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Time-derivative of integral over sub-level set 𝑠(𝑡):=∫𝑓−1((−∞,𝑡])𝑝(𝑥)𝑑𝑥 Asked today Active today Viewed 18 timesadded 96 characters in body
dohmatob
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Update: Solution for Example 1

It seems the right apparatus for studying such problems is Mallavian calculus, though I'm not at all yet familiar with the tool.


So, let $G = (g_1,\ldots,g_d)$ be a standard Gaussian random vector in $\mathbb R^d$ and consider the Gaussian process defined by $Z(a) := \langle G,a\rangle = \sum_j a_j g_j$, for any $a \in \mathbb R^d$. The random variable "$f(x)$, $x \sim N(m,\sigma^2 I_d)$" in the question can be written as $X := \sigma Z(w)-c+\sum_{j=1}^d w_j m_j$, where $w$ is a deterministic unit-vector in $\mathbb R^d$ and $c$ is a deterministic scalar.

Let $S(X):=\delta(DX/\|DX\|^2)$, where $D$ (resp. $\delta$) is the Mallavian derivative (resp. Skorohod integral) operator. A simply computation gives $DX = \sigma w$, and so $$ S(X) = \sum_j w_j Z_j / (\sigma \|w\|^2) = (1/\sigma)Z(w) \overset{Law}{=}(1/\sigma)g_1. $$

Therefore, thanks to Proposition 2.1.1 of Introduction to Mallavian Calculus, one obtains

$$ s_f'(t) = \mathbb E[1_{X \le t} S(X)] = \frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{(t-f(m))/\sigma}ze^{-z^2/2}dz = \varphi(\frac{t-f(m)}{\sigma}), $$ where $\varphi$ is the standard Gaussian pdf. This exactly matches the computation carried out in the comments section to the question above.

dohmatob
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