I am reading a book that in one page it talks about cdf of a random vector. This is from the book:

Given $X=(X_1,...,X_n)$, each of the random variables $X_1, ... ,X_n$ can be characterized from a probabilistic point of view by its cdf.

However the cdf of each coordinate of a random vector does not completely describe the probabilistic behaviour of the whole vector. For instance, if $U_1$ AND $U_2$ are two independent random variables with the same cdf $G(x)$, the vectors $X=(X_1, X_2)$ defined respectively by $X_1=U_1$, $X_2=U_2$ and $X_1=U_1$, $X_2=U_1$ have each of their coordinates with the same cdf, and they are quite different.

My question is:

From the very last paragraph, it says $U_1$ and $U_2$ are coming from the same c.d.f. And then they define $X=(X_1, X_2)$, but they say $X=(X_1, X_2)$ is different from $X=(X_1, X_1)$. I don't really understand why the two $X$ are different.

(i.e. I don't understand why $X=(X_1, X_2)$ and $X=(X_1, X_1)$ are different). Isn't $X_1$ the same as $X_2$, so it doesn't matter whether you put two $X_1$ to form $X=(X_1, X_1)$ or put one $X_1$ and one $X_2$ to form $X=(X_1, X_2)$. Shouldn't they be the same? why does the author says they are "quite different"?

Could someone explain why they are different?

  • 6
    $\begingroup$ I throw a fair coin and record its outcome with a binary indicator ($X_1$). You throw a fair coin and similarly record its outcome ($X_2$). Is it not obvious those two random variables differ? Nature will not guarantee they always produce the same result, that's for sure! Yet they have identical distributions and are independent. $\endgroup$
    – whuber
    Jan 23, 2014 at 23:34
  • 2
    $\begingroup$ There are two simple but quite different ways to approach this (well, there are more than two, but I'll mention two). (i) by actually sampling $U_1,U_2$ and just looking at the density of points for the two definitions of $X$ (what software do you have available? Excel? Matlab? R? C?); (ii) Proceeding directly from definitions (i.e. do you know what the definition of the cdf is?). $\endgroup$
    – Glen_b
    Jan 23, 2014 at 23:42
  • $\begingroup$ The components of $(X_1, X_1)$ are not independent (unless $X$ is almost surely constant) while you have defined the components of $(X_1, X_2)$ to be independent. So their joint distributions must be different $\endgroup$
    – Henry
    Feb 5, 2023 at 12:02

3 Answers 3


Let us take the simplest example of Bernoulli random variables with parameter $\frac12$. The value of the (joint) CDF $F_{X_1,X_2}(x,y)$ of $X_1$ and $X_2$ is the total probability mass in the southwest quadrant with northeast corner $(x,y)$

  • If $X_1$ and $X_2$ are two independent Bernoulli random variables, then we have four probability masses of $\frac14$ sitting at $(0,0), (1,0), (0,1)$, and $(1,1)$. Hence $$F_{X_1,X_2}\left(\frac12,\frac12\right) = \frac14.$$

  • If $X_2 = 1-X_1$, then we have two probability masses of $\frac12$ sitting at $(1,0)$ and $(1,0)$. Hence $$F_{X_1,X_2}\left(\frac12,\frac12\right) = 0.$$

  • If $X_2 = X_1$, then we have two probability masses of $\frac12$ sitting at $(0,0)$ and $(1,1)$. Hence $$F_{X_1,X_2}\left(\frac12,\frac12\right) = \frac12.$$

Thus, the joint CDF of $X_1$ and $X_2$ does depend on what kind of relationship (if any) they have with each other, and just knowing the common CDF of $X_1$ and $X_2$ (these are marginal CDFs) tells us nothing about the behavior the joint CDF.

  • $\begingroup$ thanks. I think I understand now. Sorry it has been a year lol haha. $\endgroup$
    – john_w
    Feb 25, 2015 at 18:34

Random objects can have the same distribution and be almost surely different. Take a look:

Can two random variables have the same distribution, yet be almost surely different?


The key is the difference between joint and marginal distributions. When the book talks about the cdf of a random vector $X=(X_1,X_2)$ they mean the function $$F_X(x_1,x_2):=\mathbb{P}(X_1\leq x_1,X_2\leq x_2 ).$$ As you probably know, if $X_1$ and $X_2$ are independent, this equals $$F_{X_1}(x_1)F_{X_2}(x_2)=\mathbb{P}(X_1\leq x_1)\mathbb{P}(X_2\leq x_2),$$ but, on the other hand if they are not independent, we cannot separate the joint probability into their product like that per definition of dependence/independence.

Indeed, in your example where $X_1=X_2$, we have that $$F_X(x_1,x_2):=\mathbb{P}(X_1\leq x_1,X_2\leq x_2 )=\mathbb{P}(X_1\leq x_1,X_1\leq x_2 )=\mathbb{P}(X_1\leq \min(x_1, x_2))\neq\mathbb{P}(X_1\leq x_1)\mathbb{P}(X_1\leq x_2).$$

Again, the key is that just because two variables have the same marginal distribution, this does not mean their joint distributions with another variable are the same.


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