# CDF for R.V. uniformly distributed over boundary of square?

Let $X(x,y)=x$, and $Y(x,y)=y$, and let $P$ be a probability measure (on an appropriate space) that distributes mass uniformly over the boundary of the square defines by vertices $(0,0), (0,1),(1,0),(1,1)$.

What would be the joint $PDF$ of $X,Y$?

What is the CDF of $Y$?

For the first question, since the mass is distributed uniformly the joint PDF should be a constant, than integrates to one over the boundary. I'm not quite sure how to write this out, but my attempt is $$2\int_{0}^1cdx + 2\int_{0}^1cdy = 1 \implies c= \frac14$$ which makes intuitive sense to me.

I'm not sure how to find the CDF of $Y$, though, as normally I would integrate the joint PDF to get the marginal pdf of Y, then integrate this over the relevant region w.r.t $Y$. However, integrating over the square seems wrong since the distribution is not uniform over the square, only the boundary.

My best guess is that $$P(Y\leq y) = \frac14 + 2\int_0^y \frac14 dy \quad 0\leq y <1$$ (and then the obvious cases when $y\leq 0$ and $y\geq 1$

If this is correct, is there a nicer way to do it (or a way where I can integrate over the whole square and not just the boundary?)

You are right to approach this over the boundary, not the two-dimensional plane since the probability is concentrated there. Since the probability is squeezed into one dimension, the joint distribution of $(X,Y)$ is singular. So it has no joint PDF. We can still write an expression in terms of delta functions: $$f_{X,Y}(x,y) = \frac{1}{4}\big(\delta(x-0)+\delta(x-1)+ \delta(y-0)+\delta(y-1)\big)$$ where $0\le x,y\le 1.$ Each of these four pieces represents one side of the square and the delta functions enforce that the density is only on the square boundary.
$Y$ should be a mixture of half uniformly distributed on $[0,1]$, one quarter atom at $0$ and one quarter atom at $1.$ So your answer for $Y$ looks right.
• The $x$ and $y$ factors in the joint "PDF" are a mystery. Actually, it is as if the abuse of notations you suggest was coming back with a vengeance... A correct formula for the common distribution $P_X=P_Y$ of $X$ and $Y$ is $$P_X(dx)=\frac14\delta_0(dx)+\frac14\delta_1(dx)+\frac12\mathbf 1_{0<x<1}dx$$ The distribution $P_{X,Y}$ of $(X,Y)$ has a similar, fully rigorous, expression. – Did Jun 28 '17 at 5:48
• @did don't see how it's still wrong. My first term would be $\delta_0(dx)1_{0\le y \le 1}dy$ in your less abusive lingo. – spaceisdarkgreen Jun 28 '17 at 18:26
• Where is $\mathbf 1_{0<y<1}$ in your formula? Coming and going between 1D and 2D notations is killing you here... – Did Jun 28 '17 at 18:40