| 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 |
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| Content-Type: | text/plain; charset="us-ascii" |
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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.
Regards,
John
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