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# Elementary analysis of an economic model with errors in some variables

Paul van Der Laan

► To cite this version:

Paul van Der Laan. Elementary analysis of an economic model with errors in some variables. Annales de l'ISUP, 1970, XIX (3), pp.179-190. hal-04083406

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Distributed under a Creative Commons Attribution 4.0 International License
# ELEMENTARY ANALYSIS OF AN ECONOMIC MODEL WITH ERRORS IN SOME VARIABLES

## P. VAN DER LAAN *

### 1. The Model.

The model with which we shall deal in this paper is the following. Given are three variables, $\xi$, $\eta$, and $z$, which are considered to be linearly related by the expression

$$ \eta = \alpha + \beta \xi + \gamma z $$

in which $\alpha$, $\beta$ and $\gamma$ are unknown parameters.

The variable $z$ can be determined exactly, but the variables $\xi$ and $\eta$ are observed with errors. This implies that instead of $\xi$ and $\eta$ we observe the random variables

$$ x = \xi + u $$

and

$$ y = \eta + y, $$

in which it is assumed that the random variable $u$ has a normal distribution, with expectation 0 and variance $\sigma^2$, and the random variable $v$ has a normal distribution, with expectation 0 and variance $\tau^2$. It is furthermore assumed that the random variables $u$ and $v$ do not depend on each other.

Now the problem is to estimate the parameters $\alpha$, $\beta$ and $\gamma$ on the basis of $n$ observations:

$(x_1, y_1, z_1), (x_2, y_2, z_2), \dots, (x_n, y_n, z_n)$, on the assumption that both $y_i = \eta_i + v_i$ ($i = 1, 2, \dots, n$) and $x_i = \xi_i + u_i$ ($i = 1, 2, \dots, n$) are mutually independent.

The purpose is to determine the functional relation $\eta = \alpha + \beta \xi + \gamma z$.

<footnote>* PHILIPS Information Systems and Automation (Research Dept.), Eindhoven, Netherlands.</footnote>
For the sake of completeness it should be noted that, if it is desired to find a prediction-relation, in order later with observed values of x and z to predict the corresponding values of η, one can straight away use the method of least squares, that is one can choose α, β and γ in such a way that

$$ \sum_{i=1}^{n} (y_i - \alpha - \beta x_i - \gamma z_i)^2 $$

is minimal.

The problem dealt with in the present paper (the estimation of the functional relation η = α + βξ + γz, where η and ξ are observed with errors) occurs in the investigation of the functional approach, or reality content, of the following economic model:

$$ \eta = \alpha + \beta \xi + \gamma z $$

in which

* η = market share of a certain product

* ξ = degree of penetration of this product

* z = price of this product

and in which the n observations (with estimates of ξ and η, based on enquiries) are obtained in n regions. The purpose of this paper is to give an elementary analysis of the estimation problem for this functional relation.

## 2. Method of Estimation.

For the estimation of the parameters α, β and γ we can make use of the maximum likelihood method. This implies that for the parameters α, β, γ, σ², and τ² as well as for the unknown ξᵢ (i = 1, 2, ..., n) we select those values which maximize the likelihood function

$$ L = (2\pi)^{-\frac{n}{2}} (\tau^2)^{-\frac{n}{2}} \exp \left\{ -\frac{1}{2\tau^2} \sum_{i=1}^{n} (y_i - \alpha - \beta\xi_i - \gamma z_i)^2 \right\} $$

$$ . (2\pi)^{-\frac{n}{2}} (\sigma^2)^{-\frac{n}{2}} \exp \left\{ -\frac{1}{2\sigma^2} \sum_{i=1}^{n} (x_i - \xi_i)^2 \right\} $$

Now it turns out that, if both σ² and τ² are unknown, we obtain
the following relation between the maximum likelihood estimators of $\beta$, $\sigma^2$, and $\tau^2$:

$$ \hat{\beta}^2 \quad \hat{\tau}^2 = \hat{\sigma}^2 $$

This relation shows the impossibility of finding consistent estimators for the three parameters $\beta$, $\sigma^2$ and $\tau^2$ simultaneously. That is why we assume the ratio of the variances $\sigma^2$ and $\tau^2$ to be known. Hence:

$$ \frac{\sigma^2}{\tau^2} = \lambda $$

is known ($\lambda > 0$).

How we can write down the following expression for $\ln L$, by making use of the equality $\sigma^2 = \lambda \tau^2$:

$$ \ln L = \text{constant} - n \ln \tau^2 - \frac{1}{2 \tau^2} \sum_{i=1}^{n} (y_i - \alpha - \beta \xi_i - \gamma z_i)^2 - \frac{1}{2 \lambda \tau^2} \sum_{i=1}^{n} (x_i - \xi_i)^2 $$

In order to find the values of $\alpha$, $\beta$, $\gamma$ and $\tau^2$ which maximize $\ln L$, we differentiate $\ln L$ with respect to these four variables as well as to the unknown $\xi_i$ ($i = 1, 2, ..., n$). This gives us the following set of equations*

$$ \frac{\partial \ln L}{\partial \alpha} = 0 \Rightarrow \sum_{i=1}^{n} (y_i - \hat{\alpha} - \hat{\beta} \hat{\xi}_i - \hat{\gamma} z_i) = 0 $$

$$ \text{or:} \quad \bar{y} - \hat{\alpha} - \hat{\beta} \hat{\xi} - \hat{\gamma} \bar{z} = 0 \quad (1) $$

$$ \frac{\partial \ln L}{\partial \beta} = 0 \Rightarrow \sum_{i=1}^{n} \hat{\xi}_i (y_i - \alpha - \beta \hat{\xi}_i - \gamma \hat{z}_i) = 0 $$

(2)

$$ \frac{\partial \ln L}{\partial \gamma} = 0 \Rightarrow \sum_{i=1}^{n} z_i (y_i - \alpha - \beta \hat{\xi}_i - \gamma z_i) = 0 \quad (3) $$

(3)

<footnote>* We introduce the following notation: $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$, $m_{xy} = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})(y_i - \bar{y})$ and in a similar way we define: $\bar{y}, \bar{z}, ...$ and $m_{xz}, m_{\xi\xi}, ...$</footnote>
$$ \frac{\partial \ln L}{\partial \xi_i} = 0 \Rightarrow \lambda \hat{\beta} (y_i - \hat{\alpha} - \hat{\beta} \hat{\xi}_i - \hat{\gamma} z_i) + (x_i - \xi_i) = 0 \quad (4) $$
$$ (i = 1, 2, \dots, n) $$

$$ \frac{\partial \ln L}{\partial \tau^2} = 0 \Rightarrow \rho \tau^2 = \frac{1}{2n} \left\{ \sum_{i=1}^{n} (y_i - \hat{\alpha} - \hat{\beta} \hat{\xi}_i - \hat{\gamma} z_i)^2 + \right. \\ \left. + \frac{1}{\lambda} \sum_{i=1}^{n} (x_i - \hat{\xi}_i)^2 \right\} \quad (5) $$

Summation over i of (4) and equation (1) implies

$$ \hat{\xi} = \bar{x} $$

so that equation (1) leads to the following expression of $\hat{\alpha}$:

$$ \hat{\alpha} = \bar{y} - \beta \bar{x} - \hat{\gamma} \bar{z} \quad (6) $$

From equation (2) we obtain, by the use of equation (1):

$$ \sum_{i=1}^{n} (\hat{\xi}_i - \bar{\hat{\xi}}) (y_i - \hat{\alpha} - \hat{\beta} \hat{\xi}_i - \hat{\gamma} z_i) = 0 $$

After substituting equation (6) in this we can write the equality as follows (for notation see the footnote on page 4).

$$ m_{\xi y} - \hat{\beta} m_{\xi \xi} - \hat{\gamma} m_{\xi z} = 0 \quad (7) $$

Similarly we obtain from eq. (3):

$$ m_{yz} - \hat{\beta} m_{\xi z} - \hat{\gamma} m_{zz} = 0 \quad (8) $$

Obviously we must first eliminate the $\hat{\xi}_i$ terms. Here it should be noted that, in consequence of eq. (4):

$$ \lambda \hat{\beta} \{ (y_i - \bar{y}) - \hat{\beta} (\hat{\xi}_i - \bar{\hat{\xi}}) - \hat{\gamma} (z_i - \bar{z}) \} + (x_i - \bar{x}) \cdot (\hat{\xi}_i - \bar{\hat{\xi}}) = 0 $$

and therefore
$$ \hat{\xi}_i - \bar{\hat{\xi}} = \frac{1}{1 + \lambda \hat{\beta}^2} \{ \lambda \hat{\beta}^2 (y_i - \bar{y}) - \lambda \hat{\beta} \hat{\gamma} (z_i - \bar{z}) + (x_i - \bar{x}) \} $$

By means of this expression for $\hat{\xi}_i - \bar{\hat{\xi}}$ it is a simple matter to express $m_{\xi\xi}$, $m_{\xi y}$, and $m_{\xi z}$ in the parameters $\hat{\alpha}$, $\hat{\beta}$ and $\hat{\gamma}$, and in the observations $x_i$, $y_i$, and $z_i$ ($i = 1, 2, ..., n$). The following equations are then found:

$$ m_{\xi\xi} = \frac{1}{(1 + \lambda \hat{\beta}^2)^2} \left( \lambda^2 \hat{\beta}^2 m_{yy} + \lambda^2 \hat{\beta}^2 \hat{\gamma}^2 m_{zz} + m_{xx} - 2 \lambda^2 \hat{\beta}^2 \hat{\gamma} m_{yz} + \right. \\ \left. \qquad + 2 \lambda \hat{\beta} m_{xy} - 2 \lambda \hat{\beta} \hat{\gamma} m_{xz} \right) \quad (9) $$

$$ m_{\xi y} = \frac{1}{1 + \lambda \hat{\beta}^2} \left( \lambda \hat{\beta} m_{yy} - \lambda \hat{\beta} \hat{\gamma} m_{yz} + m_{xy} \right) \quad (10) $$

$$ m_{\xi z} = \frac{1}{1 + \lambda \hat{\beta}^2} \left( \lambda \hat{\beta} m_{yz} - \lambda \hat{\beta} \hat{\gamma} m_{zz} + m_{xz} \right) \quad (11) $$

Eq. (8) leads to

$$ m_{yz} - \frac{\hat{\beta}}{1 + \lambda \hat{\beta}^2} \left( \lambda \hat{\beta} m_{yz} - \lambda \hat{\beta} \hat{\gamma} m_{zz} + m_{xz} \right) - \hat{\gamma} m_{zz} = 0 $$

so that we have

$$ \hat{\gamma} = \frac{m_{yz}}{m_{zz}} - \frac{\hat{\beta} m_{xz}}{m_{zz}} \quad (12) $$

We can now substitute this expression for $\hat{\gamma}$ in eqs. (9), (10), and (11). The result is then

$$ m_{\xi\xi} = \frac{1}{(1 + \lambda \hat{\beta}^2)^2} \left\{ \begin{aligned} & \lambda^2 \frac{m_{xz}^2}{m_{zz}} \hat{\beta}^4 + (\lambda^2 m_{yy} - \lambda^2 \frac{m_{yz}^2}{m_{zz}}) + \\ & + 2 \lambda \frac{m_{xz}^2}{m_{zz}} \hat{\beta}^2 + (2 \lambda m_{xy} - 2 \lambda \frac{m_{xz} m_{yz}}{m_{zz}}) \hat{\beta} + m_{xx} \end{aligned} \right\} \quad (13) $$
$$ m_{\xi y} = \frac{1}{1 + \lambda \hat{\beta}^2} \left\{ \lambda \frac{m_{xz} m_{yz}}{m_{zz}} \hat{\beta}^2 + \left( \lambda m_{yy} - \lambda \frac{m_{yz}^2}{m_{zz}} \right) \hat{\beta} + m_{xy} \right\} \quad (14) $$

$$ m_{\xi z} = m_{xz} \qquad (15) $$

If these expressions are inserted in eq. (7) we finally obtain, with the help of eq. (12), the requisite equation in $\hat{\beta}$. Cursory examination of this equation in $\hat{\beta}$ makes it seem one of the fifth degree. The coefficients corresponding to $\beta^5, \beta^4$, and $\beta^3$, however turn out to be zero. The quadratic equation in $\hat{\beta}$ finally obtained is

$$ c_2 \hat{\beta}^2 + c_1 \hat{\beta} + c_0 = 0 \qquad (16) $$

with

$$ \left. \begin{aligned} c_o &= m_{xy} - \frac{m_{xz} m_{yz}}{m_{zz}} \\ c_1 &= \lambda (m_{yy} - \frac{m_{yz}^2}{m_{zz}}) - (m_{xx} - \frac{m_{xz}^2}{m_{zz}}) \\ c_2 &= -\lambda (m_{xy} - \frac{m_{xz} m_{yz}}{m_{zz}}) = -\lambda c_o \end{aligned} \right\} \quad (17) $$

For the possible roots of this quadratic equation we find:

$$ \begin{aligned} \hat{\beta}_{1,2} &= \frac{-c_1 \pm \sqrt{c_1^2 - 4 c_o c_2}}{2 c_2} \\ &= \frac{c_1 \mp \sqrt{c_1^2 + 4 \lambda c_o^2}}{2 \lambda c_o} \end{aligned} \qquad (18) $$

In appendix 1 it will be proved that $\hat{\beta}^2 = \frac{c_1 + \sqrt{c_1^2 + 4 \lambda c_o^2}}{2 \lambda c_o}$ is the
correct root which maximizes the likelihood function. It is easy to see that in the case of

$$m_{xy} - \frac{m_{xz} m_{yz}}{m_{zz}} > 0$$

the estimator $\hat{\beta}_2$ is always positive and in the case of

$$m_{xy} - \frac{m_{xz} m_{yz}}{m_{zz}} < 0$$

the estimator $\hat{\beta}_2$ is always negative. Notice that in the normal linear regression model (in which $\xi$ is then observed also without error) the magnitude

$$m_{xy} - \frac{m_{xz} m_{yz}}{m_{zz}}$$

is simply, except for a positive constant $\frac{1}{n}$, the inner product of y and the orthogonal complement of x with respect to z (after making x,y and z orthogonal to the space of levels). Thus it seems also intuitively right to choose $\hat{\beta}_2$ as the correct estimator of $\beta$.

### 3. Results

The maximum likelihood estimators of the parameters $\alpha, \beta$, and $\gamma$ are:

$$\hat{\beta} = \frac{c_1 + \sqrt{c_1^2 + 4 \lambda c_o^2}}{2 \lambda c_o}$$

(where $c_o$ and $c_1$ are defined by (17))

$$\hat{\gamma} = \frac{m_{yz}}{m_{zz}} - \hat{\beta} \frac{m_{xz}}{m_{zz}} \quad (19)$$

$$\text{and} \quad \hat{\alpha} = \bar{y} - \hat{\beta} \bar{x} - \hat{\gamma} \bar{z}$$
Note:

i) About the consistency of the estimators thus obtained and about a consistent estimator of $\tau^2$ a few remarks will be found in appendix 2.

ii) With $\lambda \to 0$, eq. (16) leads to the following equation for $\hat{\beta}$:

$$ - (m_{xx} - \frac{m_{xz}^2}{m_{zz}}) \hat{\beta} + m_{xy} - \frac{m_{xz} m_{yz}}{m_{zz}} = 0 $$

$$ \Rightarrow \hat{\beta} = \frac{\frac{m_{xy} - \frac{m_{xz} m_{yz}}{m_{zz}}}{m_{xx} - \frac{m_{xz}^2}{m_{zz}}}} $$

which in a normal regression model is the regression coefficient of $y - \bar{y}$ in that part of $x - \bar{x}$ that is orthogonal to $z - \bar{z}$.

iii) With $\lambda \to \infty$, eq. (16) leads to the following equation for $\hat{\beta}$:

$$ - (m_{xy} - \frac{m_{xz} m_{yz}}{m_{zz}}) \hat{\beta} + m_{yy} - \frac{m_{yz}^2}{m_{zz}} = 0 $$

$$ \Rightarrow \hat{\beta} = \frac{\frac{m_{yy} - \frac{m_{yz}^2}{m_{zz}}}{m_{xy} - \frac{m_{xz} m_{yz}}{m_{zz}}}} $$

which with normal regression analysis is the reciprocal of the regression coefficient of $x - \bar{x}$ on that part of $y - \bar{y}$ that is orthogonal to $z - \bar{z}$ in the case of regression of $x$ to $y$ and $z$.

## APPENDIX 1. Selection of the Estimator for $\beta$

Maximization of the likelihood function with respect to $\alpha, \beta$, and
$\gamma$ is in fact no more than the minimizing, with respect to $\alpha$, $\beta$, and $\gamma$, of

$$ S = \lambda \sum_{i=1}^{n} (y_i - \alpha - \beta \hat{\xi}_i - \gamma z_i)^2 + \sum_{i=1}^{n} (x_i - \hat{\xi}_i)^2 $$

After substitution of $\hat{\alpha}$, $\hat{\beta}$, and $\hat{\gamma}$ in $S$ we get, using eq. (4):

$$ 
\begin{aligned}
S &= \lambda (1 + \lambda \hat{\beta}^2) \sum_{i=1}^{n} (y_i - \hat{\alpha} - \hat{\beta} \hat{\xi}_i - \hat{\gamma} z_i)^2 \\
&= \lambda (1 + \lambda \hat{\beta}^2) \sum_{i=1}^{n} \{(y_i - \bar{y}) - \hat{\beta} (\hat{\xi}_i - \hat{\xi}) - \hat{\gamma} (z_i - \bar{z})\}^2 \\
&= \lambda (1 + \lambda \hat{\beta}^2) (m_{yy} + \hat{\beta}^2 m_{\xi\xi} + \hat{\gamma}^2 m_{zz} - 2 \hat{\beta} m_{\xi y} - 2 \hat{\gamma} m_{yz} + \\
& \qquad \qquad \qquad \qquad + 2 \hat{\beta} \hat{\gamma} m_{\xi z})
\end{aligned}
$$

From this we can eliminate $m_{\xi\xi}$ and $m_{\xi z}$ by means of eq. (7) and (8).
The result is

$$ S = \lambda (1 + \lambda \hat{\beta}^2) (m_{yy} - \hat{\beta} m_{\xi y} - \hat{\gamma} m_{yz}) $$

By means of eq. (14) we now find

$$ 
\begin{aligned}
S &= \lambda (1 + \lambda \hat{\beta}^2) \left[ m_{yy} - \frac{\hat{\beta}}{1 + \lambda \hat{\beta}^2} \right] \left\{ \lambda \frac{m_{xz} m_{yz}}{m_{zz}} \hat{\beta}^2 + \right. \\
& \qquad \qquad \qquad \qquad \left. + \lambda \left(m_{yy} - \frac{m_{yz}^2}{m_{zz}}\right) \hat{\beta} + \lambda \frac{m_{xy}^2}{m_{zz}} \hat{\beta} \right\} \\
& \qquad \qquad \qquad \qquad - \frac{m_{yz}^2}{m_{zz}} + \frac{m_{xz} m_{yz}}{m_{zz}} \hat{\beta} \\
&= \lambda \left\{ \left(m_{yy} - \frac{m_{yz}^2}{m_{zz}}\right) \hat{\beta} - \frac{m_{xz} m_{yz}}{m_{zz}} \hat{\beta} \right\}
\end{aligned}
$$

Noting that $\lambda$ is positive we therefore find that, in order to minimize $S$, we must take $\hat{\beta}_2$ as the estimator of $\beta$.
APPENDIX 2. Consistency of the Maximum Likelihood Estimators

We use the property that sample variances and covariances converge in probability to their expectations (indicated by $\rightarrow_p$). We assume that the requisite conditions are met in order that in respect of the variables $\xi$ and $z$ we can in limit ($n \rightarrow \infty$) speak of their means and about the variance $S_{\xi}^2$ of the non-observable $\xi$, the variance $S_z^2$ of the observable $z$, and their covariant $C_{\xi,z}$. We now can readily understand the following relations:

$$m_{xx} \rightarrow_p S_{\xi}^2 + \sigma^2 = S_{\xi}^2 + \lambda \tau^2$$

$$m_{yy} \rightarrow_p \beta^2 S_{\xi}^2 + \gamma^2 S_z^2 + 2 \beta \gamma C_{\xi,z} + \tau^2$$

$$m_{zz} \rightarrow_p S_z^2$$

$$m_{xy} \rightarrow_p \beta S_{\xi}^2 + \gamma C_{\xi,z}$$

$$m_{yz} \rightarrow_p \beta C_{\xi,z} + \gamma S_z^2$$

$$m_{xz} \rightarrow_p C_{\xi,z}$$

We have the maximum likelihood estimator of $\beta$

$$\hat{\beta} = \frac{-c_1 - \sqrt{c_1^2 - 4 c_0 c_2}}{2 c_2}$$

After a little calculation it will be readily seen that

$$ \begin{aligned} c_1^2 - 4 c_0 c_2 \rightarrow_p & \lambda^2 (\beta^2 S_{\xi}^2 - \beta^2 \frac{C_{\xi,z}^2}{S_z^2}) + 2 \lambda (S_{\xi}^2 - \frac{C_{\xi,z}^2}{S_z^2}) + \\ & + (\beta^2 S_{\xi}^2 - \beta^2 \frac{C_{\xi,z}^2}{S_z^2}) + (S_{\xi}^2 - \frac{C_{\xi,z}^2}{S_z^2})^2 \\ = & (S_{\xi}^2 - \frac{C_{\xi,z}^2}{S_z^2})^2 (\lambda \beta^2 + 1)^2, \end{aligned} $$
so that, with the help of eq. (20), the validity of the following relation is obvious :

$$ \hat{\beta} \hat{\rho} \rightarrow \frac{\left[ \lambda (\beta^2 S_{\xi,z}^2 + \tau^2 - \beta^2 \frac{C_{\xi,z}^2}{S_z^2}) - (S_{\xi,z}^2 + \lambda \tau^2 - \frac{C_{\xi,z}^2}{S_z^2}) \right] - \left[ (\lambda \beta^2 + 1) (S_{\xi,z}^2 - \frac{C_{\xi,z}^2}{S_z^2}) \right]}{\beta} = \beta $$

In other words, the maximum likelihood estimator $\hat{\beta}$ is a consistent estimator of $\beta$.

By using eq. (12) we find

$$ \hat{\gamma} \hat{\rho} \rightarrow \frac{\beta C_{\xi,z} + \gamma S_z^2}{S_z^2} - \beta \frac{C_{\xi,z}}{S_z^2} = \gamma, $$

so that $\hat{\gamma}$, too, is a consistent estimator of $\gamma$. Similarly $\hat{\alpha}$ must be a consistent estimator also.

In conclusion a brief remark about the estimator $\hat{\tau}^2$ should be made. We find with eq. (5)

$$ \begin{aligned} \hat{\tau}^2 &= \frac{1}{2n} \left[ \sum_{i=1}^{n} (y_i - \hat{\alpha} - \hat{\beta} \hat{\xi}_i - \hat{\gamma} z_i)^2 + \frac{1}{\lambda} \sum_{i=1}^{n} (x_i - \hat{\xi}_i)^2 \right] \\ &= \frac{1}{2n} (1 + \lambda \hat{\beta}^2) \sum_{i=1}^{n} (y_i - \hat{\alpha} - \hat{\beta} \hat{\xi}_i - \hat{\gamma} z_i)^2 \\ &= \frac{1}{2n} (1 + \lambda \hat{\beta}^2) \sum_{i=1}^{n} \left[ y_i - \hat{\alpha} - \frac{\hat{\beta}}{1 + \lambda \hat{\beta}^2} \left\{ \lambda \hat{\beta} (y_i - \hat{\alpha} - \hat{\gamma} z_i) + x_i \right\} - \hat{\gamma} z_i \right]^2 \\ &= \frac{1}{2n (1 + \lambda \hat{\beta}^2)} \sum_{i=1}^{n} (y_i - \hat{\alpha} - \hat{\beta} x_i - \hat{\gamma} z_i)^2 \end{aligned} $$
$$
\begin{align*}
&= \frac{1}{2n(1 + \lambda \hat{\beta}^2)} \sum_{i=1}^{n} \{(y_i - \bar{y}) - \hat{\beta}(x_i - \bar{x}) - \hat{\gamma}(z_i - \bar{z})\}^2 \\
&= \frac{1}{2(1 + \lambda \hat{\beta}^2)} (m_{yy} + \hat{\beta}^2 m_{xx} + \hat{\gamma}^2 m_{zz} - 2\hat{\beta} m_{xy} - 2\hat{\gamma} m_{yz} + 2\hat{\beta}\hat{\gamma} m_{xz}) \\
&\rightarrow \frac{1}{2(1 + \lambda \hat{\beta}^2)} (\beta^2 S_{\xi}^2 + \gamma^2 S_z^2 + 2\beta\gamma C_{\xi,z} + \tau^2 + \beta^2 S_{\xi}^2 + \\
&\qquad + \lambda \beta^2 \tau^2 + \gamma^2 S_z^2 - 2\beta^2 S_{\xi}^2 - 2\beta\gamma C_{\xi,z} - 2\beta\gamma C_{\xi,z} - \\
&\qquad - 2\gamma^2 S_z^2 + 2\beta\gamma C_{\xi,z}) = \frac{1}{2} \tau^2
\end{align*}
$$

Hence it follows that the maximum likelihood estimator $\hat{\tau}^2$ is not a consistent estimator of $\tau^2$. If we replace $2n$ in the denominator of $\hat{\tau}^2$ by the number of degrees of freedom, $n-3$, we do obtain a consistent estimator of $\tau^2$. This consistent estimator of $\tau^2$ then has the following form:

$$ \frac{2n}{n-3} \hat{\tau}^2 = \frac{1}{n-3} \frac{\sum_{i=1}^{n} (y_i - \hat{\alpha} - \hat{\beta} x_i - \hat{\gamma} z_i)^2}{1 + \lambda \hat{\beta}^2} $$

REFERENCES

[1] JONSTON (J.) (1963) *Econometric methods*, Mc Graw-Hill, New York.

[2] MALINVAUD (E.) (1966) *Statistical methods of econometrics*, North-Holland, Publishing Company, Amsterdam.

[3] GRAYBILL (F.A.) (1961) *An introduction to linear statistical models*, Vol. 1, Mc Graw-Hill, New York.

[4] KENDALL (M.G.) and STUART (A.) (1967) *The advanced theory of statistics*, Vol. 2, Charles Griffin and Company Ltd., London.

Reçu en Mars 1970

PHILIPS Statistics and Probability Group

I.S.A. - R. - EINDHOVEN - Pays-Bas.