Estimation - Measurement

Let's say we want to measure the resistance of a resistor with an ohmmeter.  The measured value is $z$ with a standard deviation $\sigma$.  Assuming the resistance is $x$.

The measurement is like $z=x+v$ and $v$ is the measurement noise with mean $0$ and standard deviation $\sigma$.

If we only take one measurement, we get an estimation $z_1$ for the $x$.  The standard deviation of the error is $\sigma$.

If we take two measurements, we get $z_1,z_2$.  We can use $(z_1+z_2)/2$ as the estimation for the $x$.  Then what's the standard deviation of the error?

If two measurements are independent and uncorrelated, the standard deviation would be $\sqrt{1/2}\sigma$.

Then how about we use different weights, say $(z_1+2*z_2)/3$.  Then what's the standard deviation of the error in this case?  It would be $\sqrt{5/9}\sigma$, and it's greater than $\sqrt{1/2}\sigma$.

We can prove that the weight $(1/2,1/2)$ can give the minimum standard deviation of the error.

Now say we have already make two measurements and get the estimate $(z_1+z_2)/2$.

We want to do another measurement and get $z_3$.  Then how are we going to merge the data?

We know the previous two measurement give us the data $((z_1+z_2)/2,\sqrt{1/2}\sigma)$ and the new information is $(z_3,\sigma)$.  If we want to merge these two data and minimize the standard deviation of the error, the weight would be $2/3$ for the first data and $1/3$ for the new data.

The result would be like $((z_1+z_2+z_3)/3,\sqrt{1/3}\sigma)$.

As we can see for estimation, the information of the standard deviation of the error of the previous estimate is quite useful in this case.  Of course, in this simple example, we can also keep the number of measurements and find the new weights used in the new estimation.