Let $Y_{1} = \min(X_{1},X_{2})$, $Y_{2} = \mid X_{1}-X_{2} \mid$, where $X_{1}$ and $X_{2}$ are two independent random variables distributed exponentially with parameter $1$. I want to know if $Y_{1}$ and $Y_{2}$ are independent.
I've found their distribution functions as follows:
$F(Y_{1} \leq y_{1})$ = $1-e^{-y_{1}/2}$ and $F(Y_{2} \leq y_{2}) = 1-\frac{1}{2}e^{-y_{2}}$, with $y_{2}$ is positive.
But I don't know how to compute the joint distribution
In order to compute the joint distribution, you can start by computing $\mathbb E\left[\phi(Y_1,Y_2)\right]$ where $\phi\colon \mathbb R^2\to\mathbb R$ is measurable and bounded. To do so, observe that $\phi(Y_1,Y_2)$ is a symmetric function of $(X_1,X_2)$, hence $$ \mathbb E\left[\phi(Y_1,Y_2)\right]=2\mathbb E\left[\phi(X_1,X_2-X_1)\mathbf{1}_{\{X_2>X_1\}}\right]. $$ One can express the previous expectation as a double integral, namely, $$ \mathbb E\left[\phi(X_1,X_2-X_1)\mathbf{1}_{\{X_2>X_1\}}\right]=\int_0^\infty\int_0^\infty \phi(x_1,x_2-x_1)\mathbf{1}_{\{x_2>x_1\}}e^{-x_1-x_2} dx_2dx_1.$$ Doing the substitution $t=x_2-x_1$ in the inner integral (for a fixed $x_1$) gives $$\mathbb E\left[\phi(Y_1,Y_2)\right]=\int_0^\infty\int_0^\infty \phi(x_1,t)f(x_1,t)dtdx_1$$ for some function $f$ which is the joint density of $(Y_1,Y_2)$.