Let $Y$ have probability density function $$f_{Y}(y)=\left\{\begin{array}{ll}
\frac{2(\theta-y)}{\theta^{2}}, & 0<y<\theta \\
0, & \text { elsewhere }
\end{array}\right.$$ a. Show that $Y$ has distribution function $$F_{Y}(y)=\left\{\begin{array}{ll}
0, & y \leq 0 \\
\frac{2 y}{\theta}-\frac{y^{2}}{\theta^{2}}, & 0<y<\theta \\
1, & y \geq \theta
\end{array}\right.$$ b. Show that $Y / \theta$ is a pivotal quantity.
c. Use the pivotal quantity from part (b) to find a $90 \%$ lower confidence limit for $\theta$