   (23)covm(y, x)varm(x)=covm(yp, xp)varm(xp) (23)

   Then equation (14) is satisfied for all λ when bx=βx∞(m).

   Proof: First, note that in this case:

   covm(x, y−βx∞(m)x−cβc(βx∞(m);m))=covm(x, y−βx∞(m)x−yp+βx∞(m)xp) =covm(x, y)−βx∞(m)varm(x)−covm(yp, xp)+βx∞(m)varm(xp) =covm(x, y)−covm(x, y)varm(x)varm(x)−covm(yp, xp)+cov(xp, yp)var(xp)var(xp) =0

   and:

   covm(x, cβc(βx∞(m);m))=covm(x, yp−βx∞(m)xp)=covm(xp, yp)−βx∞(m)varm(xp)=covm(xp, yp)−covm(xp, yp)varm(xp)varm(xp)=0

   Equation (14) thus reduces to 0=λ0, which is satisfied for all λ.

3. Suppose that neither (22) nor (23) hold. Then for any λ∈(–∞, λ^∞(m))∪(λ^∞(m), ∞) there is a b_x such that λ(b_x; m)=λ, i.e., that solves equation (14).

   Proof: First, note that since covm(xp, yp)−βx∞(m)varm(xp)=0, the existence of a solution to equation (14) when bx=βx∞(m) requires that either varm(yp−βx∞(m)xp)=0, implying (22) holds, or covm(x,y)−βx∞(m)varm(x)=0, implying (23) holds. Since neither holds, there is no λ that satisfies equation (14) for bx=βx∞(m).

   Next I characterize the behavior of λ(b_x; m) near βx∞(m). Since var_m(x^p)>0, p₂ is positive for bx<βx∞(m), negative for bx>βx∞(m), and zero when bx=βx∞(m). Also note that covm(x,y)−βx∞(m)varm(x)=covm(x, y)−covm(xp, yp)varm(xp)varm(x), so p₁ is strictly positive for all bx≈βx∞(m) if covm(x, y)varm(x)>covm(xp, yp)varm(xp), and strictly negative for all bx≈βx∞(m) if covm(x, y)varm(x)<covm(xp, yp)varm(xp). This implies that:

   limbx↑βx∞(m)λ(bx; m)={∞if covm(y, x)varm(x)>covm(yp, xp)varm(xp)−∞if covm(y, x)varm(x)<covm(yp, xp)varm(xp)

   and

   limbx↓βx∞(m)λ(bx; m)={−∞if covm(y, x)varm(x)>covm(yp, xp)varm(xp)∞if covm(y, x)varm(x)<covm(yp, xp)varm(xp)

   I have thus established that limbx→−∞λ(bx; m)=λ∞(m), that limbx↑βx∞λ(bx; m) is either –∞ or ∞, and that λ(b_x; m) is continuous on (−∞, βx∞(m)). Suppose for the moment that limbx↑βx∞(m)λ(bx; m)=−∞. By the intermediate value theorem, for any λ∈(–∞(m)), there exists some bx∈(−∞, βx∞(m)) such that λ(b_x; m)=λ. This is a sufficient condition for b_x to solve equation (14). Since limbx↑βx∞=−∞, then limbx↓βx∞(m)λ(bx; m)=∞. Again, since λ(b_x; m) is continuous on (βx∞(m), ∞), the intermediate value theorem implies that for any λ∈(λ^∞(m), ∞) there exists some bx∈(βx∞(m), ∞) such that λ(b_x; m)=λ. Therefore, for any λ∈(–∞, λ^∞(m))∪(λ^∞(m), ∞) there is a b_x such that λ(b_x; m)=λ, i.e., that solves equation (14). The same argument can be duplicated for the case limbx↑βx∞λ(bx)=∞. Note that there may or may not be a b_x such that λ(b_x; m)=λ^∞(m).

To prove result 4, pick any b_x and consider two cases. First, suppose that bx=βx∞(m). Then bx∉B˜(Λ; m) since λ(b_x; m) does not exist. Next, suppose that bx≠βx∞(m). Then λ(b_x; m) exists (by result 1 of this proposition) and provides the unique λ that solves equation (14) for that λ. Therefore,

bx∈B˜(Λ; m)if and only if bx∈Bx(Λ; m) and bx≠βx∞(m)

which is another way of stating the result.        □

A.2 Proposition 2

Proof: Since Λ is nonempty, λ^∞(m₀)∉Λ implies that Λ must contain some λ≠λ^∞(m₀). Result 3 of Proposition 1 says that there exists some b_x such that (λ, b_x) satisfy equation (14). Therefore, the identified set is nonempty.

Since Λ is closed, λ^∞(m₀)∉Λ implies that there is some ε>0 such that (λ_∞(m₀)–ε, λ^∞(m₀)+ε) is disjoint from Λ. Result 2 of Proposition 1 says that limbx→∞λ(bx; m0)=limbx→−∞λ(bx; m0)=λ∞(m0). This means that given such an ε, there is some finite B_ε such that Bε>βx∞(m0) and:

|bx|>Bε⇒λ(bx;m0)∈(λ∞(m0)−ε, λ∞(m0)+ε)  (by result 2 of Proposition1)⇒λ(bx;m0)∉Λ  (since(λ∞(m0)−ε, λ∞(m0)+ε)is disjoint from Λ)⇒bx∉B˜(Λ, m0)  (by definition of B˜)⇒bx∉B˜(Λ, m0)∪{βx∞(m0)}  (since Bε>βx∞(m0))⇒bx∉Bx(Λ, m0)  (by result 4 of Proposition 1)

Therefore, the identified set is bounded.        □

A.3 Proposition 3

Proof: Both βx∞(m) and λ^∞(m) are continuous in m by the quotient rule, given that var_m(x^p)>0. Result 1 of Proposition 1 says that λ(b_x; m) is continuous in m for all bx≠βx∞(m)). So the first set of results follows from the straightforward application of Slutsky’s theorem.

For the second result, note that the implicit function theorem implies that βxL(Λ; m) is continuously differentiable in m if dλ(bx; m)dbx|bx=βxL(Λ; m)≠0. In that case, consistency of b^xL(Λ) follows from Slutsky’s theorem. The same argument applies to b^xH(Λ).

For the third result, note that if B_x(Λ; m₀)=ℝ, then result 2 of Proposition 1 implies λ^∞(m₀) is in the interior of Λ. Therefore, there exists an ε>0 and B₁<B such that [λ(B₁; m₀)–ε, λ(B₁; m₀)+ε]⊂Λ. Since λ^(B1)→pλ(B1; m0):

limn→∞ Pr(b^xL<B)≥limn→∞ Pr(λ^(B1)∈Λ)=1

The same argument applies to b^xH(Λ), with a change of sign.        □

A.4 Proposition 4

Proof: Both βxL(Λ; m) and βxH(Λ; m) are differentiable in m under these conditions, so the result follows from direct application of the delta method, where:

The expression for A given in the proposition comes from applying the implicit function theorem:

and substituting. While mathematically unnecessary, this substitution is important computationally. Derivatives of λ(b_x; m) – a closed form function with closed form derivatives – can be calculated much more accurately than derivatives of βxL(Λ; m) – an implicit function that must be approximated by iterative methods.        □

A.5 Proposition 5

Proof: If var(x^p)=0, then cov(x, y^p–b_xx^p)=0 for all b_x. This implies that (14) holds if and only if cov(x, y–b_xx)=0, i.e., if b_x=cov(x, y)/var(x).        □

A.6 Proposition 6

Proof: First, rewrite:

λ(bx; m)=corrm(x, y−bxx−yp+bxxp)corrm(x, yp−bxxp)=covm(x, y−bxx−yp+bxxp)corrm(x, yp−bxxp)varm(x)varm(y−bxx−yp+bxxp)=q1(bx; m)q2(bx; m)

The numerator of λ(bx; m^n) is:

q1(bx; m^n)→pcov(x, y)−bxvar(x)

while the denominator is

q2(bx; m^n)→p0

In a given finite sample, q2(bx; m^n) will be nonzero with probability one if x or any of c is continuously distributed, and probability approaching one as n→∞ (WPA1) otherwise. So λ(bx; m^n) will exist even though λ(b_x; m₀) does not. Let βxOLS(m) be the value of b_x that implies q₁(b_x; m)=0, or equivalently:

βxOLS(m)=covm(x−xp, y−yp)varm(z−xp)

Note that βxOLS(m^n) is just the coefficient on x from the OLS regression of y on x and c, and that:

Since q1(βxOLS(m^n))=0 by construction and q2(βxOLS(m^n))≠0 WPA1:

λ(βxOLS(m^n);m^n)=0∈Λ WPA1

Therefore:

Pick any ε>0. The event (|βxOLS(m^n)−βx|<ε) clearly implies (βxOLS(m^n)>βx−ε), which itself implies (β^H(Λ)>βx−ε) by equation (27). Therefore:

Pr(|βxOLS(m^n)−βx|<ε)≤Pr(β^H(Λ)>βx−ε)≤1

By (26), Pr(|βxOLS(m^n)−βx|<ε)→1, so by the sandwich theorem:

Let λ^max satisfy |λ|≤λ^max for all λ∈Λ. Then λ∈Λ implies |λ|≤λ^max. Therefore:

Now, for any δ≠0

q1(βx+δ;m^n)→pcov(x, y)−(βx+δ)var(x)=−δvar(x)≠0q2(βx+δ;m^n)→p0

Therefore:

By the sandwich theorem (29) and (30) imply Pr(β^H(Λ)≥βx+ε)→0, or equivalently that:

Taking (28) and (31) together produces:

which is the result stated in the proposition. The same argument applies to βxL.        □