# 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!

Michael

Contents

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]\$

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