Let $X_1$ and $X_2$ two random variable with CDF \begin{align} F_{X_1}(x)&=(1-e^{-ax})^{N_1}\\ F_{X_2}(x)&=(1-e^{-ax})^{N^2} \end{align} Let $Z$ random variable with CDF $$ F_Z(z)=[1-(1-F_{X_1}(x))(1-F_{X_2}(x))]^M $$ The PDF is the derivation of CDF given by $$ f_Z(x)=M[1-(1-F_{X_1}(x))(1-F_{X_2}(x))]^{M-1}(f_{X1}(x)+f_{X2}(x)-f_{X1}(x)F_{X2}(x)-f_{X2}(x)F_{X1}(x) $$ My Problem is to compute the Moment generatrice function of $Z$.
I would like to write $f_Z(x)$ in term of sum exponential, because it easy to compute the MGF.
The difficult part in $f_Z(x)$ is $$ A=[1-(1-F_{X_1}(x))(1-F_{X_2}(x))]^{M-1} $$ After i use binomial theorem i arrive to the flowing result
$$A=\sum_{z=0}^{M-1}\binom{M-1}{z}(-1)^z \begin{pmatrix} 1-\sum_{m=0}^{N_1}\binom{N_1}{m}(-1)^me^{-xm} \end{pmatrix}^z\begin{pmatrix} 1-\sum_{m=0}^{N_2}\binom{N_2}{m}(-1)^me^{-xm} \end{pmatrix}^z$$
I found in paper the following simplification $$ A=\sum_{z=0}^{N-1}\sum_{a=0}^{z}\sum_{b=0}^{z}\sum_{n=0}^{aN_1}\sum_{m=0}^{bN_2}\binom{N-1}{z}\binom{z}{a}\binom{z}{b}\binom{aN_1}{n}\binom{aN_2}{m}(-1)^{z+a+b+n+m} me^{-x(m+n)} $$ Can any one explain to me how the simplify it.
Let me use the notation $\mu=M-1$. Define $X=e^{-ax}$ and $F_i=(1-X)^{N_i}$. Then
$$ \begin{aligned} &\left[1-(1-F_1)(1-F_2)\right]^\mu \\ \xrightarrow{\text{expand }(\cdots)^\mu} &\sum_{z=0}^\mu \binom{\mu}{z}(-1)^z (1-F_1)^z(1-F_2)^{z} \\ \xrightarrow{\text{expand }(1-F_i)^z}& \sum_{z=0}^\mu \binom{\mu}{z}(-1)^z \sum_{a=0}^z \binom{z}{a}(-1)^a F_1^a \sum_{b=0}^z \binom{z}{b}(-1)^b F_2^b\\ \xrightarrow{F_i=(1-X)^{N_i}}& \sum_{z=0}^\mu \sum_{a=0}^z \sum_{b=0}^z \binom{\mu}{z}\binom{z}{a}\binom{z}{b}(-1)^{z+a+b} (1-X)^{aN_1} (1-X)^{bN_2}\\ \xrightarrow{\text{expand }(1-X)^{\cdots}}& \sum_{z=0}^\mu \sum_{a=0}^z \sum_{b=0}^z \binom{\mu}{z}\binom{z}{a}\binom{z}{b}(-1)^{z+a+b} \sum_{m=0}^{aN_1}\binom{aN_1}{m}(-1)^m X^m \sum_{n=0}^{bN_2} \binom{bN_2}{n}(-1)^n X^n \end{aligned} $$