What is the

easiestway to see that the following statement is true?Suppose Y_1, \dots, Y_n \overset{\text{iid}}{\sim} \text{Exp}(1).

Show \sum_{i=1}^{n}(Y_i – Y_{(1)}) \sim \text{Gamma}(n-1, 1).Note that Y_{(1)} = \min\limits_{1 \leq i \leq n}Y_i.

By X \sim \text{Exp}(\beta), this means that f_{X}(x) = \dfrac{1}{\beta}e^{-x/\beta} \cdot \mathbf{1}_{\{x > 0\}}.

It is easy to see that Y_{(1)} \sim \text{Exponential}(1/n). Furthermore, we also have that \sum_{i=1}^{n}Y_i \sim \text{Gamma}(\alpha = n, \beta = 1) under the parametrization

f_{Y}(y) =\dfrac{1}{\Gamma(\alpha)\beta^{\alpha}}x^{\alpha-1}e^{-x/\beta}\mathbf{1}_{\{x > 0\}}\text{, }\qquad \alpha, \beta> 0\text{.}

Solution given Xi’an’s Answer: Using the notation in the original question: \begin{align}

\sum_{i=1}^{n}[Y_i – Y_{(1)}] &= \sum_{i=1}^{n}[Y_{(i)}-Y_{(1)}] \\

&= \sum_{i=1}^{n}Y_{(i)}-nY_{(1)}\\

&= \sum_{i=1}^{n}\{Y_{(i)}-Y_{(i-1)}+Y_{(i-1)}-\cdots-Y_{(1)}+Y_{(1)}\}-nY_{(1)}\\

&= \sum_{i=1}^n\sum_{j=1}^{i}\{Y_{(j)}-Y_{(j-1)}\}-nY_{(1)}\text{ where } Y_{(0)} = 0 \\

&= \sum_{j=1}^n\sum_{i=j}^{n}\{Y_{(j)}-Y_{(j-1)}\}-nY_{(1)}\\

&= \sum_{j=1}^{n}(n-j+1)[Y_{(j)}-Y_{(j-1)}]-nY_{(1)}\\

&= \sum_{i=1}^{n}(n-i+1)[Y_{(i)}-Y_{(i-1)}]-nY_{(1)}\\

&= \sum_{i=2}^{n}(n-i+1)[Y_{(i)}-Y_{(i-1)}]+nY_{(1)}-nY_{(1)} \\

&= \sum_{i=2}^{n}(n-i+1)[Y_{(i)}-Y_{(i-1)}]\text{.}

\end{align}

From this, we get that \sum_{i=2}^{n}(n-i+1)[Y_{(i)}-Y_{(i-1)}] \sim \text{Gamma}(n-1, 1).

**Answer**

The proof is given in the Mother of All Random Generation Books, Devroye’s Non-uniform Random Variate Generation, on p.211 (and it is a very elegant one!):

Theorem 2.3 (Sukhatme, 1937)If we define E_{(0)}=0 then the normalised exponential spacings (n-i+1)(E_{(i)}-E_{(i-1)})

derived from the order statistics E_{(1)}\le\ldots\le E_{(n)} of an i.i.d. exponential sample of size n

are themselves i.i.d. exponential variables

**Proof.** Since

\begin{align*}

\sum_{i=1}^n e_i &= \sum_{i=1}^n e_{(i)} =\sum_{i=1}^n \sum_{j=1}^i(e_{(j)}-e_{(j-1)})\\ &=\sum_{j=1}^n \sum_{i=j}^n(e_{(j)}-e_{(j-1)}) =\sum_{j=1}^n (n-j+1)(e_{(j)}-e_{(j-1)})

\end{align*}

the joint density of the order statistic (E_{(1)},\ldots,E_{(n)}) writes as

f(\mathbf{e})=n!\,\exp\left\{-\sum_{i=1}^ne_{(i)}\right\}=n!\,\exp\left\{-\sum_{i=1}^n (n-i+1)(e_{(i)}-e_{(i-1)})\right\}

Setting Y_i=(E_{(i)}-E_{(i-1)}), the change of variables from (E_{(1)},\ldots,E_{(n)}) to (Y_1,\ldots,Y_n) has a constant Jacobian [incidentally equal to 1/n! but this does not need to be computed] and hence the density of (Y_1,\ldots,Y_n) is proportional to

\exp\left\{-\sum_{i=1}^n y_i \right\} which establishes the result. **Q.E.D.**

An alternative suggested to me by Gérard Letac is to check that (E_{(1)},\ldots,E_{(n)})has the same distribution as\left(\frac{E_1}{n},\frac{E_1}{n}+\frac{E_2}{n-1},\ldots,\frac{E_1}{n}+\frac{E_2}{n-1}+\ldots+\frac{E_n}{1}\right) (by virtue of the memoryless property), which makes the derivation of \sum_{k=1}^n(E_k-E_{(1)})\sim \sum_{k=1}^{n-1}E_k straightforward.

**Attribution***Source : Link , Question Author : Clarinetist , Answer Author : Xi’an*