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?
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4$\begingroup$ A term that might interest you is copula. $\endgroup$– DaveAug 9, 2021 at 18:05
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4$\begingroup$ Consider the case $X=Y.$ $\endgroup$– whuber ♦Aug 9, 2021 at 21:32
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1$\begingroup$ See Is there a bivariate β distribution I can fit to my data? for examples! (choose the parameters so that the marginals are uniform). $\endgroup$– kjetil b halvorsen ♦Aug 10, 2021 at 0:43
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5$\begingroup$ Consider a chessboard, and imagine the probability is uniform on the white squares $\endgroup$– Thomas LumleyAug 10, 2021 at 1:29
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2 Answers
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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.
answered Aug 9, 2021 at 17:25
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2$\begingroup$ +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)$\endgroup$– DaveAug 9, 2021 at 18:25
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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 $X$.
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$\begingroup$ 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, 2021 at 15:02
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$\begingroup$ 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. $\endgroup$– aditerOct 8, 2021 at 16:16