minimizing the mean squared error?
Given measurements
$$\begin{align}
Z_1 &amp;= x + N_1\\
Z_2 &amp;= x + N_2
\end{align}$$
where $N_1$ and $N_2$ are independent zero-mean Gaussian random variables
with variances $\sigma_1^2$ and $\sigma_2^2$ respectively, it can be shown
that the minimum-mean-square-error (MMSE) estimate of $x$ in terms of $Z_1$ and
$Z_2$ is a linear function of $Z_1$ and $Z_2$.  Note that this linearity
is a consequence of the Gaussian assumption; for other kinds of distributions
that might be assumed for the noise variables $N_1$ and $N_2$, the MMSE
estimate is usually not a linear function.

So, let us consider the estimate $\hat{X} = k_0 + k_1Z_1 + k_2Z_2$ for the
unknown parameter $x$.  The mean-square error is thus
$$\begin{align}
E[(\hat{X}-x)^2] &amp;=  E[(k_0 + k_1Z_1 + k_2Z_2 - x)^2]\\
&amp;= E[(k_0 + (k_1+k_2-1)x + k_1N_1+k_2N_2)^2]\\
&amp;= \operatorname{var}(k_1N_1+k_2N_2) + (k_0 + (k_1+k_2-1)x)^2
&amp;{\scriptstyle{\text{because}~E[(Y+a)^2=\operatorname{var}(Y)+(\mu_Y+a)^2}}\\
&amp;\geq \operatorname{var}(k_1N_1+k_2N_2)
\end{align}$$
where equality holds in that last inequality if we choose $k_0=0$ and $k_1+k_2=1$. With a slight change of notation to incorporate these
constraints into our calculations,
let us write 
$$\begin{align}
E[(\hat{X}-x)^2] &amp;= \operatorname{var}(\alpha N_1+(1-\alpha)N_2)\\
&amp;= \alpha^2\sigma_1^2 + (1-\alpha)^2 \sigma_2^2
&amp;{\scriptstyle{\text{because}~N_1, N_2 ~\text{are independent random variables}}}\\
&amp;= \alpha^2(\sigma_1^2 + \sigma_2^2) -2\alpha\sigma_2^2 +  \sigma_2^2
\end{align}$$
where it is easily shown that the minimum value of that quadratic function
of $\alpha$ occurs when we choose
$$\alpha = \frac{\sigma_2^2}{\sigma_1^2 + \sigma_2^2} = k_1, \qquad
1-\alpha = \frac{\sigma_1^2}{\sigma_1^2 + \sigma_2^2} = k_2,$$
that is, when the two measurements are afflicted with Gaussian
noise of different variances (say with $\sigma_2^2 &gt; \sigma_1^2$),
then the measurement with the smaller variance ($N_1$ in this
example) is given greater weight 
$k_1=\frac{\sigma_2^2}{\sigma_1^2 + \sigma_2^2}$
than the measurement with the larger variance ($N_2$ in this case)
which has smaller weight $k_2 = \frac{\sigma_1^2}{\sigma_1^2 + \sigma_2^2}$.

When $\sigma_2^2 = \sigma_1^2$, the measurements are given equal weight
$\frac 12$ which is, according to the OP's insistence, the weight that
must be
assigned in all circumstances so as to minimize the mean-square error
because that is the only weighting that makes sense to the OP.