I have the following problem:
Let $(X,Y)$ be a uniformly random point in the triangle in the plane with vertices $(0,0),(0,1),(1,0)$. Find the joint PDF of $X$ and $Y$, the marginal PDF of $X$, and the conditional PDF of $X$ given $Y$.
I found the joint PDF as follows:
$$\begin{align} p_{X,Y}(x, y) &= \int_{0}^1 \int_0^{1 - x} p_{X,Y}(x, y) \ dy dx \\ &= \int_{0}^1 \int_0^{1 - x} c \ dy dx \ \ \text{(Since the region $(X, Y)$ is uniformly distributed.)} \\ &= 1 \ \ \text{(Since the probability over the entire region must equal 1.)} \end{align}$$
$$\begin{align} \Rightarrow c = 2 \end{align}$$
I found the marginal PDF of $X$ as follows:
$$\begin{align} p_X (x) &= \int_0^{1 - x} 2 \ dy \\ &= 2(1 - x) \end{align}$$
I found the conditional PDF of $X$ given $Y$ as follows:
$$\begin{align} p_{X | Y} (x | y) &= \dfrac{p_{X, Y} (x, y)}{p_Y (y)} \\ &= \dfrac{2}{2(1 - y)} \\ &= \dfrac{1}{1 - y} \end{align}$$
The provided solution agrees with my work. However, the solution makes an additional claim for conditional PDF:
Since $\dfrac{1}{1 - y}$ is a constant with respect to $x$, we have $X | Y \sim \text{Unif}(0, 1 - Y)$.
I have a couple of questions regarding this last point:
How specifically did the author conclude that $X | Y \sim \text{Unif}(0, 1 - Y)$? They did say that "Since $\dfrac{1}{1 - y}$ is a constant with respect to $x$ ...," but I don't see how this statement alone means that we can conclude have $X | Y \sim \text{Unif}(0, 1 - Y)$. I have the following thoughts on the question:
Does it have to do with the fact that all points $(X, Y)$ are Uniformly random in the triangle?
Does it have to do with the fact that the conditional PDF $\dfrac{1}{1 - y}$ resembles the PDF of a continuous uniform random variable $\dfrac{1}{b - a}$? BUT, if this were true, then we would have $a = y$ and $b = 1$, and therefore $X | Y \sim \text{Unif}(y, 1)$, so I don't understand where $X | Y \sim \text{Unif}(0, 1 - Y)$ came from?
I would greatly appreciate it if people could please take the time to clarify/explain this.