Let $X$ be a Normal random variable of mean $\mu$ and variance $\sigma^2$. Also let $g\colon \mathbb{R}\to\mathbb{R}_{>0}$ be a positive strictly increasing bijective function.

I would like to find the symmetric 0.95 "confidence interval" of $g(X)$ centred by its mean, that is, find $a$ such that

$$ \mathbb{P}\big(z - a \leq g(X) \leq z + a\big)=0.95, $$

where $z = \mathbb{E}[g(X)]$ is the mean. Suppose that such interval exists.

However, I failed to compute the interval in closed-form. Observe that

$$ \mathbb{P}\big(z - a \leq g(X) \leq z + a\big) = \mathbb{P}\big(g^{-1}(z - a) \leq X \leq g^{-1}(z + a)\big), $$

so essentially, we want to find the $a$ by solving

$$ \mathrm{CDF}(g^{-1}(z + a)) - \mathrm{CDF}(g^{-1}(z - a)) = 0.95, $$

where $\mathrm{CDF}$ is the CDF of $X$. But how to find a closed-form solution/approximation to this equation? By "approximation", I mean an analogy of $\mu \pm 1.96\sigma$ for $\approx0.95$ of Normal.

  • $\begingroup$ Your observation implicitly assumes that $\ g\ $ is increasing. However, bijectivity of $\ g\ $ is insufficient to guarantee that. In general, $$ \mathbb{P}\big( z-a\le g(X)\le z+a\big)=\mathbb{P}\big(X\in g^{-1}\big([z-a,z+a] \big)\big)\ , $$ but $\ g^{-1}\big([z-a,z+a]\big)\ $ is not even necessarily an interval, let alone the interval $\ \big[g^{-1}(z-a),g^{-1}(z+a)\big] \ $. $\endgroup$ Apr 5, 2022 at 1:48
  • $\begingroup$ *If $\ g\ $ is continuous*, it must be *either* strictly increasing *or* strictly decreasing, but if it's the latter, you get $$ \mathbb{P}\big( z-a\le g(X)\le z+a\big)=\mathbb{P}\big(g^{-1}(z+a)\le X\le g^{-1}(z-a)\big)\ , $$ rather than your identity. $\endgroup$ Apr 5, 2022 at 1:48
  • $\begingroup$ @lonzaleggiera Indeed, it's hard to verify there exists such $a$ when $g$ is only known bijective. I have now corrected the question as per your suggestion. $\endgroup$
    – null
    Apr 5, 2022 at 6:48


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