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Suppose that $X_1,X_2,\cdots ,X_n$ are iid. random variables and have pdf $$f(x) = \dfrac{1}{\sigma } e^{- \frac{x - \mu }{\sigma } } I(x \geq \mu) $$ Find out the distribution of $$Y := \sum\limits_{i=1}^n (X_i - X_{min} ) $$ where $X_{min}$ stands for the $\min \{X_1, \cdots , X_n \}$.

The book tells me that $Y$ has certain gamma distribution. My idea is to show that $X_i - X_{min}$ are independent random variables, so that since they have the exponential distribution, there sum has gamma distribution. However, I am not able to prove the independence. Can anyone offer a solution, or they are in fact not independent ?

j200932
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  • Consider the case when $n=1$. Then the only $X_i$ is $X_n=X_{\mathrm{min}}$, at which point they're definitely not independent. I'm not really sure how to tackle this problem, but one vague approach might be to start by characterising $X_\mathrm{min}$ using the the method of order statistics, and then maybe you could calculate the moment generating function of the sum (using the property that the moment generating function of a sum is the product of the individual moment generating functions) and see if you could manipulate that into the form of the mgf of the gamma distribution or something – Jack Crawford Oct 25 '19 at 13:34
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    $X_i - X_{\min}$ is $0$ for exactly one $i$ with probability $1$, and for any particular $i$ you have $\mathbb P(X_i - X_{\min}=0) = \frac1n$. So they are not independent . But if it is $i=1$ for which this is zero, then I think you can use memorylessness to show the other $X_i - X_{\min}$ are independent – Henry Oct 25 '19 at 13:39
  • I also thought of the memoryless property but I didn't know how to make use of it. – j200932 Oct 25 '19 at 13:47
  • You can use a specific change of variables to show this (https://math.stackexchange.com/questions/2764443/for-a-random-sample-from-the-distribution-fx-e-x-theta-x-theta-sh/). – StubbornAtom Oct 25 '19 at 15:06

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