joint pdf of $(X,X)$

Question

Let $X$ be a continuous random variable with uniform distribution on $[0,1]$, i.e, the probability density function of $X$ is $f(x)=1$ on $[0,1]$. Let $Y=X$. Then what is the joint probability density function $h(x,y)$ of $X$ and $Y$? It seems that $h$ is supported on the line $y=x$, which has $0$ measure. I guess the joint pdf does not exist or equals to Dirac delta function. Can anyone confirm?

Answer 1

There is a measure on $\mathbb{R}^2$ which is lifted from the Lesbegue measure on the line $y=x$ (so the measure of $\{(x,x)|x\in [a,b]\}$ is equal to $b-a$) and the measure of the rest of $\mathbb{R}^2$ is $0$. Denote this measure $\nu$.

Then, given that the distribution of $X$ has density of $f$ with respect to the one-dimensional Lesbegue measure, the distribution of $(X,X)$ has density $g(x,y)$ with respect to $\nu,$ were $g(x,y)=f(x)$ if $x=y$ and $0$ else.

The computation/proof should is straight forward.

Answer 2

$(X,X)$ does not have a density w.r.t. Lebesgue measure on $\mathbb R^2$. Indeed if $D=\{(x,x), x\in [0,1]\}$, then $P_{(X,X)}(D)=1$ whereas $\lambda_2(D)=0$.

Besides, $P_{(X,X)}(\{(a,b)\})=0$ for any $(a,b)\in \mathbb R^2$, so the distribution of $(X,X)$ is not discrete either.