Let $X_1$ and $X_2$ be independent with uniform distribution over $(0,1)$ and set $X=\max\{X_1,X_2\}$. What is the joint density of $(X,X_1)$?
For $0<s,t<1$ we have \begin{align} F_{X,X_1}(s,t) &= \mathbb P(X\leqslant s,X_1\leqslant t)\\ &= \mathbb P(X_1\leqslant s,X_2\leqslant s, X_1\leqslant t)\\ &= \mathbb P(X_1\leqslant s\wedge t)\mathbb P(X_2\leqslant s)\\ &= (s\wedge t)s\\ &= \begin{cases} s^2,& s<t\\ st,& s\geqslant t. \end{cases} \end{align} Then the joint density should be obtained by differentiating: $$ f_{X,X_1}(s,t) = \frac{\partial^2}{\partial s\partial t} F_{X,X_1}(s,t). $$ But this yields $f_{X,X_1}(s,t) = \mathsf 1_{\{0<t\leqslant s<1\}}(s,t)$ which doesn't integrate to $1$. What am I doing wrong?
There is no density function. Note that $P(X=X_1) = P(X_2 \leq X_1)=\frac 1 2$. Since the set $\{(x,y): x=y\}$ has Lebesgue measure $0$ the distribution of $(X,X_1)$ is not absolutely continuous w.r.t. Lebesgue measure and so there is no density joint function for $(X,X_1)$.