# Complicated change of variables formula with CDF of normal distribution instead of diffeomorphism

I have a parameter $$\theta\in[0, 10]^4$$ and a variable $$z\in\mathbb{R}^m$$. Consider the following integral $$I(\theta, z) := \int F(\theta, z) \mathbb{I}(\|f(\theta, z)\| \leq 1) p(\theta, z) dz d\theta$$ Is it possible to rewrite it using the variables $$(\vartheta, z)$$ where $$\theta = G(\vartheta)$$ and $$G(\vartheta) = 10\cdot (\Phi(\vartheta_1), \Phi(\vartheta_2), \Phi(\vartheta_3), \Phi(\vartheta_4))$$ with $$\Phi$$ being the CDF of a standard normal distribution.

# Attempt

Informally, one could go ahead and simply replace $$\theta$$ with $$G(\vartheta)$$ and multiply the integrand by $$|J_G(\vartheta)|$$, the absolute determinant of the Jacobian of $$G$$ $$I(\theta, z) = \int F(G(\vartheta), z) \mathbb{I}(\|f(G(\vartheta), z)\| \leq 1) p(G(\vartheta), z) |J_G(\vartheta)| dz d\vartheta =: I(\vartheta, z)$$ However, this only works informally. The substitution I have just done is not correct according to the Change of Variables formula (e.g. see Billingsley). This is because the CDF of a normal distribution is not a diffeomorphism: it has no closed-form inverse.

Firstly, the lack of a closed form for the inverse does not preclude you from having a diffeomorphism. And indeed, the CDF $$\Phi$$ of a standard normal distribution is in fact a diffeomorphism, with $$\Phi'(x)=\frac{e^{-\frac{x^2}{2}}}{\sqrt{2\pi}}$$ and $$(\Phi^{-1})'(y)=\frac{1}{\Phi'(\Phi^{-1}(y))}\text{.}$$ Notice that both of these derivatives are continuous, and for that matter, nonzero.
The change of variables formula in Billingsley (Theorem 17.2) requires the mapping $$G$$ to continuously differentiable with nonzero Jacobian. Since the Jacobian matrix of $$G$$ is simply diagonal with entries equal to $$10\Phi'(\vartheta_i)$$, this follows.
If I were to nitpick I'd say maybe you should mention that you are restricting $$\theta$$ in your original integral to the open cube $$(0,10)^4$$, which is perfectly fine, since you are throwing away a set of $$0$$ measure.
Now, depending on what you are trying to do with this formula, you might have additional problems, since you don't have a closed form for $$\Phi$$, hence not for $$G$$, so to calculate the actual integral you might need to use numerical methods, depending on what $$F$$ and $$f$$ and $$p$$ are. But the formula itself is perfectly valid.
One final remark - it's a little shady notation-wise for you to define $$I(\vartheta,z)$$ as your new formula since you already used $$I$$ for your original integral $$I(\theta,z)$$.