Pivots for exponential distribution In my mathematical statistics course I got the following problem:
Let $X_{1}, ... , X_{n}$ be an i.i.d. sample from the Exp($\lambda$) distribution. Construct two different pivots
and two confidence intervals for $\lambda$ (of confidence level $1-\alpha$) based on these pivots. Use that if $X \sim Exp(\lambda) \Rightarrow \lambda X \sim Exp(1)$ in combination with the following two facts (do not prove them):
(1) $X_{(1)}$ $\sim$ Exp($n\lambda$),
(2) if $Y_{1},...,Y_{n}$ are i.i.d. with Exp($1$) distribution, then $2\sum\limits_{i=1}^{n}Y_{i} \sim \chi^{2}_{2n}$.
So far this is what I got:
We will first use $2 \sum\limits_{i=1}^{n} Y_{i} \sim \chi^{2}_{2n}$. We now use $y=\lambda x$, so we get
\begin{equation*}
\begin{split}
    2 \sum\limits_{i=1}^{n} Y_{i} \sim \chi^{2}_{2n} &\Rightarrow 2 \lambda \sum\limits_{i=1}^{n} x_{i} = 2 \lambda \bar{x} n \sim \chi^{2}_{2n}\\
\end{split}
\end{equation*}
Thus
\begin{equation*}
    \begin{split}
        \chi^{2}_{2n,\alpha / 2} &\leq 2\lambda \bar{x} n \leq \chi^{2}_{2n, 1-\alpha /2}\\
        \frac{\chi^{2}_{2n,\alpha / 2}}{2 \bar{x} n} &\leq  \lambda \leq \frac{\chi^{2}_{2n, 1-\alpha /2}}{2 \bar{x} n}
    \end{split}
\end{equation*}
So this is my first pivot. I have no idea whether this is okay, and I also have no clue how to proceed for the second pivot. Any suggestions are welcome!
 A: Your derivation of the confidence interval for rate $\lambda$ using the sample mean
and the chi-squared distribution is correct. You can verify the result at Wikipedia under 'Confidence intervals'. This is a commonly used CI for $\lambda$ using the sufficient statistic.
The derivation of the CI for $\lambda$ using the minimum $X_{(1)} = V$ is similar.
The quantity $nV\lambda \sim \mathsf{Exp}(1).$ For example, 
$$P(L_e < nV\lambda < U_e) = P\left(\frac{L_e}{nV} < \lambda < \frac{U_e}{nV}\right) = 0.95,$$ 
where $L_e$ and $U_e$ cut probability 0.25 from the lower and upper tails of 
$\mathsf{Exp}(1),$ respectively.

Below is a simulation in R with a million iterations, for $n = 20, \lambda = 0.2.$
m = 10^6;  n = 20;  lam = .2;  x = rexp(m*n, lam)
MAT = matrix(x, nrow=m)  # m x n matrix, each row a sample of size n

s = rowSums(MAT)         # one million sample sums
mean(s); sd(s)
[1] 100.0436             # aprx E(S) = 100
[1] 22.36627
L.c = qchisq(.025,2*n); U.c = qchisq(.975,2*n); L.c; U.c
[1] 24.43304
[1] 59.34171
mean(L.c/(2*s) < lam & U.c/(2*s) > lam)
[1] 0.950114             # CI covers rate with probability 95% 

v = apply(MAT, 1, min)   # one million sample minimums
mean(v); sd(v)
[1] 0.2501739            # aprx E(V) = 1/4
[1] 0.2503302
L.e = qexp(.025); U.e = qexp(.975); L.e; U.e
[1] 0.02531781
[1] 3.688879
mean(L.e/(n*v) < lam & U.e/(n*v) > lam)
[1] 0.950174             # CI covers rate with probability 95%

The histogram in the left panel below shows the simulated chi-squared distribution of $2n\bar X\lambda = 2\lambda\sum_iX_i$ and the right panel
shows the simulated exponential distribution of $nV\lambda = n\lambda X_{(1)}.$
Exact density functions are shown as black curves.

R code to make the figure follows:
par(mfrow=c(1,2))
  hist(2*s*lam, prob=T, col="skyblue2", main="Pivot Using Sum")
    curve(dchisq(x,2*n), add=T, lwd=2); abline(v=c(L.c, U.c), col="red")
  hist(n*lam*v, prob=T, ylim=0:1, col="skyblue2", main="Pivot Using Minimum")
    curve(dexp(x), add=T, lwd=2); abline(v=c(L.e, U.e), col="red")
par(mfrow=c(1,1))

