Joint distribution of random variable with an order statistic Let $X_1,...,X_n$ be i.i.d. with the distribution of the $X_i$ being nice and continuous.  I'm interested in the expression of the CDF $F_{X_{(1)},X_j}(u,v)$. To be clear $X_{(1)} = min(X_i)$ and $X_j$ just one of the n i.i.d. random variables. 
I'm wondering if the following derivation is correct?
$$F_{X_{(1)},X_j}(u,v) = P(X_{(1)}\leq u, X_j\leq v)$$
Partition the probability by whether or not $X_j$ is the minimum 
$$=P(X_{(1)}\leq u,X_j\leq v,X_{(1)}= X_j)+P(X_{(1)}\leq u,X_j\leq v,X_{(1)}\neq X_j)$$
The first probability reduces down as follows
$$P(X_{(1)}\leq u,X_j\leq v,X_{(1)}= X_j)$$
$$=P(X_{(1)}\leq min(u,v), X_{(1)}= X_j)$$
The above is the probability that $X_j$ is minimum of the n i.i.d. random variables and it is less than both $u$ and $v$. Since the distribution of the $X_i$ is continuous we can compute this directly as
$$=\int_{-\infty}^{min(u,v)}{f_X(x)[1-F_X(x)]^{n-1}dx}$$
$$=\frac{1-[1-F_X(min(u,v))]^{n}}{n}$$
The last equality comes from integrating by substitution.  
Returning to the other probability we have 
$$P(X_{(1)}\leq u,X_j\leq v,X_{(1)}\neq X_j)$$
$$=P(X_{(1)}\leq min(u,v),X_j\leq v,X_{(1)}\neq X_j)$$
Since the $X_i$ are i.i.d., the above is equivalent to the probability that
$$P(X_j\leq v, X'_{(1)}\leq min(u,v),X'_{(1)}\leq X_j )$$ 
where $X'_{(1)}$ is the minimum of the other $n-1$ random variables, so that $X_j$ and $X'_{(1)}$ are independent. The probability above can be written as
$$\int_{-\infty}^{min(u,v)}{[F_X(v)-F_X(x)]f_{X'_{(1)}}(x)dx}$$
Using the fact that $f_{X'_{(1)}}(x)=(n-1)f_X(x)[1-F_X(x)]^{n-2}$ we have the the above probability can be computed as
$$\int_{-\infty}^{min(u,v)}{[F_X(v)-F_X(x)](n-1)f_X(x)[1-F_X(x)]^{n-2}dx}$$
To compute the above integral, let $s=min(u,v), b=F_X(v),$ and use the substitution $y=1-F_X(x)$ so that the above integral becomes
$$\int_{1}^{s}{-(n-1)(b-1+y)(y)^{n-2}dy}$$
$$=-(n-1)\int_{1}^{s}{[(b-1)y^{n-2}+y^{n-1}]dy}$$
$$=-(n-1)\int_{1}^{s}{[(b-1)y^{n-2}+y^{n-1}]dy}$$
$$=(n-1)\int_{1}^{s}{[(1-b)y^{n-2}-y^{n-1}]dy}$$
$$=(1-b)y^{n-1}-y^{n}\frac{(n-1)}{n}\Big|_1^s$$
$$=[1-F_X(v)]([1-F_X(s)]^{n-1}-1)+\frac{(n-1)}{n}(1-[1-F_X(s)]^{n})$$
So, both parts together give us
$$=\frac{1-[1-F_X(s)]^{n}}{n}+[1-F_X(v)]([1-F_X(s)]^{n-1}-1)+\frac{(n-1)(1-[1-F_X(s)]^{n})}{n}$$
$$=[1-F_X(v)]([1-F_X(s)]^{n-1}-1)+(1-[1-F_X(s)]^{n})$$
and finally, with $s=min(u,v)$:
$$F_{X_{(1)},X_j}(u,v) = [1-F_X(v)]([1-F_X(s))]^{n-1}-1)+(1-[1-F_X(s)]^{n})$$
 A: It is obvious that for every $j\in\{1,\dots,n\}$:$$P\left(X_{\left(1\right)}\leq u,X_{j}\leq v\right)=P\left(X_{\left(1\right)}\leq u,X_{1}\leq v\right)$$ 
If $v\leq u$ then $\left\{ X_{\left(1\right)}\leq u,X_{1}\leq v\right\} =\left\{ X_{1}\leq v\right\} $
so that in that case: $$P\left(X_{\left(1\right)}\leq u,X_{1}\leq v\right)=P\left(X_{1}\leq v\right)=F_{X}\left(v\right)$$
If $v>u$ then we can go for:
$$\begin{aligned}P\left(X_{\left(1\right)}\leq u,X_{1}\leq v\right) & =P\left(X_{\left(1\right)}\leq u,X_{1}\leq u\right)+P\left(X_{\left(1\right)}\leq u,u<X_{1}\leq v\right)\\
 & =P\left(X_{1}\leq u\right)+P\left(\min\left(X_{2},\dots,X_{n}\right)\leq u,u<X_{1}\leq v\right)\\
 & =P\left(X_{1}\leq u\right)+P\left(\min\left(X_{2},\dots,X_{n}\right)\leq u\right)P\left(u<X_{1}\leq v\right)\\
 & =F_{X}\left(u\right)+\left(1-\left(1-F_{X}\left(u\right)\right)^{n-1}\right)\left(F_{X}\left(v\right)-F_{X}\left(u\right)\right)
\end{aligned}
$$
A: From @drhab's answer we see that the joint distribution of $(X_{(1)},X_j)$ is given by
$$ 
G(u,v) = F(v)\mathsf 1_{\{v\leqslant u\}} + \left(1-(1-F(u))^{n-1}\right)(F(v)-F(u))\mathsf 1_{\{v>u\}}.
$$
For a concrete example, let $X_n\stackrel{\mathrm{i.i.d.}}{\sim}\mathrm{Expo}(\lambda)$, that is, $F(t) = \left(1 - e^{-\lambda t}\right)\mathsf 1_{(0,\infty)}(t)$. Then the joint distribution of $(X_{(1)},X_j)$ is given by
\begin{align}
G(u,v) =& F(v)\mathsf 1_{\{v\leqslant u\}} + \left(1-(1-F(u))^{n-1}\right)(F(v)-F(u))\mathsf 1_{\{v>u\}}\\
&=\begin{cases}
1-e^{-\lambda v},& v\leqslant u\\
\left(1-e^{-(n-1)\lambda u}\right)\left(e^{-\lambda u}-e^{-\lambda v}\right),& v>0.
\end{cases} 
\end{align}
