PDF of $Y=\max(x_{11},x_{12},x_{13}, \ldots)+\max(x_{21},x_{22},x_{23},\ldots)+\cdots$ needed I need to find the probability density function (pdf) of the following variable 
\begin{align}
Y= \max(x_{11},x_{12},x_{13}, \ldots, x_{x1N})+ {} & \max(x_{21},x_{22},x_{23},\ldots, x_{2N})+\cdots \\
& \cdots+\max(x_{K1},x_{K2},x_{K3},\ldots, x_{KN})
\end{align}
where all $x_{ij}$ are i.i.d exponential random variables with same parameters (mean). I know the pdf of individual terms in the summation but I do not know whether there is an expression for the pdf of sum available or not. I will be very thankful if somebody could provide me the desired pdf or could provide me some reference where I can find it. Thanks in advance.
 A: We can get the moment-generating function of $Y$, but I'm not aware of a distribution with such an MGF.
Let $X_{ij}$ for $i \in 1, 2, \ldots K$ and $j \in 1, 2, \ldots N$ be iid exponential random variables with rate parameter $\theta$.  Define $X_{i(N)} = \max(x_{i1}, x_{i2}, \ldots, x_{iN})$ for each $i = 1, 2, \ldots, K$.  Then it is obvious that $$F_{X_{i(N)}}(x) = \Pr[X_{i(N)} \le x] = \left(\Pr[X_{ij} \le x]\right)^N = (1 - e^{-\theta x})^N$$ is the CDF of the maximum order statistic and the PDF is simply $$f_{X_{i(N)}}(x) = N \theta e^{-\theta x} (1 - e^{-\theta x})^{N-1}, \quad x > 0.$$  Consequently the MGF of $X_{i(N)}$ is $$M_{X_{i(N)}} = \operatorname{E}[e^{tX_{i(N)}}] = N \theta \int_{x=0}^\infty  e^{(t-\theta) x} (1 - e^{-\theta x})^{N-1} \, dx.$$  Since $N$ is a positive integer, we can expand the integrand and integrate term by term:
$$\begin{align*} M_{X_{i(N)}} (t) &= N\theta \sum_{n=0}^{N-1} \binom{N-1}{n} \int_{x=0}^\infty e^{(t-\theta)x} (-e^{-\theta x})^n \, dx \\
&= N \theta \sum_{n=0}^{N-1} \binom{N-1}{n} (-1)^n \int_{x=0}^\infty e^{(t-\theta(n+1))x} \, dx \\
&= N \sum_{n=0}^{N-1} \binom{N-1}{n} \frac{(-1)^n}{(n+1)-t/\theta} \\ 
&= \frac{\Gamma(N+1) \Gamma(1 - t/\theta)}{\Gamma(N + 1 - t/\theta)}, \quad t < \theta.
 \end{align*}$$  This is also expressible as a beta function; i.e., $$M_{X_{i(N)}} (t) = N B(N,1-t/\theta).$$  Then the MGF of $Y = \sum_{i=1}^K X_{i(N)}$ is $$M_Y(t) = \prod_{i=1}^K M_{X_{i(N)}} (t) = N^K (B(N,1-t/\theta))^K.$$  But I do not recognize such an MGF.
