Pivot Transformation for $N(\mu, \mu^2)$ So I'm currently learning about pivot transformations in my statistics course (so far for the purpose of constructing confidence intervals), and was curious as to what pivot transformation you'd use for a sequence of iid random variables that are $N(\mu, \mu^2)$. Because I know that in some applied scenarios it would be more realistic for the measurement variability to increase as the measurement itself increases. Like a scientist might take more care when measuring the size of a smaller unit (say, 1cm in length) but less care when measuring the size of a larger unit (say, 10cm in length). So instead of using a $N(\mu, \sigma^2)$ distribution to model such measurements, it might be better to consider the model $N(\mu, \mu^2)$ in which the variance increases with the mean. I know there'd be multiple different pivots you could use, but is there one that is easier to construct a confidence interval with?
 A: Let $X_9, X_2, \dots, X_n$ be a random sample from $\mathsf{Norm}(\mu, \mu),$ where the second argument is the population standard deviation.
Consider $Z = \frac{\bar X -\mu}{\sqrt{\mu^2/n}} \sim \mathsf{Norm}(0,1),$ so that $P(-1.96 \le Z \le 1.96) = 0.95.$
Then, by manipulating inequalities,
$$P\left(\frac{\bar X}{1+1.96/\sqrt{n}} \le\mu\le \frac{\bar X}{1-1.96/\sqrt{n}}\right) = 0.95,$$
So that a 95% confidence interval (CI) for $\mu$ is of the form
$$\left(\frac{\bar X}{1+1.96/\sqrt{n}},\; \frac{\bar X}{1-1.96/\sqrt{n}}\right).$$
Check using simulation: Try using such CIs on a million samples of size $n = 25$ from
$\mathsf{Norm}(50,50).$ Very nearly 95% of them cover $\mu=50.$
set.seed(2020)
m = 10^6;  n = 25;  x = rnorm(m*n, 50,50)
DTA = matrix(x, nrow=m)    # each ros a sample of 25
a = rowMeans(DTA)          # m sample means
c1 = 1 + 1.96/sqrt(n);  c2 = 1 - 1.96/sqrt(n)
lcl = a/c1; ucl = a/c2     # confidence limits
mean(ucl-lcl)
[1] 46.31934               # mean length of CIs
mean(lcl < 50 & ucl > 50)  # coverage probability
[1] 0.949769               # aprx 0.95


Note: Exponential distributions have mean and standard deviation equal. Let $\mu$ be the mean (rate $\lambda = 1/\mu.$
Then a pivotal quantity is $\bar X/\mu \sim \mathsf{Gamma}(n,n),$
where arguments are shape and rate. Then a 95% CI for $\mu$ is
of the form $(\bar X/U, \bar X/L),$ where $L$ and $U$ cut,
respectively, probability $0.025$ from the lower and upper tails
of $\mathsf{Gamma}(n,n).$
For example, if x is a random sample of $n=25$ from $\mathsf{Exp}(\lambda = 1/50),$ then a 95% CI $(36.30,80.12)$ for $\mu = 1/\lambda$ can be found as follows:
set.seed(929)
x = rexp(25, 1/50)
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  2.026  13.819  28.192  51.852  45.281 367.177 
mean(x)/qgamma(c(.975,.025), 25, 25)
[1] 36.30071 80.12408

