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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.

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1 Answer 1

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The substitution in your attempt looks fine, and your concerns about it are misplaced.

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 $I(\theta,z)$.

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