两个服从Gamma分布的随机变量的和的pdf和cdf
随机变量\(X\)服从参数\((\alpha,\beta)\)的Gamma分布，则其概率密度函数(pdf)可以表示为
\[f(x)=\frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x},(x > 0)\tag{1}
\]
其对应的矩母函数为\({\rm M}_x(t)=(1+t/\beta)^{-\alpha}\)，所以若\(X_1 \backsim G(\alpha_1,\beta_1),X_2 \backsim G(\alpha_2,\beta_2),X=X_1+X_2\)，且\(\beta_1=\beta_2=\beta\)，则由矩母函数可以很容易得到\({\rm M}_x(t)={\rm M}_{x_1}(t){\rm M}_{x_2}(t)=(1+t/\beta)^{-(\alpha_1+\alpha_2)}\)，所以\(X \backsim G(\alpha_1+\alpha_2,\beta)\)，该性质称为Gamma分布的可加性，其更具体的推导过程可以查看《不同方法推导Gamma分布可加性产生的矛盾》。对于更一般的情况，即两个随机变量服从参数不同的Gamma分布，则这两个随机变量和的随机变量的矩母函数可以表示\({\rm M}_x(t)={\rm M}_{x_1}(t){\rm M}_{x_2}(t)=(1+t/\beta_1)^{-\alpha_1}(1+t/\beta_2)^{-\alpha_2}\)，此时不再服从上述可加性性质。本文将就上述\(\alpha_1,\alpha_2 \in \Bbb Z^+\)情况下的两个Gamma分布随机变量的和的pdf和cdf进行推导。
根据概率论相关知识，我们知道，两个随机变量的和的pdf等于各自pdf的卷积，因此
\[\begin{equation}
\begin{aligned}
f_x(x)=\int_{-\infty}^\infty f_{x_1}(y)f_{x_2}(x-y)dy
\end{aligned}
\end{equation}\tag{2}
\]
易知，仅当\(y>0,x-y>0\)，即\(y>0,y<x\)时，上述积分的被积函数不等于0，将相关表达式带入(2)可以得到
\[\begin{equation}
\begin{aligned}
f_x(x)&=\frac{\beta_1^{\alpha_1}\beta_2^{\alpha_2}}{\Gamma(\alpha_1)\Gamma(\alpha_2)} \int_0^x y^{\alpha_1-1}e^{-\beta_1 y}(x-y)^{\alpha_2-1}e^{-\beta_2(x-y)}dy\\
&=\frac{\beta_1^{\alpha_1}\beta_2^{\alpha_2}}{\Gamma(\alpha_1)\Gamma(\alpha_2)} e^{-\beta_2 x} \int_0^x y^{\alpha_1-1}(x-y)^{\alpha_2-1}e^{-(\beta_1-\beta_2)y}dy
\end{aligned}
\end{equation}\tag{3}
\]
由二项展开定理有:\((x-y)^{\alpha_2-1}=\sum_{i=0}^{\alpha_2-1}(-1)^{\alpha_2-1-i} \begin{pmatrix}\alpha_2-1\\ i\end{pmatrix}x^i y^{\alpha_2-1-i}\)，将其代入(3)并化简有
\[\begin{equation}
\begin{aligned}
f_x(x)&=\frac{\beta_1^{\alpha_1}\beta_2^{\alpha_2}}{\Gamma(\alpha_1)\Gamma(\alpha_2)} e^{-\beta_2 x} \sum_{i=0}^{\alpha_2-1}(-1)^{\alpha_2-1-i} \begin{pmatrix}\alpha_2-1\\ i\end{pmatrix}x^i \int_0^x y^{\alpha_1+\alpha_2-i-2}e^{-(\beta_1-\beta_2)}dy
\end{aligned}
\end{equation}\tag{4}
\]
上述积分式的求导，需要用到下面的定理[1]：
\[\begin{equation}
\begin{aligned}
\int_0^\mu{x^m e^{-tx}dx}=m!t^{-(m+1)}[1-\sum_{k=0}^m{\frac{t^k}{k!} \mu^k e^{-t\mu}}] \quad\underrightarrow{\mu \rightarrow \infty} \quad \int_0^\mu{x^m e^{-tx}dx}=m!t^{-(m+1)}
\end{aligned}
\end{equation}\tag{5}
\]
所以
\[\int_0^x y^{\alpha_1+\alpha_2-i-2}e^{-(\beta_1-\beta_2)}dy=(\alpha_1+\alpha_2-i-2)!(\beta_1-\beta_2)^{-(\alpha_1+\alpha_2-i-1)}[1-\sum_{k=0}^{\alpha_1+\alpha_2-i-2}{\frac{(\beta_1-\beta_2)^k}{k!} x^k e^{-(\beta_1-\beta_2) x}}]\tag{6}
\]
将(6)代入(4)整理得到
\[\begin{equation}
\begin{aligned}
f_x(x)=A_0\sum_{i=0}^{\alpha_2-1}A_1(i)[x^i e^{-\beta_2 x}-\sum_{k=0}^{\alpha_1+\alpha_2-i-2}{\frac{(\beta_1-\beta_2)^k}{k!} x^{k+i} e^{-\beta_1 x}}]
\end{aligned}
\end{equation}\tag{7}
\]
其中
\[\begin{equation}
\begin{aligned}
A_0 &\overset{\Delta}{=} A_0(\alpha_1,\alpha_2,\beta_1,\beta_2)=\frac{\beta_1^{\alpha_1}\beta_2^{\alpha_2}}{\Gamma(\alpha_1)\Gamma(\alpha_2)}\\
A_1(i)&\overset{\Delta}{=}A_1(\alpha_1,\alpha_2,\beta_1,\beta_2,i)=(-1)^{\alpha_2-1-i} \begin{pmatrix}\alpha_2-1\\ i\end{pmatrix}(\alpha_1+\alpha_2-i-2)!(\beta_1-\beta_2)^{-(\alpha_1+\alpha_2-i-1)}
\end{aligned}
\end{equation}\tag{8}
\]
\(X\)的cdf可以表示为其pdf的积分，因此
\[\begin{equation}
\begin{aligned}
F_x(x)=\int_0^x f_x(x)dx=I_0-I_1
\end{aligned}
\end{equation}\tag{9}
\]
其中
\[\begin{equation}
\begin{aligned}
I_0&=A_0\sum_{i=0}^{\alpha_2-1}A_1(i)\int_0^x x^i e^{-\beta_2 x}dx\\
&=A_0\sum_{i=0}^{\alpha_2-1}A_1(i) i!\beta_2^{-(i+1)}[1-\sum_{k=0}^i{\frac{\beta_2^k}{k!} x^k e^{-\beta_2 x}}]\\
I_1&=A_0\sum_{i=0}^{\alpha_2-1}A_1(i)\sum_{k=0}^{\alpha_1+\alpha_2-i-2}{\frac{(\beta_1-\beta_2)^k}{k!} \int_0^x x^{k+i} e^{-\beta_1 x}}dx\\
&=A_0\sum_{i=0}^{\alpha_2-1}A_1(i)\sum_{k=0}^{\alpha_1+\alpha_2-i-2}\frac{(\beta_1-\beta_2)^k}{k!}(k+i)!\beta_1^{-(k+i+1)}[1-\sum_{j=0}^{k+i}{\frac{\beta_1^j}{j!} x^j e^{-\beta_1 x}}]
\end{aligned}
\end{equation}\tag{10}
\]
将(10)带入(9)并整理得到
\[\begin{equation}
\begin{aligned}
F_x(x)&=A_0\sum_{i=0}^{\alpha_2-1}A_1(i)[i!\beta_2^{-(i+1)}-\sum_{k=0}^{\alpha_1+\alpha_2-i-2}\frac{(\beta_1-\beta_2)^k}{k!}(k+i)!\beta_1^{-(k+i+1)}]\\
&-A_0\sum_{i=0}^{\alpha_2-1}A_1(i)[i!\sum_{k=0}^i{\frac{\beta_2^{-(i-k+1)}}{k!} x^k e^{-\beta_2 x}}-\sum_{k=0}^{\alpha_1+\alpha_2-i-2}\frac{(\beta_1-\beta_2)^k}{k!}(k+i)!\sum_{j=0}^{k+i}{\frac{\beta_1^{-(k+i-j+1)}}{j!} x^j e^{-\beta_1 x}}]
\end{aligned}
\end{equation}\tag{11}
\]
通过数值仿真发现(11)式中的前一部分满足
\[A_0\sum_{i=0}^{\alpha_2-1}A_1(i)[i!\beta_2^{-(i+1)}-\sum_{k=0}^{\alpha_1+\alpha_2-i-2}\frac{(\beta_1-\beta_2)^k}{k!}(k+i)!\beta_1^{-(k+i+1)}]\equiv 1
\]
但是目前还不知道如何证明？显然此时pdf和cdf的表达式均很复杂，也有可能还没有化简到最简形式，但是暂时不知道怎样化简下去了。
参考文献
[1] Table of Integrals, Series, and Products
