$X_1,X_2,\dots,X_n \overset{\text{iid}}{\sim} N(\mu,\sigma^2)$. Derive a confidence interval for $\mu+\sigma$ and $\mu/\sigma$ I know how to find confidence interval for each of the parameters  $\mu$ and $\sigma$ resectively but stuck in finding for the parametric functions above.
 A: Suppose $\mu$ is estimated by $\bar{X}$ and $\sigma$ is estimated with $S=\sqrt{\sum_{i=1}^n(X_i-\bar{X})^2/(n-1)}$. For $\mu/\sigma$, the estimator $\bar{X}/S$ happens to follow a scaled non-central $t$-distribution, with d.f. $n-1$, ncp $\sqrt{n}\mu/\sigma$, and scale $1/\sqrt{n}$. With some simple algebra we can work out a $95\%$ CI as: $$\frac{1}{\sqrt{n}}\left(t_{0.025,n-1,\sqrt{n}\mu/\sigma},\quad t_{0.975,n-1,\sqrt{n}\mu/\sigma}\right).$$
For $\mu+\sigma$ it's much trickier, although $\bar{X}\sim N(\mu, \sigma^2/n)$ and $S\sim(\sigma/\sqrt{n-1})\chi_{n-1}$ are independent so in principle you can work out the sampling distribution "straightforward", but the derivation will be messy I imagine. Bootstrapping is recommended.
Here I will give a R example in computing CIs with bootstrap:
library(boot)

# statistics to bootstrap
SUM = function(data, ind) {
  mean(data[ind]) + sd(data[ind])
}
RATIO = function(data, ind) {
  mean(data[ind]) / sd(data[ind])
}

# simulate data
mu = 2
sigma = 9
n = 1000
X = rnorm(n, mu, sigma)

# bootstrap
X.boot1 = boot(data = X, statistic = SUM, R = 10000)
X.boot2 = boot(data = X, statistic = RATIO, R = 10000)

boot.ci(X.boot1, type = c("norm", "basic", "bca"))
boot.ci(X.boot2, type = c("norm", "basic", "bca"))

# theoretical result
ncp = sqrt(n) * mu / sigma
c(qt(0.025, n - 1, ncp) / sqrt(n), qt(0.975, n - 1, ncp) / sqrt(n))

Comparison for the CIs of $\mu/\sigma$, kinda close:
> boot.ci(X.boot2, type = c("norm", "basic", "bca"))
BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
Based on 10000 bootstrap replicates

CALL : 
boot.ci(boot.out = X.boot2, type = c("norm", "basic", "bca"))

Intervals : 
Level      Normal              Basic                BCa          
95%   ( 0.1742,  0.3004 )   ( 0.1731,  0.3005 )   ( 0.1739,  0.3009 )  
Calculations and Intervals on Original Scale
> c(qt(0.025, n - 1, ncp), qt(0.975, n - 1, ncp)) / sqrt(n)
[1] 0.1598871 0.2855220

A: Assuming both $\mu,\sigma$ are unknown, we can construct a Bonferroni confidence region for $(\mu,\sigma)$:
With usual notation let $$L_1(\mathbf X)=\overline X-t_{\alpha/4,n-1}\frac{S}{\sqrt n}\quad,\quad L_2(\mathbf X)= \overline X+t_{\alpha/4,n-1}\frac{S}{\sqrt n}$$ and 
$$L_3(\mathbf X)=\sqrt{\frac{(n-1)S^2}{\chi^2_{\alpha/4,n-1}}}\quad,\quad L_4(\mathbf X)=\sqrt{\frac{(n-1)S^2}{\chi^2_{1-\alpha/4,n-1}}}$$
Then, $$P_{\mu}\left[\mu\in (L_1,L_2)\right]=1-\frac{\alpha}{2}\quad\forall \,\mu\quad\text{ and }\quad P_{\sigma}\left[\sigma\in (L_3,L_4)\right]=1-\frac{\alpha}{2}\quad\forall\,\sigma$$
Combining both probabilities, we get $$P_{\mu,\sigma}\left[L_1<\mu<L_2,L_3<\sigma<L_4\right]\ge 1-\alpha\quad\forall\,\mu,\sigma\tag{*}$$
That is, $(L_1,L_2)\times (L_3,L_4)$ is a two-sided (joint) confidence region for $(\mu,\sigma)$ with confidence coefficient at least $1-\alpha$.
It follows from $(*)$ that $$P_{\mu,\sigma}\left[L_1+L_3<\mu+\sigma<L_2+L_4\right]\ge 1-\alpha\quad\forall\,\mu,\sigma$$
I guess something similar can be said for a probability statement involving $\mu/\sigma$.
