# Joint PDF of two uniform distributions on $[0,1]$ [closed]

Question: Given two random variables $$X$$ and $$Y$$ uniformly distributed on $$[0,1]$$, what is their joint density function $$f_{X,Y}?$$

In this post, it seems that @Hans Parshall mentioned that the joint PDF is given by $$f(x,y) = 6.$$

Why is this the case?

• The post you link is answering a different question. Assuming independence, the joint density is simply $$f_{X,Y}(x,y) = f_X(x)f_Y(y) = 1 \cdot 1 = 1$$ for $0 \leq x,y \leq 1$. – averagemonkey Feb 14 at 9:42
• Within the constraints the distribution is uniform so the density is constant. But you want the total probability to be $1$; in the linked question this is across a sixth of the unit square and so the density needs to be $6$ for that to happen, while in this question it is across the whole unit square so the density needs to be $1$ – Henry Feb 14 at 9:51
• @averagemonkey That is only if both $X$ and $Y$ are independent. What if they are not? – Idonknow Feb 14 at 13:29
• If the variables are not assumed independent, a million things could happen. As an example we might have $X = Y$ or $X = 1-Y$, in which case the joint density does not even exist. – averagemonkey Feb 15 at 1:47

## 1 Answer

There is exactly one segment out of the six $$(X,Y,(X-Y)1_{X-Y}, 1-X,1-Y,(Y-X)1_{Y-X})$$ that is the shortest.

Thus the expectation is $$\mathbb{E}[\text{whichever is the shortest}]$$, and the integral should be carried out over the entire square for each point within the unit square $$(x,y) \in [0,1]^2$$, using the density $$f_{XY}(x,y) = 1$$.

That is, with density $$f_{XY}(x,y) = 1$$ the integrand is not always $$x$$. There are six regions, and only within one of the regions is the integrand (shortest length) $$x$$.

At this point, one can recognize the $$6$$-fold symmetry, and integrates only over the one-sixth region $$\{x<\frac13, x as shown in the diagram there, then multiply the whole thing by $$6$$.

In that answer, stating that "the density is $$6$$" is implicitly considering the conditional expectation $$\mathbb{E}[X \mid \text{shortest is X}]$$. Again, if one were doing this unconditionally, one would need to add $$\mathbb{E}[X\cdot 1_{\text{{X is the shortest}}}]+\mathbb{E}[Y\cdot 1_{\text{{Y is the shortest}}}] + \mathbb{E}[(1-X)\cdot 1_{\text{{1-X is the shortest}}}]+\cdots$$

Here, doing this conditionally one has the conditional density $$f_{\text{conditional}} = \frac{ f_{\text{joint}} }{ \Pr_{\text{conditional}} } = \frac{ 1 }{ \frac12 \cdot \frac13 } = 6$$ where the $$1$$ on top is the usual uniform-over-square, the $$1/2$$ below is for $$X, and the $$1/3$$ comes from the fact "It's not hard to show that they all have probability $$1/3$$ of being the shortest."

Allow me to be pedantic.

Since the six segments are interchangeable, all the conditional expectations are the same. Namely, one will arrive at the same answer: $$\mathbb{E}[X \mid \text{shortest is X}] = \mathbb{E}[Y \mid \text{shortest is Y}] = \mathbb{E}[1-X \mid \text{shortest is 1-X}] =\cdots$$

The reason that anyone of them gives the desired expectation (that answer by Hans Parshall picked to calculate $$\mathbb{E}[X \mid \text{shortest is X}]$$) is because \begin{align} &\hphantom{{}={}}\mathbb{E}[\text{whichever is the shortest}] \\ &= \frac16\mathbb{E}[X \mid \text{shortest is X}] + \frac16 \mathbb{E}[Y \mid \text{shortest is Y}] + \frac16 \mathbb{E}[1-X \mid \text{shortest is 1-X}] + \cdots \\ &= 6\cdot\frac16 \mathbb{E}[X \mid \text{shortest is X}] \\ &= \mathbb{E}[X \mid \text{shortest is X}] \end{align}