Why does the pivotal quantity $g(X,θ)$ have to be a monotonic function of $θ$ for a fixed $X$?

Question

I'm learning about pivotal quantities, which are functions $g(X,θ)$ of the data $X$ and the parameter $θ$. One property they must satisfy is that $g(X,θ)$ is a monotonic function of $θ$ for a fixed $X$.

I'm told the reason for this, is to be able to manipulate inequalities.

For example, if we have a random sample

$X_1, X_2, ... , X_n$ from $N(θ,σ^2)$, we can construct a confidence interval for $θ$ by creating a pivotal quantity: $\frac{\bar X - \theta}{\sigma/\sqrt n}$

So that for example, $P(-1.96 \leq \frac{\bar X - \theta}{\sigma/\sqrt n} \leq 1.96) = 0.95$

Now, it is apparently this property of $\frac{\bar X - \theta}{\sigma/\sqrt n}$ being monotonic in $θ$ for a fixed $X$ that allows us to manipulate these inequalities and write this as

$P(\bar X - 1.96 \frac{σ}{\sqrt n} \leq θ \leq \bar X + 1.96 \frac{σ}{\sqrt n}) = 0.95$.

But I do not understand any part of this. Why does it need to be monotonic in $θ$? What does it mean for it to be monotonic in $θ$? Why does $X$ need to be fixed? Why does this let us manipulate the inequalities?

Answer 1

In a sense you want a sort of inverse to go from $\mathbb P(c_1 \le g(X, \theta) \le c_2) \le k$ to one of:

• (if monotonic increasing): $\mathbb P(h(X,c_1) \le \theta \le h(X,c_2)) \le k$ or
• (if monotonic decreasing): $\mathbb P(h(X,c_2) \le \theta \le h(X,c_1)) \le k$

and having $g(X,\theta)$ monotonic in $\theta$ for given $X$ enables such an inverse $h(X,c)$ to exist