Difference of random variables If $X_1,\ldots,X_n$ iid $Ga(1,\lambda)$, then would $\sum X_i - nX_{(1)} \sim Ga(n-1,\lambda)$ ? 
$X_{(1)}=\min(X_1,\ldots,X_n)$. 
Any help would be appreciated.
P.S.: This is an edit of a previous question of mine.
 A: This time the answer is yes. 
This is because $Ga(1, \lambda)$ is actually the exponential variable with parameter $\lambda$, which have the property of "forgetting the past". This is to say that, for any $y$, the distribution of $X_i -y $ knowing $X_i \geq y$ is still $Ga(1,\lambda)$.
Since this is true for any fixed $y$, we can say that the distribution of $X_i - Y$ knowing $X_i \geq Y$ is $Ga(1, \lambda)$ for any random variable $Y$.
Thus in the sum $\sum (X_i - X_{(1)})$, you have $(n-1)$ independent $Ga(1, \lambda)$ and one $0$. And we know that the sum of $(n-1)$ independent $Ga(1,\lambda)$ is $Ga(n-1,\lambda)$.
Indeed, this is not a rigorous proof. It only explains why the conclusion should be true.
Added:
Firstly remark almost surely all $X_i$ take distinct values
\begin{align}
E(f(\sum (X_i - X_{(1)}))) &= \sum_{j=1}^n E(f(\sum_{i=1}^n (X_i - X_{(1)})) | X_{(1)} = X_j)P(X_{(1)} = X_j)\\
&= \frac{1}{n}\sum_{j=1}^nE(f(\sum_{i=1}^n (X_i - X_{(1)})) | X_{(1)} = X_j)
\end{align}
For a fixed $j_0$
\begin{align}
&E(f(\sum_{i=1}^n (X_i - X_{(1)})) | X_{(1)} = X_{j_0}) \\
=& E(f(\sum_{i=1}^n (X_i - X_{j_0})) | X_i \geq X_{j_0}, \forall i \neq j_0)\\
=&\int_0^{+\infty}E(f(\sum_{i=1}^n (X_i - y)) | X_{j_0}=y, X_i \geq y, \forall i \neq j_0) P(X_{j_0}=dy) \\
= & \int_0^{+\infty}E(f(\sum_{i\neq j_0}X_i - y) | X_i \geq y)P(X_{j_0}=dy)\\
= & \int_0^{+\infty}E(f(Ga(n-1,\lambda))P(X_{j_0}=dy) \\
= & E(f(Ga(n-1,\lambda))\int_0^{+\infty}P(X_{j_0}=dy) \\
= & E(f(Ga(n-1,\lambda))
\end{align}
Thus 
\begin{align}
E(f(\sum (X_i - X_{(1)}))) &= \frac{1}{n}\sum_{j=1}^nE(f(\sum_{i=1}^n (X_i - X_{(1)})) | X_{(1)} = X_j) \\
& = E(f(Ga(n-1,\lambda))
\end{align}
