Random Variable X and Y has a joint probability density function.
$$f_{X, Y} (x, y) =\begin{cases}
c(x + 3y)& 5 \leq x \leq y, 6 \leq y \leq10\\
0 & \text{otherwise }\\
\end{cases}
$$
(a) Find $f_{X | Y}(x | y)$
(b) $P(x \leq 5 | Y = 9)$
My attempt:
$f_{X | Y}(x | y) = \frac{f_{X, Y}(x, y)}{f_Y(y)}$
$$f_Y(y) = c\int_{5}^{y}(x+3y)dx = c/2 (-25 – 30 y + 7 y^2)$$
for $f_Y(y)$ has support $6 \leq y \leq 10$, 0 otherwise
$$f_{X | Y}(x | y) = \frac{0.5c(x+3y)}{(-25 – 30 y + 7 y^2)}$$
$f_{X | Y}(x | y)$ has support the same as the joint probability function
(b)
$f_{X | Y}(x | y = 9) = \frac{x+27}{272}$
$$P(x \leq 5 | Y = 9) = \int_{5}^{?} f_{X | Y}(x | y = 9)dx = $$
Not sure
Best Answer
(a) Everything is done okay, save that the final answer should be $$f_{X\mid Y}(x\mid y)=\dfrac{2(x+3y)}{7y^2-30y-25}\mathbf 1_{5\leqslant x\leqslant y, 6\leqslant y\leqslant 10}$$
(b) Whatever value of $Y$, the support for $X$ indicates that it is impossible to have $X\leqslant 5$. $$\mathsf P(X{\leqslant}5\mid Y{=}9)=0$$