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Date:         Thu, 13 May 1999 15:28:06 +0100
Reply-To:     John Whittington <medisci@POWERNET.COM>
Sender:       "SAS(r) Discussion" <SAS-L@UGA.CC.UGA.EDU>
From:         John Whittington <medisci@POWERNET.COM>
Subject:      Re: Maximum likelihood estimation
Comments: To: "Berryhill, Timothy" <TWB2@PGE.COM>
Content-Type: text/plain; charset="us-ascii"

At 18:16 11/05/99 -0700, Berryhill, Timothy wrote:

>Seems to me the median is zero. For any theta < 0, f(x) is 0 for all x. >For any positive theta, f(x) is 0 for all negative x and for all x>theta. >If x is distributed uniformly, the median is zero regardless of the value of >theta. >> f(x)=2x/theta**2 , 0<x<theta >> =0, otherwise.

Tim, much as I usually keep away from 'homework' questions, I couldn't help noticing what you'd written! I'm addressing only your comment, not the original question ....

If x is negative (for any theta), then f(x)=0. Similarly, if theta is negative (for any x), then f(x)=0, and, in any event theta**2 must be positive. It therefore follows that f(x) cannot be negative. However, f(x) can have any number of positive values, as well as zero. If f(x) is either zero or positive, but never negative, it seems a bit rash to suggest that the median is inevitably zero. I'm not saying (without considering the actual function!) that it couldn't be zero, but, unless I'm missing something, it's not as inevitable as you seem to imply.



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