Bapat–Beg theorem for two order statistics Bapat–Beg theorem gives the joint probability distribution of order statistics of independent but not necessarily identically distributed random variables in terms of the cumulative distribution functions.
I have troubles formulating the case of two order statistics $X_{(1)} \le X_{(2)}$ from the general one. Is this correct?
$$F_{X_{(1)}, X_{(2)}}(x_1, x_2) = \frac{P_{n_1}(x_1) + P_{n_2}(x_2)}{n_1! (n_2 - n_1)! \times (n_2 - n_2)! }$$
 A: What you wrote doesn’t make much sense. On the left, you chose $k=2$, $n_1=1$ and $n_2=2$, but on the right you have $n_1$ and $n_2$ as free variables. (Also $(n_2-n_2)!=0!=1$.)
Substituting $k=2$, $n_1=1$ and $n_2=2$ into the definition in the Wikipedia article you linked to yields
\begin{eqnarray}
F_{X_{(1)},X_{(2)}}(x_1,x_2)
&=&
\operatorname{Pr}\left(X_{(1)}\le x_1\land\operatorname{Pr}(X_{(2)}\le x_2\right)
\\
&=&
\sum_{i_2=2}^n\sum_{i_1=1}^{i_2}\frac{P_{i_1,i_2}(x_1,x_2)}{i_1!(i_2-i_1)!(n-i_2)!}\;,
\end{eqnarray}
where
$$
P_{i_1,i_2}(x_1,x_2)=
\operatorname{per}\,
\begin{bmatrix}
F_1(x_1)                \;\cdots\;  F_1(x_1)               & 
F_1(x_2)-F_1(x_1) \;\cdots\;  F_1(x_2)-F_1(x_1) & 
1-F_1(x_2)             \;\cdots\;  1-F_1(x_2)  \\
F_2(x_1)                 \;\cdots\;  F_2(x_1)                & 
F_2(x_2)-F_2(x_1)  \;\cdots\;  F_2(x_2)-F_2(x_1)   & 
1-F_2(x_2)             \;\cdots\;  1-F_1(x_2)\\
\vdots                                                            &    
\vdots                                                            &              
\vdots                                                  \\
\underbrace{F_n(x_1)               \;\cdots\;  F_n(x_1)              }_{i_1}   & 
\underbrace{F_n(x_2)-F_n(x_1)\;\cdots\;F_n(x_2)-F_n(x_1)}_{i_2-i_1} & 
\underbrace{1-F_n(x_2) \;\cdots\; 1-F_n(x_2) }_{n-i_2}
\end{bmatrix}\;.
$$
I’m not sure whether you also intended to set $n=2$. In that case, the result is
$$
F_{X_{(1)},X_{(2)}}(x_1,x_2)
=\sum_{i=1}^2\frac{P_{i,2}(x_1,x_2)}i\;,
$$
where
$$
P_{i,2}(x_1,x_2)=
\operatorname{per}\,
\begin{bmatrix}
F_1(x_1)                \;\cdots\;  F_1(x_1)               & 
F_1(x_2)-F_1(x_1) \;\cdots\;  F_1(x_2)-F_1(x_1)\\
\underbrace{F_2(x_1)                 \;\cdots\;  F_2(x_1)}_i                & 
\underbrace{F_2(x_2)-F_2(x_1)  \;\cdots\;  F_2(x_2)-F_2(x_1)}_{2-i}
\end{bmatrix}\;.
$$
Writing out the sum yields
\begin{eqnarray}
F_{X_{(1)},X_{(2)}}(x_1,x_2)
&=&F_1(x_1)(F_2(x_2)-F_2(x_1))+F_2(x_1)(F_1(x_2)-F_1(x_1))+F_1(x_1)F_2(x_1)
\\
&=&
F_1(x_1)F_2(x_2)+F_2(x_1)F_1(x_2)-F_1(x_1)F_2(x_1)\;.
\end{eqnarray}
That makes sense, as it adds the two ways in which the order statistics can be related to the variables ($X_{(1)}=X_1$ and $X_{(2)}=X_2$ or $X_{(1)}=X_2$ and $X_{(2)}=X_1$) and then subtracts the double-counted contribution where both variables are below $x_1$.
