Solved – Parameter estimation of exponential distribution with biased sampling

I want to calculate the parameter $lambda$ of the exponential distribution $e^{-lambda x}$ from a sample population taken out of this distribution under biased conditions. As far as I know, for a sample of n values, the usual estimator is $hat{lambda} = frac{n}{sum x_i}$. However my sample is biased as follows:

From a complete population of m elements drawn i.i.d from the exponential distribution, only the n smallest elements are known. How can I estimate the parameter $lambda$ in this scenario?

A bit more formaly, if ${x_1,x_2,x_3,…,x_m }$ are iid samples drawn from $e^{-lambda x}$, such that for every $i < j$ we have $x_i leq x_j$, then how can I estimate $lambda$ from the set ${x_1,x_2,x_3,…,x_n}$ where $n < m$.

Thanks a lot!


The maximum likelihood estimator for the parameter of the exponential distribution under type II censoring can be derived as follows. I assume the sample size is $m$, of which the $n < m$ smallest are observed and the $m – n$ largest are unobserved (but known to exist.)

Let us assume (for notational simplicity) that the observed $x_i$ are ordered: $0 leq x_1 leq x_2 leq cdots leq x_n$. Then the joint probability density of $x_1, dots, x_n$ is:

$f(x_1, dots, x_n) = {m!lambda^n over {(m-n)!}}expleft{-lambdasum_{i=1}^nx_iright}expleft{-lambda(m-n)x_nright}$

where the first exponential relates to the probabilities of the $n$ observed $x_i$ and the second to the probabilities of the $m-n$ unobserved $x_i$ that are greater than $x_n$ (which is just 1 – the CDF at $x_n$.) Rearranging terms leads to:

$f(x_1, dots, x_n) = {m!lambda^n over {(m-n)!}}expleft{-lambdaleft[sum_{i=1}^{n-1}x_i+(m-n+1)x_nright]right}$

(Note the sum runs to $n-1$ as there is a "$+1$" in the coefficient of $x_n$.) Taking the log, then the derivative w.r.t. $lambda$ and so on leads to the maximum likelihood estimator:

$hat{lambda} = n / left[sum_{i=1}^{n-1}x_i+(m-n+1)x_nright]$

Similar Posts:

Rate this post

Leave a Comment