Compute pdf of a k-th order statistic How to compute the density function of the k-th order statistic of a sample of $X_1, X_2, ..., X_n$ random variables distributed independently but not identically (i.e., $X_i \sim F_i$ with $F_i\neq F_j$)? 
What would be the explicit solution if each $F_i$ is uniform on $[m_i,1]$ with $0<m_i<m_{i+1}<...<m_n<1$?
 A: Preamble
As per the above comments, order statistics from non-identical distributions typically require complicated calculations, and generally yield complicated solutions, which makes them well-suited for solving with computer algebra systems. I am not aware that one can generally derive closed-form solutions as a function of the sample size $n$ and $k$-th order statistic (though it may be possible for your example)... but one CAN certainly obtain quite neat solutions to your problem, given any arbitrary integer value for $n$ and $k$ of your own choice. I am going to pursue the computer algebra approach here, because those tools are familiar to me, and because it makes short shrift of a lot of messy algebra.
The Problem
Let $X_i$ denote a continuous random variable with pdf $f(x; m_i)$, such that $(X_1,X_2,\dots,X_n)$ are independent but not identical variables due to differing parameters $m_i$, for $i = 1,\dots,n$. 
For the OP's question, we have a $Uniform(m,1)$ parent where identicality is relaxed by replacing parameter $m$ with $m_i$, for $i = 1, \dots, n$. Thus, the pdf $f(x; m_i)$, can be written:

To illustrate, here is plot of the family of pdf's, when $n = 4$, and $m_i = \frac{i}{5}$.

Solution
If $X_i$ has pdf $f(x; m_i)$, then, for any sample size $n$, the pdf of the $k$-th order statistic is given by:
$\qquad \qquad $OrderStatNonIdentical[k, {$f_i$}, {n}] 
where OrderStatNonIdentical is a function from the mathStatica package for Mathematica, and where $n$ and $k$ are integers. For the OP's question, in a sample of size $n = 4$, the pdf of the 2nd smallest order statistic is given immediately by:

Here is a plot of the pdf of the 2nd order statistic (just derived), when the sample size is $n=4$, and $m_i = i/5$:



*

*Similarly, the pdf of the 3rd order statistic is:



... and here is a plot of same:

Monte Carlo check
When doing symbolic work, it is always a good idea to check one's work using numerical methods, to make sure no errors have crept in. Here is a quick Monte Carlo check of the $k = 2$ case solution derived above, again with $m_i = i/5$, and $n = 4$. The ragged blue line is the empirical pdf (blue), plotted on top of the theoretical solution (dashed red line) derived above:

All looks fine :)
General Solutions by Induction
It appears possible to attain general symbolic solutions by induction, at least for the 1st and 2nd order statistics. In particular:


*

*the pdf of the 1st order statistic, irrespective of the size of $n$, has form:


$$ \begin{cases}
\frac{n(1-x)^{n-1}}{\prod_{i=1}^n (1-m_i)} & m_n < x < 1 \\ 
\quad \quad \dots & \quad \quad \dots  \\
\frac{3(1-x)^2}{(1-m_1)(1-m_2)(1-m_3)} & m_3 < x \leq m_4 \\
\frac{2(1-x)}{(1-m_1)(1-m_2)} & m_2 < x \leq m_3 \\
\frac{1}{1-m_1} & m_1 < x \leq m_2 \\
0 & \text{otherwise} 
\end{cases}$$


*

*the pdf of the 2nd order statistic has form:


$$ \begin{cases}
\frac{(n-1)(1-x)^{n-2}}{\prod_{i=1}^n (1-m_i)}(n x-\sum_{i=1}^n m_i) & m_n < x < 1 \\ 
\quad \quad \dots & \quad \quad \dots  \\
\frac{2(1-x)^1}{(1-m_1)(1-m_2)(1-m_3)} (3x-m_1-m_2-m_3) & m_3 < x \leq m_4 \\
\frac{1(1-x)^0}{(1-m_1)(1-m_2)} (2x-m_1-m_2) & m_2 < x \leq m_3 \\
0 & \text{otherwise} 
\end{cases}$$
