If $X\sim U , Y\sim U$ , and $X,Y$ may be non-independent. Can we say the joint distribution of $X,Y$ is uniform?

  • 4
    $\begingroup$ A term that might interest you is copula. $\endgroup$
    – Dave
    Aug 9 at 18:05
  • 2
    $\begingroup$ Consider the case $X=Y.$ $\endgroup$
    – whuber
    Aug 9 at 21:32
  • $\begingroup$ See Is there a bivariate β distribution I can fit to my data? for examples! (choose the parameters so that the marginals are uniform). $\endgroup$ Aug 10 at 0:43
  • 3
    $\begingroup$ Consider a chessboard, and imagine the probability is uniform on the white squares $\endgroup$ Aug 10 at 1:29

No, the joint distribution is not necessarily uniform.

Consider $X$ and $Y$ with a joint pdf

$$ f(x,y) = \begin{cases} 2, \text{if } x \in (0,0.5), y \in (0,0.5)\\ 2, \text{if }x \in (0.5,1), y \in (0.5,1) \\ 0, \text{otherwise} \end{cases} $$

Then both $X$ and $Y$ have marginal $U(0,1)$ distributions, but the joint distribution is not uniform.

  • 1
    $\begingroup$ +1 I put together a visualization of your density. set.seed(2021); N <- 1000; x1 <- runif(N, 0, 0.5); x2 <- runif(N, 0.5, 1); y1 <- runif(N, 0, 0.5); y2 <- runif(N, 0.5, 1); x <- c(x1, x2); y <- c(y1, y2); par(mfrow = c(3, 1)); plot(x, y); plot(x, ecdf(x)(x), main = "CDF of x"); abline(a = 0, b = 1); plot(y, ecdf(y)(y), main = "CDF of y"); abline(a = 0, b = 1) $\endgroup$
    – Dave
    Aug 9 at 18:25

Suppose that $X\sim Unif(0,1)$ and $Y\sim Unif(0,X)$, then the joint distribution will be

$$f(X,Y)=f(Y|X)f(X)=\frac{1}{x}, \ \ \ 0\leq y \leq x\leq 1$$

If I'm not mistaken a uniform distribution gives the same importance to the whole domain of $(X,Y)$, whereas in the case displayed the importance given depends on the sampled $X$.

  • $\begingroup$ I think it's worth noting that although $Y|X \sim Unif(0,X)$, the marginal distribution of $Y$ is not uniform. $\endgroup$ Aug 10 at 15:02
  • $\begingroup$ If the parameter of interval is a random variable, it is a uniform disturbution? In fact, many definitions do not clearly indicate this problem. Anyway, copula is a better interpretation to solve this question for me. $\endgroup$
    – aditer
    Oct 8 at 16:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.