# Re: st: Joint hypothesis with one-sided alternatives

 From "Guillaume Frechette" To statalist@hsphsun2.harvard.edu Subject Re: st: Joint hypothesis with one-sided alternatives Date Fri, 21 Jun 2002 18:38:24 +0000

In response to my questions about how to test a joint inequality hypothesis "Paul Seed" <P.T.Seed@qmul.ac.uk> wrote:

```This can be bootstrapped, without any programming, using the -bs- command.

Using the auto data set (& changing the hypothesis slightly),
the following will do it:

bs "xtprobit foreign weight, re i(rep78)"
"(_b[_cons]>0)&(_b[_cons]+_b[weight]<4)"

I trust that your application makes more sense than this one.

Adapt as necessary.  No spaces are permitted when describing
the hypotheses to be tested.  A reps option is needed.
See help for the command -bs- for more details, inclusing the meanings of
the difference confidence intervals.
```
Thanks for the suggestion Paul but I am not sure this works, maybe you (or someone else) can clarify. I knew I could bootstrap, as you suggested, however, I think there is a problem you omit. The joint inequalities imply that the frontier of the acceptance region is not continuous, which is a problem for hypothesis testing (deriving the distribution for the test). To see this, notice that my original hypothesis a > 0 and a + b < 0 can be re-written as (a+b)^(sign of (-a)) < 0. However, even if bs would give me a test result if I bootstrapped (a+b)^(sign of (-a)) < 0 I do not think that result is valid. The composite hypothesis makes it quite difficult I believe. Redefine a + b = -g, then the region we are looking for, in a (a,g) graph, is the positive quadrant. Think about estimates such that you are really close to (0,0) but in the region where a is negative, now draw different sized circles around that point to represent different critical values, as you can see, there's no easy formula to represent the area of these circles that falls into the positive quadrant. I think this is more or less a way to represent the difficulty of this problem. Is there something I am missing about bootstrapping that solves this problem? Thanks again Paul, since nobody was answering I was thinking that this might just be so trivial that nobody wanted to bother answering it.

Best,

g

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