I'm trying this excercise: Let the joint density function of random variables X and Y be:
$$f_{x,y}(x,y)=\left\{\begin{array}{l} \frac{3}{16}(4-2x+y) & for \,\,\, x \geqslant 0, y \geqslant 0 \,\,\ \text{and} \,\,\, 2x+y<4\\ 0 & \text {otherwise} \end{array}\right.$$
a) Give the conditional density function of Y for every value of X;
b) Find P[Y $\geq2$ | X = $0.5$]
I'm not shure how can i solve a). To find $f_{Y|X}(x,y) = \frac{f_{x,y}(x,y)}{f_{x}(x)}$ i need to find $f_{x}(x)$, and my doubt is: that is the right way of doing this? $$f_{x}(x) = \int_{0}^{4-2x} \frac{3}{16}(4-2x+y) dy = \frac{3}{8}(x(x-4))$$ for $x<2-\frac{y}{2}$.
I'm not sure how to deal with the domains of the two functions on $f_{Y|X}(x,y) = \frac{f_{x,y}(x,y)}{f_{x}(x)}$.
And for b), the only thing to do here will be the integration below? $$\int_{2}^{4-2x} \int_{0}^{0.5} f_{Y|X}(x,y) dx dy$$.
Thanks in advance for any help!
Edit: Please dont give full answers, I'm looking for hints and insights .
Update: I got this two answers:
First, i want to apologize from Mr. heropup because i changed a "-" signal on the function, and actually the function is:
$$f_{x,y}(x,y)=\left\{\begin{array}{l} \frac{3}{16}(4-2x-y) & for \,\,\, x \geqslant 0, y \geqslant 0 \,\,\ \text{and} \,\,\, 2x+y<4\\ 0 & \text {otherwise} \end{array}\right.$$
I appreciate your answer, and i'm sorry for the mistake.
And here are the answers i got:
a) $\frac{(4-2x-y)}{(4-2x)^2 \frac{1}{2}}$ - I commited a mistake on the last attempt.
b) Zero, because there is no volume under the curve since $X=0.5$. In order to get a nonzero probability, we should ask $X-\epsilon < 0.5 < X+\epsilon$, otherwise we won't get any volume under the density function.
This question could be reformulated as the integral: $$\int_{2}^{4-2x} \int_{0.5}^{0.5} f_{Y|X}(x,y) dx dy$$.
And since $X=x=0.5 \longrightarrow y=3$, then $$\int_{2}^{3} \int_{0.5}^{0.5} f_{Y|X}(x,y) dx dy$$.
But the integral over $x$ is zero, because i'm assuming that the random variable $X=0.5$, which means that there's no general volume, so the probability is zero.
My biggest doubt here is what i'm understanding for my answer for b), i can't see why** it's not what i'm thinking.
Are them correct?
I get $$\begin{align} f_X(x) &= \frac{3}{16} \int_{y=0}^{4-2x} 4 - 2x - y \, dy \\ &= \frac{3}{16}\left[(4 - 2x)y - \frac{y^2}{2}\right]_{y=0}^{4-2x} \\ &= \frac{3}{16}\left((4-2x)^2 - \frac{(4-2x)^2}{2}\right) \\ &= \frac{3}{8}(2-x)^2. \end{align}$$
The support of this marginal density is not a function of $y$: it is simply $X \in [0,2)$. This is because the projection of the region defined by $x \ge 0$, $y \ge 0$, $2x + y < 4$ onto the $x$-axis is the interval $[0,2)$.
The conditional density $$f_{Y \mid X}(y \mid x) = \frac{f_{X,Y}(x,y)}{f_X(x)} = \frac{4-2x-y}{2(2-x)^2}$$ whose support is a function of $x$; specifically, it is the range of $y$-values in the joint support for a given fixed $x \in [0,2)$; i.e., $$Y \mid X = x \in [0, 4-2x).$$
As for part (b), you already obtained the conditional density $f_{Y\mid X}$. So use that to compute the cumulative probability: you want $$\Pr[Y \ge 2 \mid X = 1/2] = 1 - F_{Y \mid X}(2 \mid \tfrac{1}{2}) = 1 - \int_{y=0}^2 \frac{4 - 2(1/2) - y}{2(2 - 1/2)^2} \, dy.$$