"Magical" variance reduction problem I recently came across this toy problem:
You have two sticks of unknown lengths $a>b$ and a measuring device with constant variance $1$ that you can only use twice. How can you construct estimators $\hat a,\hat b$ with minimal total MSE?
The solution involves measuring $\hat x=a+b,\hat y=a-b$ (sum and difference of the sticks) and combining them as $\hat a=\frac{\hat x+\hat y}2,\hat b=\frac{\hat x-\hat y}2$, and then each of $\hat a$,$\hat b$ have variance $1/2$. This is the same as what you would get by separately measuring each of $a,b$ twice, so it's optimal. In addition, if you make some boilerplate Gaussian assumptions, it's easy to show that 
$$(\hat a,\hat b)\sim MVN((a,b),
  \begin{bmatrix}
    \frac12 & 0 \\
    0 & \frac12
  \end{bmatrix})$$
so they actually have the same distribution!
I have two follow-up questions:


*

*What's a good intuitive explanation of why we can get a "free" variance reduction with no trade-offs? This is open-ended and multiple answers/analogies to other concepts are welcome! Bonus points for tying this in with sufficient statistics.

*Can this be generalized to $n$ sticks (ordering is not important) and $n$ uses of the device?
 A: Interesting example. I think some key intuition is right there in your post: You get to measure each stick twice. The magic is not so much about statistics or probability but about how you cleverly arrange the measurements so that you get the nuisance terms to cancel:
Let's simply say that the measured quantity will differ from the true one by some amount $\epsilon_t$ that differs for every measurement.
Measuring $a$ twice and taking the average gives $\hat{a} = \frac{a + a + \epsilon_1 + \epsilon_2}{2}$
Measuring $a+b$ and $a-b$ adding and taking the average gives $\hat{a} =  \frac{a + b + \epsilon_1 + a - b +  \epsilon_2}{2} = \frac{a + a + \epsilon_1 + \epsilon_2}{2}$
A: *

*The most convincing "intuitive" explanations I was able to come up with point out the non-intuitive aspects of the assumptions in the problem. For example, the constant variance assumption might be unintuitive for large or small $a,b$, but we do measure $a-b$ so it's not clear how constant variance helps us reduce MSE. Another answer is that this is sort of like a "mini-James-Stein" but this is not really enlightening and more of a whataboutism. Finally, there may be some prior we can put on $a,b$ to assign some blame to our frequentist assumptions but I haven't worked out the details.

*It would be nice to get estimators with individual variance $1/n$ or less, but I wasn't able to do this with even $n=3$. One interesting observation is that we can use sticks at angles to get linear combinations with scalars in the range $[-1,1]$, or if that doesn't work then maybe try extending the problem with complex numbers and use roots of unity to cancel out the $+-$ that we get with $n=2$.
A: It seems the real answer to (1) is a lot simpler than I imagined: we have 
$$\begin{bmatrix}\hat a \\\hat b\end{bmatrix}=A\begin{bmatrix}\hat x \\\hat y\end{bmatrix}$$
where $A=\begin{bmatrix}
           \frac12 & \frac12 \\
           \frac12 & -\frac12
         \end{bmatrix}$ and $Var(\begin{bmatrix}\hat x \\\hat y\end{bmatrix})=I$, so
$$Var(\begin{bmatrix}\hat a \\\hat b\end{bmatrix})=AA^T=\begin{bmatrix}
           \frac12 & 0 \\
           0 & \frac12
         \end{bmatrix}$$
So it really just comes down to $\det(A)<1$, which means it's a shrinkage of constant variance measurements, which reduces the variance.
