How to Average Equating Functions, If You Must

Abstract

An interest in averaging two or more equating functions can arise in various settings. As the motivation for the angle bisector method described later in this paper, Angoff (1971) mentioned situations with multiple estimates of the same linear equating function for which averaging the different estimates may be appropriate. In the nonequivalent groups with anchor test (NEAT) equating design, several possible linear and nonlinear equating methods are available. These are based on different assumptions about the missing data in that design (von Davier, Holland, & Thayer, 2004b). It might be useful to average the results of some of the options for a final compromise method. Other recent proposals include averaging an estimated equating function with the identity transformation to achieve more stability in small samples (Kim, von Davier, & Haberman, 2008) as well as creating hybrid equating functions that are averages of linear and equipercentile equating functions, putting more weight on one than on the other (von Davier, Fournier-Zajac, & Holland, 2006). In his discussion of the angle bisector, Angoff implicitly weighted the two linear functions equally. The idea of weighting the two functions differently is a natural and potentially useful added flexibility to the averaging process that we use throughout our discussion.

Chapter 6 Appendix

1.1 A.1 Computing the Swave for Two Equating Functions

The key to computing e _w is Equation 6.35. This equation for t(x) is nonlinear in general, so computing t(x) requires numerical methods. A derivative-free approach that is useful in this situation is Brent’s method. To use this method to solve Equation 6.35 for t we first define g(t) as follows:

$$g\left( t \right){\rm } = t-{e_{\rm 1}}\left( {x-\left( {{\rm 1}-w} \right)t} \right) + {e_{\rm 2}}\left( {x + w\,t} \right).$$

(6.A.1)

If t ₀ solves Equation 6.35, then t ₀ is a zero of g(t) in Equation 6.A.1. Brent’s method is a way of finding the zeros of functions. It requires that two values of t are known, one for which g(t) is positive and one for which g(t) is negative. Theorem 1 summarizes several useful facts about g(t) and provides the two needed values of t for use in Brent’s method.

Theorem 1

If e ₁ and e ₂ are strictly increasing continuous functions, then g(t) defined in Equation 6.A.1 is a strictly increasing continuous function that has a unique zero at t ₀ . Furthermore, t ₀ is positive if and only if e ₁ (x) – e ₂ (x) is positive. Consequently, if e ₁ (x) – e ₂ (x) is positive, then g(0) is negative and g(e ₁ (x) – e ₂ (x)) is positive; furthermore, if e ₁ (x) – e ₂ (x) is negative, then g(0) is positive and g(e ₁ (x) – e ₂ (x)) is negative.

Proof

The functions t, – e ₁(x – (1 – w) t), and e ₂(x + w t) are all strictly increasing continuous functions of t so that their sum, g(t), is also a strictly increasing continuous function of t. Hence, if g(t) has a zero at t ₀, this is its only zero. In order to show that g(t) does have a zero at some t ₀ it suffices to show that, for large enough t, g(t) > 0 and, for small enough t, g(t) < 0. But if t > 0, it follows from the strictly increasing (in t) nature of – e ₁(x – (1 – w) t) and of e ₂(x + w t) that

$$g\left( t \right){\rm } \,>\, t-{\rm }\left[ {{e_{\rm 1}}\left( x \right){\rm }-{e_{\rm 2}}\left( x \right)} \right].$$

(6.A.2)

The right side of Equation 6.A.2 is greater than 0 if t is larger than e ₁(x) – e ₂(x). Similarly, if t < 0, it also follows that

$$g\left( t \right){\rm } \,<\, t-{\rm }\left[ {{e_{\rm 1}}\left( x \right){\rm }-{e_{\rm 2}}\left( x \right)} \right].$$

(6.A.3)

The right side of Equation 6.A.3 is less than 0 if t is less than e ₁(x) – e ₂(x). Hence, these two inequalities show that g(t) always has a single zero at a value we denote by t ₀.

Now, suppose that t ₀ > 0. Then g(t ₀) = 0 by definition so that

$$0{\rm } \,<\, {t_0} = {e_{\rm 1}}\left( {x-{\rm }\left( {{\rm 1}-w} \right)\,{t_0}} \right){\rm }-{e_{\rm 2}}\left( {x + w\,{t_0}} \right)$$

(6.A.4)

But by the strict monotonicity of e ₁ and e ₂, we have

$${e_{\rm 1}}\left( {x-{\rm }\left( {{\rm 1}-w} \right)\,{t_0}} \right){\rm } <\, {e_{\rm 1}}\left( x \right),{\rm and}-{e_{\rm 2}}\left( {x + w{\,t_0}} \right){\rm } < {\rm }-{e_{\rm 2}}\left( x \right)$$

so that

$${e_{\rm 1}}\left( {x-{\rm }\left( {{\rm 1}-w} \right)\,{t_0}} \right){\rm }-{e_{\rm 2}}\left( {x + w{t_0}} \right)\,{\rm } < \,{e_{\rm 1}}\left( x \right){\rm }-{e_{\rm 2}}\left( x \right).$$

(6.A.5)

Combining Equations 6.A.4 and 6.A.5 shows that if t ₀ > 0, then e ₁(x) – e ₂(x) > 0.

A similar argument shows that if t ₀ < 0, then e ₁(x) – e ₂(x) < 0. Hence t ₀ is positive if and only if e ₁(x) – e ₂(x) is positive. Note that we can always compute e ₁(x) – e ₂(x) because it is assumed that these functions are given to us. Thus, from the relative sizes of e ₁(x) and e ₂(x) we can determine the sign of the zero, t ₀.

Because g(t) is strictly increasing we have the following additional result. If e ₁(x) – e ₂(x) is positive, then t ₀ is also positive and therefore g(0) is negative. Also, if e ₁(x) – e ₂(x) is negative, then t ₀ is also negative and therefore g(0) is positive.

Now suppose again that e ₁(x) – e ₂(x) is positive so that t ₀ is also positive. However, from Equation 6.19, for any positive t, g(t) > t – [e ₁(x) – e ₂(x)], so let t = t ₀. Hence,

$$0{\rm } = g\left( {{t_0}} \right){\rm } \,>\, {t_0}-{\rm }\left[ {{e_{\rm 1}}\left( x \right){\rm }-{e_{\rm 2}}\left( x \right)} \right],$$

(6.A.6)

so that

$$0{\rm } \,<\, {t_0} \,<\, {e_{\rm 1}}\left( x \right){\rm }-{e_{\rm 2}}\left( x \right).$$

(6.A.7)

Hence, g(e ₁(x) – e ₂(x)) is positive as well. Thus, whenever e ₁(x) – e ₂(x) is positive, then g(0) is negative and g(e ₁(x) – e ₂(x)) is positive. When e ₁(x) – e ₂(x) is negative, a similar argument shows that

$${e_{\rm 1}}\left( x \right){\rm }-{e_{\rm 2}}\left( x \right){\rm } \,<\, {t_0} \,<\, {\rm }0.$$

(6.A.8)

Hence g(e ₁(x) – e ₂(x)) is negative. This finishes the proof of Theorem 1.

1.2 A.2 Properties of the Swave

Theorem 2

The swave, e_w(x), satisfies Property 2 and lies strictly between e ₁ (x) and e ₂ (x), for all x.

Proof

Consider the case when e ₁(x) > e ₂(x) (the reverse case is proved in a similar way). We wish to show that e ₁(x) > e _w (x) > e ₂(x). Because e ₁(x) > e ₂(x), from Theorem 1 it follows that t(x) > 0 as well. From the strictly increasing natures of e ₁ and e ₂, it follows that

$${e_{\rm 1}}\left( {{{x}_{\rm 1}}} \right) \,<\, {e_{\rm 1}}\left( {x} \right),{\rm and}\,{e_{\rm 2}}\left( {{x_2}} \right) \,>\, {e_2}(x).$$

We wish to show that e ₁(x) > e _w (x) > e ₂(x), so consider first the upper bound. By definition,

$${e_w}\left( x \right){\rm } = w{e_{\rm 1}}\left( {{{x}_{\rm 1}}} \right){\rm } + {\rm }\left( {{\rm 1}-w} \right){e_{\rm 2}}\left( {{x_{\rm 2}}} \right)\,{\rm } <\, w{e_{\rm 1}}\left( {x} \right){\rm } + {\rm }\left( {{\rm 1}-w} \right){e_{\rm 2}}\left( {{x_{\rm 2}}} \right).$$

However,

$$0{\rm } \,<\, {\rm }t\left( x \right){\rm } = {\rm }e_1\left( {x_1} \right){\rm }-{\rm }e_2\left( {x_2} \right),{\rm so\ that\ }e_2\left( {x_2} \right){\rm } <\, {\rm }e_1\left( {x_1} \right){\rm } <\, {\rm }e_1\left( x \right).$$

Combining these results give us

$${e_w}\left( x \right){\rm } \,<\, {w}{{e}_{\rm 1}}\left( {\rm x} \right){\rm } + {\rm }\left( {{\rm 1}-w} \right){e_{\rm 1}}\left( x \right){\rm } = {e_{\rm 1}}\left( x \right),$$

the result we wanted to prove. The lower bound is found in an analogous manner.

Theorem 3

The swave is strictly increasing if e ₁ and e ₂ are.

Facts: e(x) monotone implies c(x) = x + e(x) is strictly monotone (since it is a sum of a monotone and a strictly monotone function). Also, c(x*) > c(x) implies x* > x and e(x*) > e(x).

Let c _i (x) = x + e _i (x), i = 1, 2. Also let e _w (x) = we ₁(x ₁) + (1 – w)e ₂(x ₂), where x = w x ₁ + (1 – w)x ₂ and c ₁(x ₁) = c ₂(x ₂), i.e., x ₁ + e ₁(x ₁) = x ₂ + e ₂(x ₂) so that (x _i , e _i (x _i )) are on same orthogonal line.

Assume e _i (x) are both monotone increasing. Now suppose x* > x where x = wx ₁ + (1 – w)x ₂ and x* = wx ₁* + (1 – w)x ₂* and suppose further that c ₁(x ₁) = c ₂(x ₂) and that c ₁(x ₁*) = c ₂(x ₂*). Then, (x _i , e _i (x _i)) are both on the same orthogonal line and (x _i *, e _i (x _i *)) are too (but possibly a different line). We want to conclude that x ₁* > x ₁ and x ₂* > x ₂. This will allow us to conclude that e _i (x _i *) > e _i (x _i ) and hence that e _w (x*) > e _w (x), thereby proving the monotonicity of e _w .

Proof

Assume to the contrary that x ₁* ≤ x ₁. Then c ₂(x ₂*) = c ₁(x ₁*) ≤ c ₁(x ₁) = c ₂(x ₂), so that x ₂* ≤ x ₂. This in turn implies that x* = wx ₁* + (1 – w)x ₂* ≤ wx ₁ + (1 – w)x ₂ = x, or x* ≤ x, contradicting the assumption that x* > x. A similar argument shows that x ₂* > x ₂. Hence, e _i (x _i *) > e _i (x _i ) and e _w (x*) > e _w (x), thereby proving the monotonicity of e _w .