How Accurate Can Instrumental Variable Models Become?

Abstract

This chapter presents and discusses various aspects of what theory predicts in terms of accuracy of instrumental variable estimates. A general derivation of the covariance matrix of the parameter estimates is presented. This matrix is influenced by a number of user choices in the identification method, and it is further discussed how these user choices can be made in order to make the covariance matrix as small as possible in a well-defined sense. The chapter includes also a comparison with the prediction error method, and a discussion of in what situations an optimal instrumental variable method can be statistically efficient.

Appendix: Proofs and Derivations

1.1 A.1 Proof of Lemma 1.1

Using the assumption on joint Gaussian distribution we can apply the general rule for product of Gaussian variables

$$\mathsf {E}\{ x_1 x_2 x_3 x_4 \}=\mathsf {E}\{ x_1 x_2 \} \mathsf {E}\{ x_3 x_4 \}+ \mathsf {E}\{ x_1 x_3 \} \mathsf {E}\{x_2 x_4 \}+ \mathsf {E}\{ x_1 x_4 \} \mathsf {E}\{ x_2 x_3 \}. $$

(1.82)

Using the result (1.82) in (1.34) leads to

(1.83)

Recall that the covariance function r _v (τ) decays exponentially with τ. Therefore we can write

$$\Biggl\|\lim_{N \rightarrow\infty}\frac{1}{N}\sum_{\tau=-N}^N| \tau| R_{z}(\tau) r_v(\tau) \Biggr\|\le \lim_{N \rightarrow\infty}\frac{1}{N}\sum_{\tau=0}^N2 \tau C \alpha^{\tau} = 0 $$

(1.84)

for some |α|<1. Using this result, we get

$$S =\sum_{\tau= -\infty}^{\infty}\bigl[ R_{z}(\tau) r_v(\tau) +r_{z v}(\tau) r_{z v}^T(-\tau)\bigr]. $$

(1.85)

Now use the conventions

$$h_0 = 1,\qquad h_i = 0\quad\mbox{for } i < 0 $$

(1.86)

The assumption (1.36) now implies that

$$r_{z v}(\tau) r_{z v}^T(-\tau) = 0\quad\forall\tau $$

(1.87)

as at least one of the factors is zero. Therefore

(1.88)

which is (1.37).

1.2 A.2 Proof of Lemma 1.2

Using the definitions, one find that an arbitrary element of the matrix S is given by

(1.89)

To proceed we need to evaluate the expectation of products of the white noise sequence. Set m _e =E{e ⁴(t)}. As e(t) has zero mean, the expected value of a product of four factors of the noise is nonzero if either the time arguments are pairwise equal, or all are equal. This principle gives

(1.90)

Inserting this into (1.89) leads to

(1.91)

Comparing the calculations in the proof of Lemma 1.3.1, we find that the first term in (1.91) is precisely

$$T_1 =\mathsf {E}\bigl\{ [ H(q^{-1})z(t) ][ H(q^{-1})z(t) ]^T \bigr\}. $$

(1.92)

Further, the second term turns out to be

$$T_2 =\sum_{\tau= - \infty} ^{\infty}r_{z v} (\tau) r_{zv}^T (-\tau) $$

(1.93)

which vanishes due to the assumption (1.44).

The last term can be written as

(1.94)

However, we know that

(1.95)

This implies that

$$T_3 = 0 $$

(1.96)

and the lemma is proven.

1.3 A.3 Proof of Lemma 1.3

We first write from (1.47)

$$P(S^{-1}) = \lambda^2 \bigl( R^T S^{-1} R \bigr)^{-1}. $$

(1.97)

The inequality (1.54) can equivalently be written as

$$\lambda^2 \bigl( R^T Q R \bigr)^{-1}R^T Q S Q R\bigl( R^T Q R \bigr)^{-1}\ge \lambda^2 \bigl( R^T S^{-1} R \bigr)^{-1} $$

(1.98)

which can be rewritten as

$$\bigl( R^T Q R \bigr)\bigl( R^T Q S Q R \bigr)^{-1}\bigl( R^T Q R \bigr)\le R^T S^{-1} R. $$

(1.99)

This in turn follows from the theory of partitioned matrices, cf. Lemma A.3 of [12], as

$$\left(\arraycolsep=5pt\begin{array}{@{}cc@{}}R^T S R & R^T Q R\\R^T Q R & R^T S Q Q R\end{array}\right)=\left(\begin{array}{c}R^T S^{-1} \\ R^T Q\end{array}\right)S\left(\begin{array}{cc}S^{-1} R & Q R\end{array}\right)\ge0. $$

(1.100)

1.4 A.4 Proof of Lemma 1.4

Using the theory of partitioned matrices, see for example Lemma A.2 in [12] and (1.56)

(1.101)

Equality in (1.101) applies if and only if

$$-S_{21} S^{-1} R + \tilde{R} = 0. $$

(1.102)

The condition (1.59) is equivalent to

$$R = S \alpha,\qquad \tilde{R} = S_{21} \alpha $$

(1.103)

for some matrix α. As S is nonsingular, this is in turn equivalent to α=S ⁻¹ R, and

$$\tilde{R} = S_{21} S^{-1} R $$

(1.104)

which is (1.102).

1.5 A.5 Answer to Exercise 1.3

Use the notation

$$r_k = \mathsf {E}\left\{ y(t+k) y(t) \right\}= \frac{\lambda^2}{1-a^2} (-a)^{|k|}. $$

(1.105)

Then

(1.106)

(1.107)

(1.108)

1.6 A.6 Proof of Lemma 1.5

Using the definition (1.50) of K(z) introduce the notations

(1.109)

(1.110)

Then it holds

(1.111)

Using (1.48) leads to

$$\lambda^2 R^T Q S Q R=\mathsf {E}\bigl\{ \alpha(t) \alpha^T (t) \bigr\}. $$

(1.112)

The stated inequality (1.62) then reads

(1.113)

Now, (1.113) is equivalent to

$$\bigl( \mathsf {E}\bigl\{ \beta(t) \alpha^T(t) \bigr\} \bigr)\bigl( \mathsf {E}\bigl\{ \alpha(t) \alpha^T(t) \bigr\} \bigr)^{-1}\bigl( \mathsf {E}\bigl\{ \alpha(t) \beta^T(t) \bigr\} \bigr)\le \bigl( \mathsf {E}\bigl\{ \beta(t) \beta^T(t) \bigr\} \bigr), $$

(1.114)

which follows from the theory of partitioned matrices, cf. Lemma A.4 in [12], as

$$\mathsf {E}\left\{\left(\arraycolsep=5pt\begin{array}{@{}cc@{}}\beta(t) \beta^T(t) & \beta(t) \alpha^T(t)\\ \alpha(t) \beta^T(y) & \alpha(t) \alpha^T(t)\end{array}\right) \right\}=\mathsf {E}\left\{\left(\begin{array}{c} \beta(t) \\ \alpha(t)\end{array}\right)\left(\begin{array}{c@{\ }c} \beta^T(t) & \alpha^T (t)\end{array}\right)\right\}\ge0. $$

(1.115)

Further, we see that with the specific choice

$$z(t) = H^{-1}(q^{-1}) \varphi_0(t),\qquad F(q^{-1}) = H^{-1}(q^{-1}),\qquad Q = I $$

(1.116)

it holds that

$$R = \mathsf {E}\bigl\{[ H^{-1}(q^{-1}) \varphi_0(t) ][ H^{-1}(q^{-1}) \varphi_0^T(t) ]\bigr\}=\mathsf {E}\bigl\{ \beta(t) \beta^T(t) \bigr\}= S $$

(1.117)

from which the equality in (1.62) follows.