# PDF of the sum of truncated exponential distributions

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.

$$T^*$$ 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 $$x < T-1$$
• 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 $n \gt 1$ – Henry Apr 3 at 1:18
• 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? – DataXplorer Apr 3 at 8:53
• @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 $n$ – Henry Apr 3 at 9:19
• Indeed! How brainless from me... Many thanks again! Concerning, the estimator $T^+$, I am going to ask another question on Stackexchange. – DataXplorer Apr 3 at 9:23
• My suggestion for $T^+$ was related to sufficient statistics and minimum-variance unbiased estimators – Henry Apr 3 at 9:25