Pivot Transformation for $N(\mu, \mu^2)$

Question

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?

Answer 1

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

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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