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Assuming $A_1, A_2, \ldots, A_n$ are independent exponential random variables (each having the same parameter and, for the sake of simplicity, let's assume the value of the parameter is 1). Define $B_i = A_i + k$ (where $k$ is a constant of unknown value).

What would the maximum likelihood estimate (MLE) of $k$ be, if we're also provided with a sequence of observations $(b_1, b_2,\ldots, b_n)$?

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  • $\begingroup$ If this is a self-study question, please add the self-study tag. $\endgroup$
    – Andy
    May 19, 2014 at 16:01
  • $\begingroup$ For some intuition--and a generalization--note that over its support $[0,\infty),$ the exponential PDF strictly decreases. Let us assume only that the $A_i$ are independent variables having any distributions on $[0,\infty)$ (possibly varying among the $(A_i)$) whose PDFs strictly decrease. Nevertheless (1) the MLE cannot exceed any of the $b_i$, making $\min\{b_i\}$ a lower bound for the MLE, while (2) decreasing $k$ increases every one of the probabilities of the $b_i$, thereby increasing the likelihood. Now draw the (only possible) conclusion. $\endgroup$
    – whuber
    May 19, 2014 at 20:37

1 Answer 1

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Note that for each $i = 1, 2, \ldots, n$, $B_i$ is a shifted/non-central exponential distribution with location parameter $k$. So write the joint density $$f(\boldsymbol b \mid k) = \prod_{i=1}^n e^{-(b_i - k)} \boldsymbol 1[b_i > k].$$ Do a little algebra and write the log-likelihood $\ell(k \mid \boldsymbol b)$.

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  • $\begingroup$ Because I misinterpreted the question to be $B_i = A_i + ki$ instead of just $k$. I will edit. $\endgroup$
    – heropup
    May 19, 2014 at 20:12
  • $\begingroup$ thanks a lot for the pdf; I eventually got to the right answer (which was also suggested by @whuber) $\endgroup$
    – Joe
    May 20, 2014 at 0:45

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