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.)