Random Variable X and Y has a joint probability density function.
$$f_{X, Y} (x, y) =\begin{cases}
c(x + 3y)& 3 \leq x \leq 7, 4 \leq y \leq10\\
0 & \text{otherwise }\\
\end{cases}
$$
(a) Find $f_{X | Y}(x | y)$
(b) $P(x \leq 5 | Y = 9)$
My attempt:
$f_{X | Y}(x | y) = \frac{f_{X, Y}(x, y)}{f_Y(y)}$
$$f_Y(y) = c\int_{3}^{7}(x+3y)dx = c\left(\frac{x^2}{2} + 3xy \right)\bigg|_{3}^{7} = c(20+12y)$$
for $f_Y(y)$ has support $4 \leq y \leq 10$, 0 otherwise
$$P_{X | Y}(x | y) = \frac{c(x+3y)}{c(20+12y)} = \frac{(x+3y)}{(20+12y)}$$
$P_{X | Y}(x | y)$ has support …
(b)
$f_{X | Y}(x | y = 9) = \frac{x + 3 \cdot 9}{20 + 12 \cdot 9} = \frac{x+27}{128}$
$$P(x \leq 5 | Y = 9) = \int_{3}^{5} f_{X | Y}(x | y = 9)dx = \int_{3}^{5} \frac{x+27}{128} = \int_{3}^{5} \left(x/128 + 27/128 \right)dx = \left(\frac{x^2}{256} + \frac{27}{128}x \right)\bigg|_{3}^{5} = \frac{31}{64}$$
Not sure if I'm right, and a question if it were the joint pdf as
$$f_{X, Y} (x, y) =\begin{cases}
c(x + 3y)& 3 \leq x \leq y, 4 \leq y \leq10\\
0 & \text{otherwise }\\
\end{cases}
$$
How would the integral for the first question look like?
Best Answer
You are correct. Don't forget, however, to indicate the supports.
(Well, assuming $P_{X\mid Y}$ is conditional probability density function; why did they use $P$ rather than $f$?)
If instead $f_{X,Y}(x,y) = c(x+2y)\,\mathbf 1_{3<x<y, 4<y<10}$, then $$f_Y(y)~{=\int_3^y c(x+2y)\mathbf 1_{4<y<10}\,\mathsf d x \\= c(\tfrac 52y^2-6y-\tfrac 92)\,\mathbf 1_{4<y<10}}$$