Find the value of $F_{X_{1/6},X_1}(0,1/4)$ Where $X_t=U\cos(\pi t)$, U is uniform $(-1,1)$.
My answer is $$F_{X_{1/6},X_1}(0,1/4)=P(U\cos(\frac{\pi}{6})\leq 0, U\cos(\pi)\leq 1/4)\\\qquad\quad=P(\frac{\sqrt{3}}{2}U\leq 0,-U\leq 1/4)\\\qquad\quad=P(U\leq 0,U\geq -1/4)=P(-1/4\leq U\leq0)\\\qquad\quad=P(-1/4\leq U\leq 0)=1/8$$
Is that correct?