so in Casella and berger, yet I'm having trouble with a problem.
Let U be a random variable that counts the number $X_{1},...,X_{n} \leq u$ and let V be a r.v. that counts $X_{1},...,X_{n}$ greater than u and less than or equal to v.
the problem is b) show that the joint cdf of $X_{(i)}, X_{(j)}$ can be expressed as
\begin{equation}
\begin{split}
F_{X_{(i)},X_{(j)}}(u,v)&=P(U\geq i, U+V \geq j)
&= \sum_{k=i}^{j-1}\sum_{m=j-k}^{n-k}P(U=k,V=m) + P (U \geq j)
\end{split}
\end{equation}
For $\{u<X_{(j)}\leq v \} = \{V \geq j\} \cap \{U \leq j-1\}=\{V \geq j\}\cup \{U > j-1\}=\{V \geq j\} \cup \{U \geq j\}$
So considering $P(X_{(i)}\leq u,X_{(j)} \leq v)=P(U\geq i,(V\cup U)\geq j) ; 0<i<j<n$
=$P(U \geq i,V\geq j) + P(U\geq j)$
this work seems reasonable, and I can construct the first summation term using the fact that $0<i<j<n$ in the sum but I am not sure if this is correct but intuitively maybe this is where the second sum term $P(U\geq j)$ comes from?
any help will be very appreciated. hopefully there is not mistake my work as well.
Note that $X_{(i)}$ is the ordered statistic FYI.
The event $\{X_{(i)}\leq u\}$ tells us that the $i^{\text{th}}$ order statistic is less than or equal to $u$, which is equivalent to saying that $\{U\geq i\}$. The possible values that $U$ can take are $i,i+1,\ldots,n$.
Now, the first term (involving the double summation) addresses what happens when $i\leq U\leq j-1$. The other term caters to $U$ being larger than or equal to $j$. You may think of the latter summation as follows:
\begin{align} P(U\geq j,U+V\geq j)&=\sum\limits_{k=j}^{n}P(U=k,V\geq j-k). \end{align}
Note that over the range of summation, $j-n\leq j-k\leq 0$, and since $V\geq 0 $ with probability $1$, the event $\{V\geq j-k\}$ happens with probability $1$. Thus, we might as well drop it from the above summation, and what remains is $\sum\limits_{k=j}^{n}P(U=k)$, which in other words is same as $P(U\geq j)$.