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The problem is slightly different than finding the sum of classic exponential distributions, this is why I am asking the question.

$p(t|T) = e^{T-t}$ if $t > T$ and $0$ otherwise

The objective is to give an estimation for parameter $T$, based on $n$ observations. As the mean is $T+1$, we can use the following estimator :

$T^* = \frac{1}{n} \cdot \sum_{i=1}^{n} (x_{i} - 1)$ with $i = 1..N$

Finding the sampling distribution of $T^*$ would allow to give an estimate for the parameter $T$ with a confidence interval.

Do you have any hint to provide? Many thanks for your help!

EDIT : I found a related post where the answers of Henry concerning the sampling distribution of $T^*$ and the $T^+$ estimator could help.

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$T^*$ will have the distribution of a gamma distributed random variable (with shape parameter $n$ and scale parameter $\frac1n$ or rate parameter $n$) plus the constant $T-1$.

As you say, its expectation is $T$ while its density is $$f_{T^* \mid T}(x) = \frac{n^n}{\Gamma(n) } (x+1-T)^{n - 1} e^{-n(x+1-T)} \text{ for } x > T-1$$ and $0$ for $x < T-1$

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    $\begingroup$ A better unbiased estimator might be $T^+ = \min(x_i)-\frac1n$ which has the distribution of an exponential distributed random variable (i.e. a gamma distributed random variable with shape parameter $1$ and scale parameter $\frac1n$ or rate parameter $n$) plus the constant $T-\frac{1}{n}$. This would be a much tighter distribution for the estimator if $n \gt 1$ $\endgroup$ – Henry Apr 3 at 1:18
  • $\begingroup$ Thank you very much for your answer! From what I understand we are considering a sum of exponential distributions $exp(1)$, shifted by a constant $(T-1)$. Then from Wikipedia (first line : en.wikipedia.org/wiki/Gamma_distribution#General) I see that the sum of $n$ distributions $exp(1)$ is a Gamma distribution with shape $n$ and rate $1$. It seems to me that this would lead to : $f(x)=((x+1−T)^{n−1} * e^{−(x+1−T)})/Γ(n)$ for $x>T-1$ Where is my mistake? Also, could you please give details about the better estimator you are proposing? Maybe I should ask another question? $\endgroup$ – DataXplorer Apr 3 at 8:53
  • $\begingroup$ @DataXplorer $\sum_{i=1}^{n} (X_{i}-T)$ has Gamma shape $n$ and rate $1$, so $\frac 1n \sum_{i=1}^{n} (X_{i}-T)$ has Gamma shape $n$ and rate $n$ $\endgroup$ – Henry Apr 3 at 9:19
  • $\begingroup$ Indeed! How brainless from me... Many thanks again! Concerning, the estimator $T^+$, I am going to ask another question on Stackexchange. $\endgroup$ – DataXplorer Apr 3 at 9:23
  • $\begingroup$ My suggestion for $T^+$ was related to sufficient statistics and minimum-variance unbiased estimators $\endgroup$ – Henry Apr 3 at 9:25

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