Re: order statistics/life testing problem
- From: Jack Tomsky <jtomsky@xxxxxxxxxxxxx>
- Date: Sat, 21 Jul 2007 12:20:31 EDT
In sci.stat.math, Jack Tomsky wrote:
theIn sci.stat.math, Jack Tomsky wrote:
problem. I have a sample of n independent randomI would need to practice the following type of
vars. I fix one of the order statistics. What is
isjoint distribution of the remaining?
take a test. When 3 people are finished, the examI transcribed an example problem below. 5 people
ishalted. Those 3 students turn in their work. How
much longer will the remaining students take. In
other words: given the time at which the median
independentlyobserved, what is the additional life left for the
remaining two.
Your help is much appreciated.Let’s take the case where the n times
incorrectfollow an exponential distribution with parameter
lambda. Here lambda is the reciprocal of the mean.
The sequel is overly complicated and quite
sameat times. If each
lifetime is an independent exponential with the
initialparameter, then the
remaining times are still exponential with the same
parameter by the
memoryless property. So in the three out of five
case, the smaller
remaining time Y_(1) has exponential distribution
with parameter 2
lambda; given Y_(1), Y_(2) is an
exponential(lambda) + Y_(1).
In this case, EY_(2) = 3 / (2 lambda) = 1 / (2
lambda) + 1/ lambda. I
suppose if you wanted to estimate lambda, you could
determine the
likelihood from the joint distribution of the
is the additional life left for the remaining two" asorder statistics.
However, the OP asked about neither moments nor
estimation; rather, he
asked about the joint distribution of the remaining
times.
Steve, I interpreted the OP's last question of "what
meaning the conditional expectation of the largest
time. Note that in the OP's example, your result of
3/(2*lambda) corresponds to my result of t* + 3
/(2*lambda) since I measured the times from time
zero.
The OP literally said, "What is the joint
distribution of the
remaining?" And the point is that your method and
your formula for the
conditional expectation are unnecessarily
complicated.
Steve, as I previously pointed out, I was dealing only with the last question asked by the OP of the additional time remaining for the last two students in the special case of an exponential. The formula for the conditional expectation might look complicated, but it simplifies for particular n and k.
Incidentally, I should have said that my unbiased estimate for the mean, 1/lambda, is based on
Sum(t(i))+(n-k)t* ~ Chisq(2k)/(2*lambda).
Jack
.
My estimate of 1/lambda, given by[Sum(t(i))+(n-k)t*]/k, is an unbiased estimate, using
the fact that
For an unbiased esitmate of the mean, again use the
Sum(t(i))+(n-k)t* ~ Chisq(k)/lambda.
memoryless property
of the exponential. Let Y_i = X_(i) - X_(i-1),
where X_(0) = 0.
Then each Y_i has exponential distirbution with
parameter
n - i + 1. Thus,
EX_(i) = E sum(j=1..i, Y_j) = sum(j=1..i, EY_j) =
sum(j= n-j+i..n) / lambda
= i (2n - i + 1) / (2 lambda).
Thus, any weighted average of 2 EX_(i) / [i (2n - i
+1)] for i = 1 to
k will provide an unbiased estimate of 1 / lambda.
However, this is only the exponential case. As I
pointed out in my
first response (to the OP's post in sci.math), in
general, given
X_(k) = x, the n - k remaining residual lifetimes
are i.i.d; the
common marginal distribution is the conditional one
of X - t given X
> t, where X is an individual lifetime.
sum
Letobserve t(k) = t*.
t(1) < t(2) < … < t(n)
denote the order statistics. Suppose that we
Then the conditional expected value of the largesttime, t(n), given the k-th order statistic t* is
simply the unconditional mean of the maximum of n-k
exponentials plus t*.
E[t(n)|t(k)=t*] = t* +[(n-k)/lambda]*Sum[(-1)^j*C(n-k-1,j)/(j+1)^2],
where the sum is from j = 0 to n-k-1.3/(2*lambda).
In the example you gave, n = 5 and k = 3.
In that case, the conditional mean becomes t* +
You can estimate lambda by
1/lambdahat = [Sum(t(i))+(n-k)t*]/k, where the
is from j = 1 to k.
--
Stephen J. Herschkorn
sjherschko@xxxxxxxxxxxx
Math Tutor on the Internet and in Central New Jersey
and Manhattan
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