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(More a comment that is too long for the comment box) ... Given a sample of size $n$ from a standard Uniform parent, one can show, by induction or otherwise, that the joint pdf of the first $k$ order statistics, is:

$$f(x_{(1)}, x_{(2)}, \dots, x_{(k)}) = \frac{n!}{(n-k)!}\left(1-x_{(k)}\right)^{n-k} \quad \quad \text{for }\quad 0 < x_{(1)} < x_{(2)} < \dots <x_{(k)} < 1$$

... which has a neat functional form.

Your problem then reduces to:

Find the cdf of the sum $X_{(1)} + X_{(2)} + \dots + X_{(k)}$ given joint pdf $f(\centerdot)$

... which unfortunately will result in a $k$-part piecewise solution with kinks at $(1, 2, \dots, k)$.

Using the mathStatica package for Mathematica, I obtained the same solution you provided for $k=2$, and was also able to derive a general exact solution for the sum in the $k = 3$ case (in 3 parts). However, an exact general solution seems to get messy quite rapidly.

For illustration, here is a plot of the cdf of the sum in the $k =2$ case (as $n$ increases from 2 to 10, and then with $n =30$):

The intuition, of course, is that when $n$ becomes large, the first two order statistics become very small, and so the cdf of their sum shifts to the left.

Here is the $k=3$ case:

(More a comment that is too long for the comment box) ... Given a sample of size $n$ from a standard Uniform parent, one can show, by induction or otherwise, that the joint pdf of the first $k$ order statistics, is:

$$f(x_{(1)}, x_{(2)}, \dots, x_{(k)}) = \frac{n!}{(n-k)!}\left(1-x_{(k)}\right)^{n-k} \quad \quad \text{for }\quad 0 < x_{(1)} < x_{(2)} < \dots <x_{(k)} < 1$$

... which has a neat functional form.

Your problem then reduces to:

Given joint pdf $f(x_{(1)}, x_{(2)}, \dots, x_{(k)})$, find the cdf of the sum $X_{(1)} + X_{(2)} + \dots + X_{(k)}$

... which unfortunately will result in a $k$-part piecewise solution with kinks at $(1, 2, \dots, k)$.

Using the mathStatica package for Mathematica, I obtained the same solution the OP provided for $k=2$, and was also able to derive a general exact solution for the sum in the $k = 3$ case (in 3 parts). However, an exact general solution seems to get messy quite rapidly.

To illustrate the behaviour of the cdf, here is a plot of the cdf of the sum in the $k =2$ case (as $n$ increases from 2 to 10, and then with $n =30$):

The intuition, of course, is that when $n$ becomes large, the first two order statistics become very small, and so the cdf of their sum shifts to the left.

Here is the $k=3$ case:

(More a comment that is too long for the comment box) ... Given a sample of size $n$ from a standard Uniform parent, one can show, by induction or otherwise, that the joint pdf of the first $k$ order statistics, is:

$$f(x_{(1)}, x_{(2)}, \dots, x_{(k)}) = \frac{n!}{(n-k)!}\left(1-x_{(k)}\right)^{n-k} \quad \quad \text{for }\quad 0 < x_{(1)} < x_{(2)} < \dots <x_{(k)} < 1$$

... which has a neat functional form.

Your problem then reduces to:

Find the cdf of the sum $X_{(1)} + X_{(2)} + \dots + X_{(k)}$ given joint pdf $f(\centerdot)$

... which unfortunately will result in a $k$-part piecewise solution with kinks at $(1, 2, \dots, k)$.

Using the mathStatica package for Mathematica, I obtained the same solution you provided for $k=2$, and was also able to derive a general exact solution for the sum in the $k = 3$ case (in 3 parts). However, an exact general solution seems to get messy quite rapidly.

For illustration, here is a plot of the cdf in the $k =2$ case (as $n$ increases from 2 to 10, and then with $n =30$):

The intuition, of course, is that when $n$ becomes large, the first two order statistics become very small, and so the cdf of their sum shifts to the left.

Here is the $k=3$ case:

(More a comment that is too long for the comment box) ... Given a sample of size $n$ from a standard Uniform parent, one can show, by induction or otherwise, that the joint pdf of the first $k$ order statistics, is:

$$f(x_{(1)}, x_{(2)}, \dots, x_{(k)}) = \frac{n!}{(n-k)!}\left(1-x_{(k)}\right)^{n-k} \quad \quad \text{for }\quad 0 < x_{(1)} < x_{(2)} < \dots <x_{(k)} < 1$$

... which has a neat functional form.

Your problem then reduces to:

Find the cdf of the sum $X_{(1)} + X_{(2)} + \dots + X_{(k)}$ given joint pdf $f(\centerdot)$

... which unfortunately will result in a $k$-part piecewise solution with kinks at $(1, 2, \dots, k)$.

Using the mathStatica package for Mathematica, I obtained the same solution you provided for $k=2$, and was also able to derive a general exact solution for the sum in the $k = 3$ case (in 3 parts). However, an exact general solution seems to get messy quite rapidly.

For illustration, here is a plot of the cdf of the sum in the $k =2$ case (as $n$ increases from 2 to 10, and then with $n =30$):

The intuition, of course, is that when $n$ becomes large, the first two order statistics become very small, and so the cdf of their sum shifts to the left.

Here is the $k=3$ case:

(More a comment that is too long for the comment box) ... Given a sample of size $n$ from a standard Uniform parent, one can show, by induction or otherwise, that the joint pdf of the first $k$ order statistics, is:

$$f(x_{(1)}, x_{(2)}, \dots, x_{(k)}) = \frac{n!}{(n-k)!}\left(1-x_{(k)}\right)^{n-k} \quad \quad \text{for }\quad 0 < x_{(1)} < x_{(2)} < \dots <x_{(k)} < 1$$

... which has a neat functional form.

Your problem then reduces to:

Find the cdf of the sum $X_{(1)} + X_{(2)} + \dots + X_{(k)}$ given joint pdf $f(\centerdot)$

... which unfortunately will result in a $k$-part piecewise solution with kinks at $(1, 2, \dots, k)$.

Using the mathStatica package for Mathematica, I obtained the same solution you provided for $k=2$, and was also able to derive a general exact solution for the sum in the $k = 3$ case (in 3 parts). However, an exact general solution seems to get messy quite rapidly.

(More a comment that is too long for the comment box) ... Given a sample of size $n$ from a standard Uniform parent, one can show, by induction or otherwise, that the joint pdf of the first $k$ order statistics, is:

$$f(x_{(1)}, x_{(2)}, \dots, x_{(k)}) = \frac{n!}{(n-k)!}\left(1-x_{(k)}\right)^{n-k} \quad \quad \text{for }\quad 0 < x_{(1)} < x_{(2)} < \dots <x_{(k)} < 1$$

... which has a neat functional form.

Your problem then reduces to:

Find the cdf of the sum $X_{(1)} + X_{(2)} + \dots + X_{(k)}$ given joint pdf $f(\centerdot)$

... which unfortunately will result in a $k$-part piecewise solution with kinks at $(1, 2, \dots, k)$.

Using the mathStatica package for Mathematica, I obtained the same solution you provided for $k=2$, and was also able to derive a general exact solution for the sum in the $k = 3$ case (in 3 parts). However, an exact general solution seems to get messy quite rapidly.

For illustration, here is a plot of the cdf in the $k =2$ case (as $n$ increases from 2 to 10, and then with $n =30$):

The intuition, of course, is that when $n$ becomes large, the first two order statistics become very small, and so the cdf of their sum shifts to the left.

Here is the $k=3$ case: