Exercise
May 05'23
Answer
Solution: E
[[math]]
\begin{align*}
\operatorname{P}[1 \lt Y \lt 3 | X = 2] = \int_1^3 \frac{f(2,y)}{f_x(2)} dy \\
f(2,y) = \frac{2}{4(2-1)} y^{-(4-1)/2 -1} = \frac{1}{2}y^{-3} \\
f_x(2) = \int_1^{\infty} \frac{1}{2}y^{-3} dy = - \frac{1}{4}y^{-2} \Big |_0^{\infty} = \frac{1}{4}
\end{align*}
[[/math]]
Finally,
[[math]]
\operatorname{P}[1 \lt Y \lt 3 | X = 2] = \frac{\int_1^3 \frac{1}{2} y^{-3} dy}{\frac{1}{4}} = -y^2 \Big |_1^3 = 1- \frac{1}{9} = \frac{8}{9}.
[[/math]]