Cfrgtkky

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($nlambda$),

(2) if $Y_{1},...,Y_{n}$ are i.i.d. with Exp($1$) distribution, then $2sumlimits_{i=1}^{n}Y_{i} sim chi^{2}_{2n}$.

So far this is what I got:

We will first use $2 sumlimits_{i=1}^{n} Y_{i} sim chi^{2}_{2n}$. We now use $y=lambda x$, so we get
begin{equation*}
begin{split}
2 sumlimits_{i=1}^{n} Y_{i} sim chi^{2}_{2n} &Rightarrow 2 lambda sumlimits_{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 2lambda 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!

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 $nVlambda sim mathsf{Exp}(1).$ For example,
$$P(L_e < nVlambda < U_e) = Pleft(frac{L_e}{nV} < lambda < frac{U_e}{nV}right) = 0.95,$$
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 $2nbar Xlambda = 2lambdasum_iX_i$ and the right panel
shows the simulated exponential distribution of $nVlambda = nlambda X_{(1)}.$
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))