3. Reasons for a large interval $I_{STN}^{95}$

Paul Knottnerus

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In order to get more insight into the difference between $\mathrm{var}\left({\widehat{g}}_{OLP}\right)$ and $\mathrm{var}\left({\widehat{g}}_{STN}\right),$ we assume $n_{12}=n_{23}=n$ and $G,S_{xy}>0;$ hence, $\lambda =\mu =n_2/n.$ Then subtracting (2.4) from (2.2) yields

$$\begin{array}{ll}\mathrm{var}\left({\widehat{g}}_{STN}\right)-\mathrm{var}\left({\widehat{g}}_{OLP}\right)\hfill & \approx \frac{1}{{\overline{X}}^2}\left\{2G\left(\frac{1}{n_2}-\frac{\lambda }{n}\right)S_{xy}-\left(\frac{1}{n_2}-\frac{1}{n}\right)\left(S_y^2+G^2{S}_x^2\right)\right\}\hfill \\ \hfill & =\frac{1}{\lambda n{\overline{X}}^2}\left\{2G\left(1-{\lambda }^2\right)S_{xy}-\left(1-\lambda \right)\left(S_y^2+G^2{S}_x^2\right)\right\}(3.1)\hfill \\ \hfill & =\frac{1-\lambda }{\lambda n{\overline{X}}^2}\left(2G\lambda S_{xy}-S_{y-Gx}^2\right).\hfill \end{array}$$

In other words, $\mathrm{var}\left({\widehat{g}}_{OLP}\right)$ is smaller than $\mathrm{var}\left({\widehat{g}}_{STN}\right)$ when $\lambda >S_{y-Gx}^2/2G{S}_{xy}$ provided $S_{xy}>0.$ Assuming $S_y^2=S_x^2,$ Qualité and Tillé (2008) derive a similar result for the parameter of absolute change when $\lambda >\left(1-{\rho }_{xy}\right)/{\rho }_{xy}.$ An anonymous referee pointed out that $\lambda <\left(1-{\rho }_{xy}\right)/{\rho }_{xy}$ is a sufficient condition for $\mathrm{var}\left({\widehat{g}}_{OLP}\right)>\mathrm{var}\left({\widehat{g}}_{STN}\right)$ because (3.1) can be rewritten as

$$\frac{\left(1-\lambda \right)G{S}_x{S}_y}{\lambda n{\overline{X}}^2}\left(2\lambda {\rho }_{xy}+2{\rho }_{xy}-\frac{S_y^2+G^2{S}_x^2}{G{S}_x{S}_y}\right)\le \frac{\left(1-\lambda \right)G{S}_x{S}_y}{\lambda n{\overline{X}}^2}\left(2\lambda {\rho }_{xy}+2{\rho }_{xy}-2\right)<0,$$

provided that $\lambda <\left(1-{\rho }_{xy}\right)/{\rho }_{xy}.$

If $N$ is sufficiently large, a weaker condition can be derived under some standard model assumptions. Suppose that the data satisfy the model $Y_i=B{X}_i+u_i$ with $E(u_i)=0,$ $E(u_i^2)={\sigma }_{}^2{X}_i^{\delta }$ and $E(u_i{u}_j)=0$ $(i\ne j);$ recall $X_i$ is not random in this context. Under this model, we make the (weak) assumptions (i) $G=S_{yx}/S_x^2$ and (ii) $S_{y-Gx}^2=S_y^2\left(1-{\rho }_{xy}^2\right).$ To justify these assumptions, recall from regression theory that $\widehat{B}=S_{yx}/S_x^2$ can be seen as the unbiased, consistent estimator for $B$ from an ordinary least squares (OLS) regression of $Y_i$ on $X_i$ and a constant $(i=1,\mathrm{...},N).$ Furthermore, the corresponding OLS estimator $\left(\overline{Y}-\widehat{B}\overline{X}\right)$ for the constant has zero expectation under the above model while its variance is of order $1/N.$ Hence, $\text{0}=\text{plim}\left(\overline{Y}-\widehat{B}\overline{X}\right)=\text{plim}\left\{\overline{X}\left(G-\widehat{B}\right)\right\}$ as $N\to \infty$ and provided $\overline{X}>c>0$ for all $N,$ we get the somewhat counterintuitive result $\text{plim}\left(G-\widehat{B}\right)=0.$ In fact, it can be shown that

$$G=\overline{Y}/\overline{X}=\widehat{B}\left[1+O_p\left(1/\sqrt{N}\right)\right]=\left(S_{yx}/S_x^2\right)\left[1+O_p\left(1/\sqrt{N}\right)\right]$$

as $N\to \infty .$ This justifies assumption (i); for further details, see the end of this section. Furthermore, $S_y^2\left(1-{\rho }_{xy}^2\right)$ can be seen as the (unexplained) variance of the residuals from the OLS regression. However, under the above model assumptions, these residuals are asymptotically equal to $Y_i-G{X}_i$ from which the approximate validity of (ii) follows. In addition, noting that $S_y^2{\rho }_{xy}^2$ is the so-called explained variance of the above OLS regression, it follows from assumption (i) that $S_y^2{\rho }_{xy}^2={\widehat{B}}^2{S}_x^2\approx G^2{S}_x^2.$ Combining this with assumptions (i) and (ii), we can rewrite (3.1) as

$$\begin{array}{cc}\mathrm{var}\left({\widehat{g}}_{STN}\right)-\mathrm{var}\left({\widehat{g}}_{OLP}\right)& \approx \frac{1-\lambda }{\lambda n{\overline{X}}^2}\left\{2G^2\lambda S_x^2-\left(1-{\rho }_{xy}^2\right)S_y^2\right\}\hfill \\ & \approx \frac{\left(1-\lambda \right)S_y^2}{\lambda n{\overline{X}}^2}\left(2\lambda {\rho }_{xy}^2-1+{\rho }_{xy}^2\right)(3.2)\hfill \\ & =\frac{\left(1-\lambda \right)S_y^2}{\lambda n{\overline{X}}^2}\left\{{\rho }_{xy}^2\left(1+2\lambda \right)-1\right\}.\hfill \end{array}$$

Hence, $\mathrm{var}\left({\widehat{g}}_{OLP}\right)$ is larger than $\mathrm{var}\left({\widehat{g}}_{STN}\right)$ when

$$\lambda <\left(1-{\rho }_{xy}^2\right)/2{\rho }_{xy}^2\left[>\left(1-{\rho }_{xy}\right)/{\rho }_{xy}\right].(3.3)$$

Thus for say ${\rho }_{xy}=0.9,$ $\mathrm{var}\left({\widehat{g}}_{OLP}\right)$ is under the above model for sufficiently large $N$ larger than $\mathrm{var}\left({\widehat{g}}_{STN}\right)$ when $\lambda <0.117,$ and for say ${\rho }_{xy}=0.75$ when $\lambda <\mathrm{0.389.}$ In addition, applying (3.2) to the data in Example 2.1 with $\lambda \approx 57/73=0.78$ and ${\rho }_{xy}=0.876$ yields as approximation for the difference between both variances 0.0017 which is not very different from the actual difference of 0.0016 (=0.00324-0.00166) in the example. For Example 2.2, taking $\lambda =54/70=0.77$ and ${\rho }_{xy}=0.970,$ applying (3.2) yields 0.00226 instead of 0.00212 (=0.00251-0.00039) in the example.

Under the above assumptions, it can also be shown that the ratio, say $Q,$ of $\mathrm{var}\left({\widehat{g}}_{OLP}\right)$ and $\mathrm{var}\left({\widehat{g}}_{STN}\right)$ can be approximated by

$$Q=\frac{\mathrm{var}\left({\widehat{g}}_{OLP}\right)}{\mathrm{var}\left({\widehat{g}}_{STN}\right)}\approx \left({\lambda }^{-1}-f\right){\left(1-f+2\left(1-\lambda \right)\frac{{\rho }_{xy}^2}{1-{\rho }_{xy}^2}\right)}^{-1},(3.4)$$

irrespective of the values of $S_y^2$ and $S_x^2;$ $f$ stands for $n/N.$ For a proof of (3.4), see Appendix A.1. From (3.4) it can be seen that $Q$ and $\mathrm{var}\left({\widehat{g}}_{OLP}\right)$ tend to zero as ${\rho }_{xy}^2$ tends to unity, provided $N$ is sufficiently large and $\lambda <1.$

It should be noted that in practice the correlations ${\rho }_{xy}$ often are rather high by the very nature of the data $(Y_i,X_i).$ That is, a large (small) enterprise in period $\left(t-12\right)$ is in most cases still large (small) after 12 months; Knottnerus and Van Delden (2012, page 47) found for various strata an overall mean correlation of 0.90 and a variance of 0.0074. So it appears that $\mathrm{var}\left({\widehat{g}}_{STN}\right)$ is more affected by a decrease of $\lambda$ than $\mathrm{var}\left({\widehat{g}}_{OLP}\right)$ unless $\lambda$ is extremely low because (i) $\mathrm{var}\left({\widehat{g}}_{OLP}\right)=\mathrm{var}\left({\widehat{g}}_{STN}\right)$ when $\lambda =1$ and (ii) $Q$ is large when ${\rho }_{xy}^2$ is large. For example, when ${\rho }_{xy}=0.9$ and $f=0.1$ a decrease of $\lambda$ from 0.9 to 0.5 leads to a decrease of $Q$ from 0.58 to 0.37; recall $Q=1$ when $\lambda =1.$ This emphasizes once more the importance of avoiding panel attrition when using estimator ${\widehat{g}}_{STN}$ while $N$ is large.

A natural question that remains to be answered is when is $N$ sufficiently large. To answer this question, consider the difference $\Delta \equiv \widehat{B}-G$ and its variance, say ${\sigma }_{\Delta }^2.$ The difference $\Delta$ can be written as

$$\begin{array}{cc}\Delta & =\frac{S_{xy}^{}}{S_x^2}-\frac{\overline{Y}}{\overline{X}}=\frac{1}{N-1}{\displaystyle \sum _{i\in U}\frac{X_i-\overline{X}}{S_x^2}}{Y}_i-\frac{1}{N}{\displaystyle \sum _{i\in U}\frac{Y_i}{\overline{X}}}\hfill \\ & \approx \frac{1}{N}{\displaystyle \sum _{i\in U}\left(\frac{X_i-\overline{X}}{S_x^2}-\frac{1}{\overline{X}}\right)}{Y}_i\hfill \\ & =\frac{1}{N}{\displaystyle \sum _{i\in U}{M}_i{U}_i}\left(M_i=\frac{X_i-\overline{X}}{S_x^2}-\frac{1}{\overline{X}}\right).\hfill \end{array}$$

In the second line we assumed $N>>1$ and in the last line we used the model assumption $Y_i=B{X}_i+U_i.$ Next, assuming $\mathrm{var}\left(U_i\right)={\sigma }^2{X}_i^{\delta },$ we get

$${\sigma }_{\Delta }^2\equiv \mathrm{var}\left(\widehat{B}-G\right)=\frac{{\sigma }^2}{N^2}{\displaystyle \sum _{i\in U}{M}_i^2{X}_i^{\delta }}\text{. }$$

This variance can be estimated by

$${\widehat{\sigma }}_{\Delta }^2=\frac{{\widehat{\sigma }}^2}{N{n}_2}{\displaystyle \sum _{i\in s_2}{\widehat{m}}_i^2{X}_i^{\widehat{\delta }}}\text{, }$$

where

$${\widehat{m}}_i=\frac{X_i-{\overline{x}}_2}{s_{x2}^2}-\frac{1}{{\overline{x}}_2},{\widehat{\sigma }}^2=\frac{1}{n_2-1}{\displaystyle \sum _{i\in s_2}{\left(Y_i-\frac{{\overline{y}}_2}{{\overline{x}}_2}{X}_i\right)}^2/X_i^{\widehat{\delta }}}$$

and $\widehat{\delta }$ is an estimate from the OLS regression

$$\ln {\left(Y_i-\frac{{\overline{y}}_2}{{\overline{x}}_2}{X}_i\right)}^2=\alpha +\delta \ln X_i+w_i\left(i=1,\mathrm{...},n_2\right);$$

units with $Y_i={\overline{y}}_2{X}_i/{\overline{x}}_2$ are omitted. Based on ${\widehat{\sigma }}_{\Delta }^2,$ one may call $N$ sufficiently large if the outcome of (3.1) will not severely be affected by replacing $G$ by $G+{\widehat{\sigma }}_{\Delta }.$ In addition, it should be borne in mind that relationships for very large $N$ are probably still a reasonably appropriate indication for what may occur when $N$ is not very large.

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