Revisions to Time-derivative of integral over sub-level set $s(t) := \int_{f^{-1}((-\infty,t])}p(x)dx$

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edited Mar 3, 2022 at 12:24

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Update: Solution for Example 1

It seems the right apparatus for studying such problems is Malliavin 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 "$f(x)$, $x \sim N(m,\Sigma)$" in the question can be written as

$$ X := Z(\widetilde w)-c+\sum_{j=1}^d w_j m_j, $$ where $\widetilde w := \Sigma^{1/2}w \in \mathbb R^d$. Note that $\|\widetilde w\| = \|w\|_\Sigma := \sqrt{w^\top \Sigma w}$.

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

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

Update: Solution for Example 1

It seems the right apparatus for studying such problems is Malliavin 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 "$f(x)$, $x \sim N(m,\Sigma)$" in the question can be written as

$$ X := Z(\widetilde w)-c+\sum_{j=1}^d w_j m_j, $$ where $\widetilde w := \Sigma^{1/2}w \in \mathbb R^d$. Note that $\|\widetilde w\| = \|w\|_\Sigma := \sqrt{w^\top \Sigma w}$.

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

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

Update: Solution for Example 1

It seems the right apparatus for studying such problems is Malliavin 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 "$f(x)$, $x \sim N(m,\Sigma)$" in the question can be written as

$$ X := Z(\widetilde w)-c+\sum_{j=1}^d w_j m_j, $$ where $\widetilde w := \Sigma^{1/2}w \in \mathbb R^d$. Note that $\|\widetilde w\| = \|w\|_\Sigma := \sqrt{w^\top \Sigma w}$.

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

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

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edited Mar 3, 2022 at 9:12

dohmatob

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Update: Solution for Example 1

It seems the right apparatus for studying such problems is Malliavin 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)$$x \sim N(m,\Sigma)$" in the question can be written as

$$ X := Z(\widetilde w)-c+\sum_{j=1}^d w_j m_j, $$ where $\widetilde w := \Sigma^{1/2}w$$\widetilde w := \Sigma^{1/2}w \in \mathbb R^d$. Note that $\|\widetilde w\| = \|w\|_\Sigma := \sqrt{w^\top \Sigma w}$.

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

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

Update: Solution for Example 1

It seems the right apparatus for studying such problems is Malliavin 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

$$ X := Z(\widetilde w)-c+\sum_{j=1}^d w_j m_j, $$ where $\widetilde w := \Sigma^{1/2}w$.

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

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

Update: Solution for Example 1

It seems the right apparatus for studying such problems is Malliavin 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 "$f(x)$, $x \sim N(m,\Sigma)$" in the question can be written as

$$ X := Z(\widetilde w)-c+\sum_{j=1}^d w_j m_j, $$ where $\widetilde w := \Sigma^{1/2}w \in \mathbb R^d$. Note that $\|\widetilde w\| = \|w\|_\Sigma := \sqrt{w^\top \Sigma w}$.

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

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

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edited Mar 2, 2022 at 17:57

dohmatob

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Update: Solution for Example 1

It seems the right apparatus for studying such problems is MallavianMalliavin 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

$$ X := Z(\widetilde w)-c+\sum_{j=1}^d w_j m_j, $$ where $\widetilde w := \Sigma^{1/2}w$.

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

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

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

$$ X := Z(\widetilde w)-c+\sum_{j=1}^d w_j m_j, $$ where $\widetilde w := \Sigma^{1/2}w$.

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

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

Update: Solution for Example 1

It seems the right apparatus for studying such problems is Malliavin 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

$$ X := Z(\widetilde w)-c+\sum_{j=1}^d w_j m_j, $$ where $\widetilde w := \Sigma^{1/2}w$.

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

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

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edited Mar 2, 2022 at 16:09

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edited Mar 2, 2022 at 15:59

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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

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edited Mar 2, 2022 at 15:51

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added 96 characters in body

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edited Mar 2, 2022 at 15:39

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created Mar 2, 2022 at 15:33

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Update: Solution for Example 1

Update: Solution for Example 1

Update: Solution for Example 1

Update: Solution for Example 1

Update: Solution for Example 1

Update: Solution for Example 1

Update: Solution for Example 1

Update: Solution for Example 1

Update: Solution for Example 1