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$\newcommand\ep\epsilon\newcommand\si\sigma\newcommand\th\theta$In your concrete example, $$p_X(t_0,\ep)=P\Big(m_1<\frac VU<m_2\Big),$$ where $$m_1:=\min_{t\in[t_0-\ep,t_0+\ep]}r(t),\quad m_2:=\max_{t\in[t_0-\ep,t_0+\ep]}r(t),\quad r(t):=-\frac t{(1-t^2)^{1/2}};$$ this follows because any continuous function maps any compact interval onto a compact interval.

Letting now $$\th_j:=\arctan m_j$$ and using the rotational symmetry of the distribution of $(U,V)$, we get $$p_X(t_0,\ep)= \left\{ \begin{aligned} \frac{|\th_2-\th_1|}{\pi} &\text{ if }\th_1\th_2\ge0, \\ \frac{|\th_2+\th_1|}{\pi} &\text{ if }\th_1\th_2\le0. \end{aligned} \right. $$

$\newcommand\ep\epsilon\newcommand\si\sigma\newcommand\th\theta$In your concrete example, $$p_X(t_0,\ep)=P\Big(m_1<\frac VU<m_2\Big),$$ where $$m_1:=\min_{t\in[t_0-\ep,t_0+\ep]}r(t) =r(t_0+\ep),\quad m_2:=\max_{t\in[t_0-\ep,t_0+\ep]}r(t) =r(t_0-\ep),\quad r(t):=-\frac t{(1-t^2)^{1/2}};$$ this follows because $r$ is a continuous decreasing function on the interval $(-1,1)$.

Letting now $$\th_j:=\arctan m_j$$ and using the rotational symmetry of the distribution of $(U,V)$, we get $$p_X(t_0,\ep)=\frac{\th_2-\th_1}{\pi}. $$

$\newcommand\ep\epsilon\newcommand\si\sigma\newcommand\th\theta$In your concrete example, $$p_X(t_0,\ep)=P\Big(m_1<\frac VU<m_2\Big),$$ where $$m_1:=\min_{t\in[t_0,t_0+\ep]}r(t),\quad m_2:=\max_{t\in[t_0,t_0+\ep]}r(t),\quad r(t):=-\frac t{(1-t^2)^{1/2}};$$ this follows because any continuous function maps any compact interval onto a compact interval.

Letting now $$\th_j:=\arctan m_j$$ and using the rotational symmetry of the distribution of $(U,V)$, we get $$p_X(t_0,\ep)= \left\{ \begin{aligned} \frac{|\th_2-\th_1|}{\pi} &\text{ if }\th_1\th_2\ge0, \\ \frac{|\th_2+\th_1|}{\pi} &\text{ if }\th_1\th_2\le0. \end{aligned} \right. $$

$\newcommand\ep\epsilon\newcommand\si\sigma\newcommand\th\theta$In your concrete example, $$p_X(t_0,\ep)=P\Big(m_1<\frac VU<m_2\Big),$$ where $$m_1:=\min_{t\in[t_0-\ep,t_0+\ep]}r(t),\quad m_2:=\max_{t\in[t_0-\ep,t_0+\ep]}r(t),\quad r(t):=-\frac t{(1-t^2)^{1/2}};$$ this follows because any continuous function maps any compact interval onto a compact interval.

Letting now $$\th_j:=\arctan m_j$$ and using the rotational symmetry of the distribution of $(U,V)$, we get $$p_X(t_0,\ep)= \left\{ \begin{aligned} \frac{|\th_2-\th_1|}{\pi} &\text{ if }\th_1\th_2\ge0, \\ \frac{|\th_2+\th_1|}{\pi} &\text{ if }\th_1\th_2\le0. \end{aligned} \right. $$

$\newcommand\ep\epsilon\newcommand\si\sigma\newcommand\th\theta$In your concrete example, $$p_X(t_0,\ep)=P\Big(m_1<\frac VU<m_2\Big),$$ where $$m_1:=\min_{t\in[t_0,t_0+\ep]}r(t),\quad m_2:=\max_{t\in[t_0,t_0+\ep]}r(t),\quad r(t):=-\frac t{(1-t^2)^{1/2}};$$ this follows because any continuous function maps any compact interval onto a compact interval.

Letting now $$\th_j:=\arctan m_j$$ and using the rotational symmetry of the distribution of $(U,V)$, we get $$p_X(t_0,\ep)= \left\{ \begin{aligned} \frac{|\th_2-\th_1|}{\pi} &\text{ if }\th_1\th_2\ge0, \\ \frac{|\th_2+\th_1|}{\pi} &\text{ if }\th_1\th_2\le0. \end{aligned} \right. $$ Of course, the result does not depend on $\si$.

$\newcommand\ep\epsilon\newcommand\si\sigma\newcommand\th\theta$In your concrete example, $$p_X(t_0,\ep)=P\Big(m_1<\frac VU<m_2\Big),$$ where $$m_1:=\min_{t\in[t_0,t_0+\ep]}r(t),\quad m_2:=\max_{t\in[t_0,t_0+\ep]}r(t),\quad r(t):=-\frac t{(1-t^2)^{1/2}};$$ this follows because any continuous function maps any compact interval onto a compact interval.

Letting now $$\th_j:=\arctan m_j$$ and using the rotational symmetry of the distribution of $(U,V)$, we get $$p_X(t_0,\ep)= \left\{ \begin{aligned} \frac{|\th_2-\th_1|}{\pi} &\text{ if }\th_1\th_2\ge0, \\ \frac{|\th_2+\th_1|}{\pi} &\text{ if }\th_1\th_2\le0. \end{aligned} \right. $$