# Primitive length measurements

A measuring rod has length $u$ (for "unit") and a long object has length $x$. Suppose $u$ is laid end to end $i$ times and $x$ is laid end to end $j$ times. We want to observe whether $iu\ \left\{\begin{array}{c} < \\ = \\ > \end{array}\right\}\ jx$. But for each iteration of $u$ and of $x$ there is a random error---say we observe whether $$iu + \varepsilon_1+\cdots+\varepsilon_i\ \left\{\begin{array}{c} < \\ > \end{array}\right\}\ jx+\delta_1+\cdots+\delta_j$$ for $i=1,\ldots,I$ and $j=1,\ldots,J$, where the $\varepsilon$s are independent and $\sim N(0,\sigma^2)$ and the $\delta$ are independent of each other and of the $\varepsilon$s and $\sim N(0,\tau^2)$. So we have $IJ$ observations, each binary, equal to either "$<$" or "$>$" (encode them with $0$s and $1$s if you like).

What is known about statistical inference about the ratio $x/u$ in this problem? Things like the MLE for $x/u$ or the MLEs for $\sigma$ and $\tau$, or confidence intervals for $x/u$, etc. For large values of $I$ and $J$, might one be able to look for things like non-normality of the distributions of $\delta$ and $\varepsilon$?

-
If you want to make statements about the ratio $x/u$ it would of course be much better to have the measurements $iu+\epsilon_1,iu+\epsilon_2,\ldots,jx+\delta_j$ rather than just binary variables (as the latter only tell you that one is bigger than the other and not how much bigger it is). With some of the parameters known, binary data might be sufficient though. Am I correct to think that $u$ is known? What about $\sigma^2$ and $\tau^2$? –  MånsT Jul 11 '12 at 7:11
@MånsT : We can't take $u$ to be known because $u$ is the unit of measurement by means of which all lengths are to be known. I'm not sure it would make any difference of $u$ were in some sense known in terms of other units. –  Michael Hardy Jul 11 '12 at 19:45
+1 This is a probit model with parameters $(u,x)$ and dependent observations determined by a multivariate normal distribution (whose covariance matrix depends on the two nuisance parameters $\sigma$ and $\tau$ but is otherwise completely determined). The integrals involved in the likelihood look formidable, suggesting a Bayesian MCMC approach would be the most practical line of attack. –  whuber Jul 11 '12 at 21:21
I think for certain values of the data there's a non-unique MLE. For example, suppose $I=J=2$ and it is observed that: \begin{align} x+\text{error} & > u+\text{error} \\ 2x+\text{error} & > u+\text{error} \\ 2x+\text{error} & > 2u + \text{error} \\ x+\text{error} & < 2u+\text{error}\end{align}. Then, if I'm not mistaken, the maximum likelihood would be obtained at every point where $\hat\sigma=\hat\tau=0$ and $1 < \widehat{x/u} < 2$. But if we get inequalities that would be logically inconsistent if not for the error terms, then I suspect the MLE would be unique. –  Michael Hardy Jul 13 '12 at 19:38
add comment

## 1 Answer

Assuming the only known thing for a combination i,j is whether it is a 1 (iu+e is bigger) or a 0 (otherwise);

Suppose you put the observations in a matrix (I,J) (top left element is 1,1 bottom left is I,1)

Here is the idea that I have. Unfortunately I cannot prove it mathematically, but it might give you some direction.

If you take the biggest possible square submatrix from the top left and calculate the average, it may be a decent estimator of x/u.

Furthermore the variance will be higher if the 'changeover' from 0 to 1 does not show a nice pattern and/or if you often see that there is no changeover (example 000110111). However, i do not even dare to suggest an estimator for the variance.

-
add comment