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!
Michael
Best Answer
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:
$f(x_1, \dots, x_n) = {m!\lambda^n \over {(m-n)!}}\exp\left\{-\lambda\sum_{i=1}^nx_i\right\}\exp\left\{-\lambda(m-n)x_n\right\}$
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:
$f(x_1, \dots, x_n) = {m!\lambda^n \over {(m-n)!}}\exp\left\{-\lambda\left[\sum_{i=1}^{n-1}x_i+(m-n+1)x_n\right]\right\}$
(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:
$\hat{\lambda} = n / \left[\sum_{i=1}^{n-1}x_i+(m-n+1)x_n\right]$