Random Variable with its own RV's joint distribution

Question

How to derive the joint distribution: $f(x, x)$, I thought it would be just $f(x)$, but if I do the double integral of $f(x)$, it would not be $1$. I can not just multiply $f(x)$ twice because $X$ and $X$ is highly dependent.

Answer 1

You can think about it this way:

consider $y=x$ now

$f(y = x|X=x) = 1,$ and zero otherwise

$f(y,x) = f(x), y=x$ zero otherwise .

Answer 2

You made valuable thoughts but your conjecture is not precise. Denote the cdf and pdf of $X$ by $F_0$ and $f_0$, respectively. In attempting to find the pdf of $(X, X)$, we shall start with looking for its cdf as follows: $$F(x, y) = P[X \leq x, X \leq y] = P[X \leq \min\{x, y\}] = F_0(\min\{x, y\}).$$

Notice that this $F$ is not differentiable (for a simple example, you may consider $X \sim U(0, 1)$, then look at its differentiability at $(0, 0)$), therefore, $f_0$ does not exist.

It is easy to understand this result from another point of view. Notice that the range of $(X, X)$ is contained in the line $L = \{(x, x): x \in \mathbb{R}\}\subset \mathbb{R}^2$. So if the density of $(X, X)$ with respect to Lebesgue measure, say $f$ existed, then we would have $$1 = P((X, X) \in L) = \iint_L f(x, y) dxdy.$$ However, since $L$ is a set of Lebesgue measure $0$, the right hand side of the above equation must be $0$, contradiction.