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.


1 Answer 1


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.

In general, remembering to think about the support of PDFs (where the PDF is nonzero) can help clarify things and avoid headaches.

  1. Well, yes the fact that the conditional distribution is also a uniform distribution is indeed a consequence of the fact that the joint distribution is uniform on the triangular region. But this is what you are asked to show rigorously.

  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 $(y, 1)$.


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