# Where did $X | Y \sim \text{Unif}(0, 1 - Y)$ come from?

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:

1. Does it have to do with the fact that all points $$(X, Y)$$ are Uniformly random in the triangle?

2. 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.

Strictly speaking the conditional PDF is $$p_{X \mid Y}(x \mid y) = \begin{cases}\frac{1}{1-y} & x \in [0, 1-y] \\ 0 & \text{otherwise} \end{cases}.$$ (Think carefully about where the joint PDF is nonzero.) This is precisely the PDF of the $$\text{Unif}(0, 1-y)$$ distribution.
2. Again, thinking about the support of the conditional PDF answers your question. In general, any uniform PDF over an interval of length $$1-y$$ will have the PDF $$\frac{1}{1-y}$$; it does not necessarily have to be the interval $$(y, 1)$$.