8
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Do you know the value of $m$? $\endgroup$ – jbowman Dec 3 '12 at 15:09
  • 3
    $\begingroup$ This is type II censoring (en.wikipedia.org/wiki/Censoring_%28statistics%29). Now, it can be shown that the usual likelihood in survival analysis also holds for a censoring mechanism of type II. $\endgroup$ – ocram Dec 3 '12 at 15:32
  • 1
    $\begingroup$ The roles of $m$ and $n$ appear to get swapped part-way through this answer. $\endgroup$ – cardinal Dec 3 '12 at 19:38
  • $\begingroup$ Thanks, you are right. I fixed the roles of m and n in the statement of the problem. $\endgroup$ – Michael Dec 5 '12 at 9:16
8
$\begingroup$

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:

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

$\endgroup$
  • 1
    $\begingroup$ Good answer. Did you switch $m$ and $n$ compared with the question by accident? $\endgroup$ – Neil G Dec 3 '12 at 18:34
  • 2
    $\begingroup$ @NeilG - thanks! I just noticed that the OP switched from "from a complete population of $m$ elements are drawn ... only the $n$ smallest are known" in the text to $m < n$ at the end. I'll clarify which notation I'm using in an edit... $\endgroup$ – jbowman Dec 3 '12 at 19:27
2
$\begingroup$

This links @jbowman's answer to my comment. Namely, under common working assumptions, one can use the 'standard survival likelihood' under type II censoring.

> #------seed------
> set.seed(1907)
> #----------------
> 
> #------some data------
> t <- sort(rexp(n=20, rate=2))        #true sample
> t[16:20] <- t[15]                    #observed sample
> delta <- c(rep(1, 15), rep(0, 5))    #censoring indicator
> data <- data.frame(t, delta)         #observed data
> #---------------------
> 
> #-----using @jbowman's formula------
> 15 / (sum(t[1:14]) + (5 + 1)*t[15])
[1] 2.131323
> #-----------------------------------
> 
> #------using the usual survival likelihood------
> library(survival)
> fit <- survreg(Surv(t, delta)~1, dist="exponential", data=data)
> exp(-fit$coef)
(Intercept) 
   2.131323 
> #-----------------------------------------------

PS1: Note that this is not restricted to the exponential distribution.

PS2: Details can be found in Section 2.2 of the book by Lawless.

$\endgroup$
1
$\begingroup$

Assuming $n$ is known, an estimate can be obtained via

$ \Phi(x_k)=1-e^{-\lambda x_k} \approx (k/n)$ where $x_k$, $0<k<m$, refers to the $k$'th smallest value in your reduced data set.

The logic is: if you had the entire set of $n$ samples, you could construct the empirical CDF, $\Phi$, from this sample. Then if you took item $k$ of this sorted array, it would correspond to the CDF value $k/n$. In many cases, $k=n/2$ is a useful choice.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.