Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math.independent

Topic: Estimate failure rate: Variable degree of freedom in chi-square
Replies: 3   Last Post: Mar 18, 2013 12:27 AM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
RGVickson@shaw.ca

Posts: 1,653
Registered: 12/1/07
Re: Estimate failure rate: Variable degree of freedom in chi-square
Posted: Mar 17, 2013 4:44 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On Saturday, March 16, 2013 10:27:21 PM UTC-7, Paul wrote:
> I've found conflicting information about the degrees of freedom to use
>
> in the chi-square distribution when estimating failure rate from the
>
> number of failures seen over a specified period of time. To be sure,
>
> the lower MTBF (upper failure rate) always uses 2n+2, where n is the
>
> number of failures. However, the upper MTBF (lower failure rate) is
>
> shown as using both 2n and 2n+2, depending on the source. I haven't
>
> found an online explanation of exactly how the chi-square distribution
>
> enters into the calculation (other than http://www.weibull.com/hotwire/issue116/relbasics116.htm,
>
> which I'm still chewing on). So I haven't been able to determine
>
> whether 2n or 2n+2 is correct from first principles at this point.
>
> Based on the reasoning in the above weibull.com page, however, I am
>
> inclined to believe that the degrees of freedom should be 2n because
>
> we're talking about the two tails of the *same* distribution for upper
>
> and lower limits. But this leaves the mystery of why 2n+2 shows up
>
> frequently. Is the reason for this straightforward enough to explain
>
> via this newsgroup?


I wanted to reply to David Jones's response below, but my browser is forcing me to reply instead to you.

With regard to David's point (1): _given_ n observed outcomes in (0,T) the outcome times are UNIFORMLY distributed in (0,T); that is, the individual arrival times are the n order statistics of the distribution U(0,T). Therefore, the observed inter-arrival times are NOT exponential and are NOT independent. (This is a fundamental and well-known property of Poisson processes.)



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.