## Partial Identification and Chi-Squared Tests

I heard Adam Rosen give his paper, “Confidence sets for partially identified parameters that satisfy a finite number of moment inequalities.” It stimulated some thoughts. (Click here to read

more.)

1. Suppose we wanted to estimate means of X and Y,(μ(x) μ(y)). Our theory says that they are distributed independently, bounded by [0,1]. But we only have data on X.

My maximum likelihood estimator will be a point estimate for μ(x) and an interval for μ(y). I have partial identification. If the sample mean of x were .6, my estimate would be (.6, [0,1]).

If I had a prior on μ(y), I could use that. Maximum likelihood or any kind of minimum distance-MOM estimator would leave every value in [0,1] equally good as an estimate of μ(y).

Another example would be if we wanted to estimate the mean of X+Y, μ(x+y), but only had data on x. If the sample mean of x was .6, our estimate for μ(x+y) would be the interval [,6, 1.6].

We would also have partial identification in a model in which y = αx1 + βx2 but x1 and x2 were endogenous and we had an instrument for x1 but not for x2.

2. Suppose we have partial identification, and our estimation has yielded us a best-estimate interval for the single parameter theta, which is thetahat = [5,10]. Our null hypothesis is that &theta &ge 6. Do we reject it?

We want to construct a confidence set C such that if we repeat the procedure, &alpha = 5% of the time we will wrongly reject the null when it is true:

(1) Prob(&theta -hat is in C)|&theta &ge 6) = .05

C will be a set of intervals.

But that probability in (1) is ill-defined, because C will differ depending on whether &theta =.6, 7, 9, 26, or whichever value greater than 6 we might pick. So we’ll be conservative, making it hard to reject the null, and pick the value of &theta for which C is biggest. That kind of conservatism problem arises even in the simplest frequentist inequality null– the problem is that the null is not “simple”.

A nice thing about the chi-squared test is that it avoids having to define C for the &theta -hat space. Instead, we just find the scalar chibarredsquared statistic, a function of the interval, and look at the confidence interval for that test statistic. This is what chi-squared tests do in general— they transform a multi-dimensional acceptance region into a one-dimensional acceptance interval. For example, we could use a Chi-squared test (or its close relation, an F-test), to test whether the pair of numbers (α, β) was close enough to (0,0).

Here, though, it’s especially neat because we’re not just doing an R-n to R mapping: we’re mapping from a set in (R, intervals on R) to R. An interval on R can be reduced to its pair of endpoints, but even then our mapping wouldn’t be as simple as a mapping from three real numbers to one.

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