I'm working through *Fifty Challenging Problems in Probability* by Frederick Mosteller. This is the problem I'm working on:

## Doubling Your Accuracy

An unbiased instrument for measuring distances makes random errors whose distribution has standard deviation $\sigma$. You are allowed two measurements all told to estimate the lengths of two cylindrical rods, one clearly longer than the other. Can you do better than to take one measurement on each rod?

## (Abridged) Solution

Here I abridge and reformat the solution to save space:

Yes. Let $A$ be the true length of the longer one, $B$ that for the shorter.

You could lay them side by side and measure their difference in length, $A - B$, and then lay them end to end and measure the sum of their lengths, $A + B$. Let $D$ be your measurement of $A - B$, $S$ of $A + B$.

Then an estimate of $A$ is $\frac{1}{2}(D + S)$, and of $B$ is $\frac{1}{2}(S - D)$.

Now $D = A - B + d$, where $d$ is a random error, and $S = A + B + s$, where $s$ is a random error.

Consequently, $\frac{1}{2}(D + S) = \frac{1}{2}(A - B + A + B + d + s) = A + \frac{1}{2}(d + s)$.

On the average, the error $\frac{1}{2}(d + s)$ is zero because both $d$ and $s$ have mean zero.

The variance of the estimate of $A$ is the variance of $\frac{1}{2}(d + s)$, which is $\frac{1}{4}(\sigma^2+\sigma^2)=\frac{\sigma^2}{2}$. This value is identical with the variance for the average of a sample of two independent measurements. ... In the same manner you can show that the variance of the estimate of B is also $\frac{\sigma^2}{2}$. Consequently, taking two measurements, one on the difference and one on the sum, gives estimates whose precision is equivalent to that where 4 measurements are used, two on each rod separately.

## My Question

The statement of the variance of $A$ in the last paragraph doesn't make sense to me. Clearly the author is arguing that $d$ and $s$ are independent. But both are functions of $A$ and $B$, so I think that they are dependent.

My question is, **are $d$ and $s$ independent?**

If they are independent, then what stops this example being generalised so that all variances can be reduced by clever adding/subtracting of measurements?

intuitivelyindependent. E.g. if I said $d=1mm$, then we can't reasonably infer a value for $s$ (e.g. $s=-1mm$) $\endgroup$ – user1108 Apr 13 '18 at 13:39