On intraclass structure estimation in the growth curve model

Abstract

In this paper the growth curve model, with the data correlated according to uniform structure, is considered. It represents a useful statistical model for a variety of areas. Our aim is to present various estimators of unknown variance parameters and compare their statistical properties. In the first part the review of known results for different estimators of $\rho$ and $\sigma ^2$ and their properties is given. The aim is to compare and order these estimators based on biasedness and mean square error.

Appendix

Proof of Lemma 1

• 1. $E({\hat{\rho }}_{M})-\rho =\frac{-(1+(p-1)\rho )(1-\rho )}{(p-1)[np-r(X)((p-r(Z))\rho +r(Z))]^{3}} \times$$\times [[(p-r(Z))r(X)n^{2}p^{2}-2(p-r(Z))r(X)^{2}npr(Z)+r(X)^{3}r(Z)^{2}(p-r(Z))]+$$+[-2(p-r(Z))^{2}r(X)^{2}np+2(p-r(Z))^{2}r(Z)r(X)^{3}+2n^{2}p^{2}(p-1)-$$-2p^{2}nr(X)(p+r(Z)-2)+2p^{2}r(X)^{2}(r(Z)-1)]\rho +[(p-r(Z))^{3}r(X)^{3}]\rho ^{2}]$ The formula is negative in case when $\rho >0.$ Thus, let us look at the case when $\rho <0.$ It can be shown that $\frac{-(1+(p-1)\rho )(1-\rho )}{(p-1)[np-r(X)((p-r(Z))\rho +r(Z))]^{3}}$ is always negative. This formula is multiplied by convex function. This function has a minimum value $c-\frac{b^{2}}{4a}$ at point $-\frac{b}{2a},$ where $a=(p-r(Z))^{3}r(X)^{3}, b=-2(p-r(Z))^{2}r(X)^{2}np+2(p-r(Z))^{2}r(Z)r(X)^{3}+2n^{2}p^{2}(p-1)-2p^{2}nr(X)(p+r(Z)-2)+2p^{2}r(X)^{2}(r(Z)-1)$ and $c=(p-r(Z))r(X)n^{2}p^{2}-2(p-r(Z))r(X)^{2}npr(Z)+r(X)^{3}r(Z)^{2}(p-r(Z)).$ If $n>\frac{r(X)}{p(p-1)}[p^{2}(r(X)-1)-p(2r(X)r(Z)+1)+r(X)r(Z)^{2}]$ the x value of vertex is not in feasible region, thus we are interested only in the opposite case. In such a case the formula can be positive for some choices of model constants.

• 2. $E({\hat{\rho }}_{MU})-\rho =\frac{2(1+(p-1)\rho )(1-\rho )((p-1)(r(X)r(Z)-np)\rho +r(X)(p-r(Z)))}{p^2(n^2(p-1)-n(p+r(Z)-2)r(X)+(r(Z)-1)r(X)^2)}$ It can be shown that the denominator is always positive. The numerator has zero points $-\frac{1}{p-1}$, 1 and v,  where $v=-\frac{r(X)(p-r(Z))}{(p-1)(np-r(X)r(Z))}\in \left( -\frac{1}{p-1};0\right) .$ Thus, the bias is positive for $\rho \in \left( -\frac{1}{p-1}, v \right)$ and negative for $\rho \in \left( v, 1\right) .$

• 3. $E({\hat{\rho }}_{O})-\rho =\frac{2(1-\rho )(1+(p-1)\rho )}{p^{2}(p-1)(n-r(X))} ((p(p-1)+r(X)(p-r(Z))(p-r(Z)-2))\rho -r(X)(p-r(Z))(p-r(Z)-2))$ Simple computation shows that formula for bias has zero points $-\frac{1}{p-1}$, 1 and w,  where $w=\frac{r(X)(p-r(Z))(p-r(Z)-2)}{p(p-1)+r(X)(p-r(Z))(p-r(Z)+2)}\in (0;1).$ The bias is positive for $\rho \in \left( -\frac{1}{p-1}, w \right)$ and negative for $\rho \in \left( w, 1\right) .$

• 4. $E({\hat{\rho }}_{U})-\rho = -\frac{2\rho (1-\rho )(1+(p-1)\rho )}{(n-r(X))p}$ Zero points for this formula are $-\frac{1}{p-1}$, 0 and 1. So we can see that under assumption $n>r(X)$ and $\rho >-\frac{1}{p-1}$ the bias is positive only for $\rho \in (-\frac{1}{p-1}, 0)$ and it is negative for $\rho \in (0,1).$

Proof of Lemma 2

$$\begin{aligned}&E({\hat{\rho }}_{U})-E({\hat{\rho }}_{O})\\&\quad =-\frac{2(1-\rho )\rho (1+(p-1)\rho )}{p(n-r(X)))}\\&\qquad + \frac{2\rho (1-\rho )(1+(p-1)\rho )}{p(n-r(X))}\left[ 1-\frac{r(X)(1-\rho )(p-r(Z))(p-r(Z)-2)}{\rho p(p-1)}\right] \\&\quad =-\frac{2(1-\rho )^{2}(1+(p-1)\rho )r(X)(p-r(Z))(p-r(Z)-2)}{p^{2}(n-r(X))(p-1)}, \end{aligned}$$

this is always negative if $n>r(X),$$p>r(Z)+2$ and $\rho >-\frac{1}{p-1}.$

$$\begin{aligned}&E({\hat{\rho }}_{MU})-E({\hat{\rho }}_{U})\\&\quad =\frac{2(1-\rho )\rho (1+(p-1)\rho )}{p(n-r(X)))}\\&\quad \quad -\frac{2(1+(p-1)\rho )(1-\rho )[np(p-1)\rho -r(X)(pr(Z)\rho -p-r(Z)\rho +r(Z))]}{p^{2}[n^{2}(p-1)-n(p+r(Z)-2)+(r(Z)-1)r(X)^{2}]}\\&\quad =-\frac{2(1-\rho )^{2}(1+(p-1)\rho )(n-r(X))r(X)(p-r(Z))}{p^{2}(n-r(X))(n^{2}(p-1)-n(p+r(Z)-2)r(X)+(r(Z)-1)r(X)^{2})}. \end{aligned}$$

This formula is always negative if $n>r(X),$$p>r(Z)+2$ and $\rho >-\frac{1}{p-1}.$ Simple computation shows that $[n^{2}(p-1)-n(p+r(Z)-2)+(r(Z)-1)r(X)^{2}]$ is always positive. $\square$

Proof of Lemma 3

$$\begin{aligned} \frac{\text {MSE } ({\hat{\sigma }}^{2}_{U})}{\text {MSE } ({\hat{\sigma }}^{2}_{O})}-1=-\frac{r(X)(1-\rho ^{2})(p-r(Z))(p-r(Z)-2)}{np(1+(p-1)\rho ^{2})+r(X)(p-r(Z))(p-r(Z)-2)} \end{aligned}$$

This function is increasing, always negative under the assumption $p>r(Z)+2.$ It assumes maximum for $\rho =1$ and this maximum is zero. Thus, the U-estimator is better than O-estimator.

$$\begin{aligned} \frac{\text {MSE } ({\hat{\sigma }}^{2}_{U})}{\text {MSE } ({\hat{\sigma }}^{2}_{MU})}-1=\frac{r(X)(p-1)(p-r(Z))(1+\rho )^{2}}{\rho ^{2}(p-1)^{2}(np-r(X)r(Z))-2r(X)(p-1)(p-r(Z))\rho +c}, \end{aligned}$$

where $c=np(p-1)-r(X)r(Z)-r(X)p^{2}+r(X)2p.$ It is obvious that the numerator is always positive. Discriminant of denominator is always negative and for the leading coefficient it holds: $(p-1)^{2}(np-r(X)r(Z))>0,$ so that the denominator is a positive convex function for $\rho \in \left( -\frac{1}{p-1}, 1\right) .$ The whole fraction is therefore always positive. Thus, MU-estimator is better then U-estimator. $\square$