Date: Mar 17, 2013 4:44 PM
Author: RGVickson@shaw.ca
Subject: Re: Estimate failure rate: Variable degree of freedom in chi-square
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.)