# Confidence interval of transformed random variable

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.

• 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] \$. Apr 5, 2022 at 1:48
• *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. Apr 5, 2022 at 1:48
• @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.
– null
Apr 5, 2022 at 6:48