Unimodal Distribution's Shortest Confidence Interval
Theorem
Definition of Unimodal Functions
If there exists a mode $x^{\ast}$ such that the function $f : \mathbb{R} \to \mathbb{R}$ does not decrease in $x \le x^{\ast}$ and does not increase in $x \ge x^{\ast}$, then $f$ is called unimodal. Especially, if the probability density function of $f$ is unimodal, we call that probability distribution a unimodal distribution.
Shortest Confidence Interval
Let’s consider $f(x)$ to be a unimodal probability density function. If the interval $[a,b]$ satisfies the following three conditions:
- (i): $\displaystyle \int_{a}^{b} f(x) dx = 1 - \alpha$
- (ii): $f(a) = f(b) > 0$
- (iii): $a \le x^{\ast} \le b$
Then, $[a,b]$ is the shortest among all intervals that satisfy $i$.
Explanation
The idea of drawing a certain line $y = k(\alpha)$ that makes $f(a) = f(b)$, to capture the confidence interval, comes from a similar concept in Bayesian highest posterior density confidence interval.
Proof
Strategy: The proof of the theorem itself does not necessarily require probabilistic concepts, but can be sufficiently covered with basic calculus knowledge, satisfying various properties of the probability density function. The formal proof employs reductio ad absurdum. Assuming $b’-a’ < b-a$ exists a shorter interval $\left[ a’, b’ \right]$ than $[a,b]$ while satisfying the condition (i) mentioned in the theorem, leads to $\int_{a’}^{b’} f(x) dx < 1-\alpha$ in all cases, indicating a contradiction.
Case 1. $b’ \le a$
$$ \begin{align*} \int_{a’}^{b’} f(x) dx \le& f \left( b’ \right) \left( b’ - a’ \right) & \because x \le b’ \le x^{\ast} \implies f(x) \le f \left( b’ \right) \\ \le& f \left( b \right) \left( b’ - a’ \right) & \because b’ \le a \le x^{\ast} \implies f \left( b’ \right) \le f \left( a \right) \\ <& f(a) (b-a) & \because b’-a’ < b-a \& f(a) > 0 \\ \le& \int_{a}^{b} f(x) dx & \because (ii), (iii), \exists x^{\ast} \implies f(x) \ge f(a) , \forall x \in [a,b] \\ =& 1 - \alpha \end{align*} $$
Case 2. $b’ > a$
If $b \le b '$, then originally $b’-a’ \ge b-a$, so only consider the case where $a’ \le a < b’ < b$. $$ \begin{align*} \int_{a’}^{b’} f(x) dx \le& f \left( b’ \right) \left( b’ - a’ \right) + \left[ \int_{a’}^{a} f(x) dx - \int_{b’}^{b} f(x) dx \right] \\ =& (1-\alpha) + \left[ \int_{a’}^{a} f(x) dx - \int_{b’}^{b} f(x) dx \right] \\ =& (1-\alpha) + R \end{align*} $$ Thus, it is sufficient to show $R := \left[ \int_{a’}^{a} f(x) dx - \int_{b’}^{b} f(x) dx \right] < 0$. $$ \begin{align*} \int_{a’}^{a} f(x) dx \le& f(a) \left( a - a’ \right) \\ \int_{b’}^{b} f(x) dx \ge& f(b) \left( b - b’ \right) \end{align*} $$ Therefore, $$ \begin{align*} R =& \int_{a’}^{a} f(x) dx - \int_{b’}^{b} f(x) dx \\ \le & f(a) \left( a - a’ \right) - f(b) \left( b - b’ \right) \\ =& f(a) \left[ \left( a - a’ \right) - \left( b - b’ \right) \right] & \because f(a) = f(b) \\ =& f(a) \left[ \left( b’ - a’ \right) - \left( b - a \right) \right] \end{align*} $$ since $f(a) > 0$ and it was assumed that $\left( b’ - a’ \right) < \left( b-a \right)$, $R < 0$ follows.
In any case, since $\int_{a’}^{b’} f(x) dx < 1 - \alpha$ follows, the assumption is contradicted, hence there does not exist a shorter interval than $[a,b]$ that satisfies condition (i).
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