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From |
wgould@stata.com (William Gould, Stata) |

To |
statalist@hsphsun2.harvard.edu |

Subject |
Re: st: Another precision problem with Mata |

Date |
Mon, 22 Aug 2005 09:55:42 -0500 |

Ben Jann <ben.jann@soz.gess.ethz.ch> asks, > in a Mata program I am using the expression > > abs((x-y)/z) < 1 > > where x, y and z are real scalars. Note that in some cases y has been > set to x + z earlier in the program. Although the expression should > evaluate to "false" in these cases, it sometimes does evaluate to "true" > due to roundoff error. > > My question now is: Would it be reasonable to code > > abs((x-y)/z) < 1-epsilon(1) > > Or how would one approach such a problem? The first step in approaching this problem is to understand the source of the error. Ben reports that when y = x + z, he can have problems. In the y=x+z case, abs((x-y)/z) = abs((x-(x+z))/z) = abs(-z/z) (inifinite precision) = 1 (inifinite precision) Ben is hoping that x-(x+z) == z, and that is not necessarily true when we perform finite-precision arithmetic. So now let's consider (x-(x+z)), the source of the problem. I am going to use the following notation + infinite-precision addition - infinite-precision subtraction PLUS finite-precision addition SUB finite-precision subtraction In this case x PLUS z = x+z + error The error can be so large so that x PLUS z = x (1) or x PLUS z = z (2) All that is required is that (1) x be very large relative to z, or (2) vice versa. The importance of that point is that we cannot bound the error by epsilon(1), as Ben was hoping. I cannot go further with this problem, in fact, without knowing something about the relative magnitudes of x and z. However, Ben has just revealed that it is important to him to identify abs((x-y)/z) == 1 when y = x+z. In that case, I suggest Ben remove y from his code and substitue (d+z) everywhere for it. Then abs((x-y)/z) == 1 is equivalent to abs(d)==abs(x). Perhaps there are other possible code-specific solutions, or perhaps, Ben does know something about magnitudes and we can bound the error, and Ben will be willing to accept abs((x-y)/z)<1 will evaluate to false in some cases where, in infinite precision, it would evaluate to true. But one thing I can guarantee: There is no reasons to think about how one might code abs((x-y)/z) < 1 so that it is guaranteed to be false when y = x PLUS z, for all values of x and z. -- Bill wgould@stata.com * * For searches and help try: * http://www.stata.com/support/faqs/res/findit.html * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/

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