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Here's a toy model that is truly linear

$$g(x) =\frac{1}{2}\sum_{i=1}^n \Lambda_i x_i^2, $$

where $\Lambda_i$ are i.i.d. $N(0,1)$ then

$$x(t)= \Big(e^{-t\Lambda_1} x_1(0),\dotsc, e^{-t\Lambda_n} x_n(0)\Big)$$

$$

$$U(t):= g(x(t))=\frac{1}{2}\sum_{i=1}^n \Lambda_ie^{-2t\Lambda_i} x_i(0)^2. $$

Denote by $N_T(U)$ the number of zeros of $U(t)$ on the interval $[0,T]$. Denote by $p_{U(t)}(u)$ the probability density of $U(t)$. Then the Kac-Rice formula state that $\newcommand{\bE}{\mathbb{E}}$

$$\bE\big[\; N_T(U)\;\big]=\int_0^T\bE\big[ \; |U'(t)|\;|\; U(t)=0\;\big] p_{U(t)}(0) dt, $$

where $\bE[-|-]$ denotes the conditional expectation.

Alternatively we have $\newcommand{\bP}{\mathbb{P}}$ $$ \bE\big[\; N_T(U)\;\big]=\underbrace{\bE\big[\; N_T(U)\;|\;g(U(T))>0\;\big]}_{=0}\bP\big[ g(U(T))>0\big]+\bE\big[\; N_T(U)\;|\;g(U(T))<0\;\big]\bP\big[ g(U(T))<0\big] $$ $$ =\bE\big[\; N_T(U)\;|\;g(U(T))<0\;\big]\bP\big[ g(U(T))<0\big] $$ $$ =\underbrace{\bE\big[\; N_T(U)\;|\;g(U(0))>0,g(U(T))<0\;\big]}_{=1}\bP\big[ g(U(0))>0,g(U(T))<0\big]+ \underbrace{\bE\big[\; N_T(U)\;|\;g(U(0))<0,g(U(T)) <0\;\big]}_{=0}\bP\big[ g(U(0))<0,g(U(T))>0\big] $$ $$ =\bP\big[ g(U(0))>0,g(U(T))<0\big]. $$

Here's a toy model that is truly linear

$$g(x) =\frac{1}{2}\sum_{i=1}^n \Lambda_i x_i^2, $$

where $\Lambda_i$ are i.i.d. $N(0,1)$ then

$$x(t)= \Big(e^{-t\Lambda_1} x_1(0),\dotsc, e^{-t\Lambda_n} x_n(0)\Big)$$

$$U(t):= g(x(t))=\frac{1}{2}\sum_{i=1}^n \Lambda_ie^{-2t\Lambda_i} x_i(0)^2. $$

Denote by $N_T(U)$ the number of zeros of $U(t)$ on the interval $[0,T]$. Denote by $p_{U(t)}(u)$ the probability density of $U(t)$. Then the Kac-Rice formula state that $\newcommand{\bE}{\mathbb{E}}$

$$\bE\big[\; N_T(U)\;\big]=\int_0^T\bE\big[ \; |U'(t)|\;|\; U(t)=0\;\big] p_{U(t)}(0) dt, $$

where $\bE[-|-]$ denotes the conditional expectation.

Alternatively we have $\newcommand{\bP}{\mathbb{P}}$ $$ \bE\big[\; N_T(U)\;\big]=\underbrace{\bE\big[\; N_T(U)\;|\;g(U(T))>0\;\big]}_{=0}\bP\big[ g(U(T))>0\big]+\bE\big[\; N_T(U)\;|\;g(U(T))<0\;\big]\bP\big[ g(U(T))<0\big] $$ $$ =\bE\big[\; N_T(U)\;|\;g(U(T))<0\;\big]\bP\big[ g(U(T))<0\big] $$ $$ =\underbrace{\bE\big[\; N_T(U)\;|\;g(U(0))>0,g(U(T))<0\;\big]}_{=1}\bP\big[ g(U(0))>0,g(U(T))<0\big]+ \underbrace{\bE\big[\; N_T(U)\;|\;g(U(0))<0,g(U(T)) <0\;\big]}_{=0}\bP\big[ g(U(0))<0,g(U(T))>0\big] $$ $$ =\bP\big[ g(U(0))>0,g(U(T))<0\big]. $$

Here's a toy model that is truly linear

$$g(x) =\frac{1}{2}\sum_{i=1}^n \Lambda_i x_i^2, $$

where $\Lambda_i$ are i.i.d. $N(0,1)$ then

$$x(t)= \Big(e^{-t\Lambda_1} x_1(0),\dotsc, e^{-t\Lambda_n} x_n(0)\Big)$$

so

$$U(t):= g(x(t))=\frac{1}{2}\sum_{i=1}^n \Lambda_ie^{-2t\Lambda_i} x_i(0)^2. $$

Denote by $N_T(U)$ the number of zeros of $U(t)$ on the interval $[0,T]$. Denote by $p_{U(t)}(u)$ the probability density of $U(t)$. Then the Kac-Rice formula state that $\newcommand{\bE}{\mathbb{E}}$

$$\bE\big[\; N_T(U)\;\big]=\int_0^T\bE\big[ \; |U'(t)|\;|\; U(t)=0\;\big] p_{U(t)}(0) dt, $$

where $\bE[-|-]$ denotes the conditional expectation.

Alternatively we have $\newcommand{\bP}{\mathbb{P}}$ $$ \bE\big[\; N_T(U)\;\big]=\underbrace{\bE\big[\; N_T(U)\;|\;g(U(T))>0\;\big]}_{=0}\bP\big[ g(U(T))>0\big]+\bE\big[\; N_T(U)\;|\;g(U(T))<0\;\big]\bP\big[ g(U(T))<0\big] $$ $$ =\bE\big[\; N_T(U)\;|\;g(U(T))<0\;\big]\bP\big[ g(U(T))<0\big] $$

Here's a toy model that is truly linear

$$g(x) =\frac{1}{2}\sum_{i=1}^n \Lambda_i x_i^2, $$

where $\Lambda_i$ are i.i.d. $N(0,1)$ then

$$x(t)= \Big(e^{-t\Lambda_1} x_1(0),\dotsc, e^{-t\Lambda_n} x_n(0)\Big)$$

$$

$$U(t):= g(x(t))=\frac{1}{2}\sum_{i=1}^n \Lambda_ie^{-2t\Lambda_i} x_i(0)^2. $$

Denote by $N_T(U)$ the number of zeros of $U(t)$ on the interval $[0,T]$. Denote by $p_{U(t)}(u)$ the probability density of $U(t)$. Then the Kac-Rice formula state that $\newcommand{\bE}{\mathbb{E}}$

$$\bE\big[\; N_T(U)\;\big]=\int_0^T\bE\big[ \; |U'(t)|\;|\; U(t)=0\;\big] p_{U(t)}(0) dt, $$

where $\bE[-|-]$ denotes the conditional expectation.

Alternatively we have $\newcommand{\bP}{\mathbb{P}}$ $$ \bE\big[\; N_T(U)\;\big]=\underbrace{\bE\big[\; N_T(U)\;|\;g(U(T))>0\;\big]}_{=0}\bP\big[ g(U(T))>0\big]+\bE\big[\; N_T(U)\;|\;g(U(T))<0\;\big]\bP\big[ g(U(T))<0\big] $$ $$ =\bE\big[\; N_T(U)\;|\;g(U(T))<0\;\big]\bP\big[ g(U(T))<0\big] $$ $$ =\underbrace{\bE\big[\; N_T(U)\;|\;g(U(0))>0,g(U(T))<0\;\big]}_{=1}\bP\big[ g(U(0))>0,g(U(T))<0\big]+ \underbrace{\bE\big[\; N_T(U)\;|\;g(U(0))<0,g(U(T)) <0\;\big]}_{=0}\bP\big[ g(U(0))<0,g(U(T))>0\big] $$ $$ =\bP\big[ g(U(0))>0,g(U(T))<0\big]. $$

Here's a toy model that is truly linear

$$g(x) =\frac{1}{2}\sum_{i=1}^n \Lambda_i x_i^2, $$

where $\Lambda_i$ are i.i.d. $N(0,1)$ then

$$x(t)= \Big(e^{-t\Lambda_1} x_1(0),\dotsc, e^{-t\Lambda_n} x_n(0)\Big)$$

so

$$U(t):= g(x(t))=\frac{1}{2}\sum_{i=1}^n \Lambda_ie^{-2t\Lambda_i} x_i(0)^2. $$

Denote by $N_T(U)$ the number of zeros of $U(t)$ on the interval $[0,T]$. Denote by $p_{U(t)}(u)$ the probability density of $U(t)$. Then the Kac-Rice formula state that $\newcommand{\bE}{\mathbb{E}}$

$$\bE\big[\; N_T(U)\;\big]=\int_0^T\bE\big[ \; |U'(t)|\;|\; U(t)=0\;\big] p_{U(t)}(0) dt, $$

where $\bE[-|-]$ denotes the conditional expectation.

Here's a toy model that is truly linear

$$g(x) =\frac{1}{2}\sum_{i=1}^n \Lambda_i x_i^2, $$

where $\Lambda_i$ are i.i.d. $N(0,1)$ then

$$x(t)= \Big(e^{-t\Lambda_1} x_1(0),\dotsc, e^{-t\Lambda_n} x_n(0)\Big)$$

so

$$U(t):= g(x(t))=\frac{1}{2}\sum_{i=1}^n \Lambda_ie^{-2t\Lambda_i} x_i(0)^2. $$

Denote by $N_T(U)$ the number of zeros of $U(t)$ on the interval $[0,T]$. Denote by $p_{U(t)}(u)$ the probability density of $U(t)$. Then the Kac-Rice formula state that $\newcommand{\bE}{\mathbb{E}}$

$$\bE\big[\; N_T(U)\;\big]=\int_0^T\bE\big[ \; |U'(t)|\;|\; U(t)=0\;\big] p_{U(t)}(0) dt, $$

where $\bE[-|-]$ denotes the conditional expectation.

Alternatively we have $\newcommand{\bP}{\mathbb{P}}$ $$ \bE\big[\; N_T(U)\;\big]=\underbrace{\bE\big[\; N_T(U)\;|\;g(U(T))>0\;\big]}_{=0}\bP\big[ g(U(T))>0\big]+\bE\big[\; N_T(U)\;|\;g(U(T))<0\;\big]\bP\big[ g(U(T))<0\big] $$ $$ =\bE\big[\; N_T(U)\;|\;g(U(T))<0\;\big]\bP\big[ g(U(T))<0\big] $$