为方便阅读，本推导采用浙大版概率论与数理统计教材的标注方法。设 X,Y 是相互独立的连续型随机变量， X\sim \text{Laplace}(\mu_1,b), Y\sim \text{Laplace}(\mu_2,b),~Z=X+Y 。即： X\sim \frac{1}{2b}e^{-\frac{|x-\mu_1|}{b}},~Y\sim \frac{1}{2b}e^{-\frac{|x-\mu_2|}{b}} 。因为 X,Y 是相互独立的，故联合分布的概率密度为：f(x,y)=f_X(x)f_Y(y)=\frac{1}{4b^2}e^{\frac{-|x-\mu_1|-|y-\mu_2|}{b}}\\ 根据相互独立的连续型随机变量 Z=X+Y 概率密度公式 (证明见教材3.5节)：\begin{align}f_{X+Y}(z)&=\int_{-\infty}^{\infty}f(z-y,y)dy=\int_{-\infty}^{\infty}f_X(z-y)f_Y(y)dy\\ &=\int_{-\infty}^{\infty}f(x,z-x)dx=\int_{-\infty}^{\infty}f_X(x)f_Y(z-x)dx \end{align} \\ 有：f_{X+Y}(z)=\int_{-\infty}^{\infty}\frac{1}{4b^2}e^{\frac{-|x-\mu_1|-|z-x-\mu_2|}{b}}dx\\ 1.~Z\geq \mu_1+\mu_2 ：\begin{align} f_{X+Y}(z)&=\int_{-\infty}^{\mu_1}\frac{1}{4b^2}e^{\frac{x-\mu_1-z+x+\mu_2}{b}}dx+\int_{\mu_1}^{z-\mu_2}\frac{1}{4b^2}e^{\frac{\mu_1-x-z+x+\mu_2}{b}}dx+\int_{z-\mu_2}^{\infty}\frac{1}{4b^2}e^{\frac{-x+\mu_1+z-x-\mu_2}{b}}dx\\
&=\int_{-\infty}^{\mu_1}\frac{1}{4b^2}e^{\frac{2x-\mu_1-z+\mu_2}{b}}dx+\int_{\mu_1}^{z-\mu_2}\frac{1}{4b^2}e^{\frac{\mu_1-z+\mu_2}{b}}dx+\int_{z-\mu_2}^{\infty}\frac{1}{4b^2}e^{\frac{-2x+\mu_1+z-\mu_2}{b}}dx\\
&=\frac{1}{4b^2}\left( \frac{b}{2}e^{\frac{\mu_1+\mu_2-z}{b}}+(z-\mu_1-\mu_2)e^{\frac{\mu_1+\mu_2-z}{b}}+\frac{b}{2}e^{\frac{\mu_1+\mu_2-z}{b}} \right)\\
&=\frac{1}{4b^2}\left( (z-\mu_1-\mu_2+b)e^{\frac{\mu_1+\mu_2-z}{b}} \right)
\end{align}\\
2.~Z< \mu_1+\mu_2 ：\begin{align} f_{X+Y}(z)&=\int_{-\infty}^{z-\mu_2}\frac{1}{4b^2}e^{\frac{x-\mu_1-z+x+\mu_2}{b}}dx+\int_{z-\mu_2}^{\mu_1}\frac{1}{4b^2}e^{\frac{x-\mu_1-x+z-\mu_2}{b}}dx+\int_{\mu_1}^{\infty}\frac{1}{4b^2}e^{\frac{-x+\mu_1+z-x-\mu_2}{b}}dx\\
&=\int_{-\infty}^{z-\mu_2}\frac{1}{4b^2}e^{\frac{2x-\mu_1-z+\mu_2}{b}}dx+\int_{z-\mu_2}^{\mu_1}\frac{1}{4b^2}e^{\frac{z-\mu_1-\mu_2}{b}}dx+\int_{\mu_1}^{\infty}\frac{1}{4b^2}e^{\frac{-2x+\mu_1+z-\mu_2}{b}}dx\\
&=\frac{1}{4b^2}\left( \frac{b}{2}e^{\frac{z-\mu_1-\mu_2}{b}}+(\mu_1+\mu_2-z)e^{\frac{z-\mu_1-\mu_2}{b}}+\frac{b}{2}e^{\frac{z-\mu_1-\mu_2}{b}} \right)\\
&=\frac{1}{4b^2}\left( (\mu_1+\mu_2-z+b)e^{\frac{z-\mu_1-\mu_2}{b}} \right)
\end{align}\\
综上：\begin{align} f_{X+Y}(z)
&=\frac{|\mu_1+\mu_2-z|+b}{4b^2}e^{\frac{-|z-\mu_1-\mu_2|}{b}}
\end{align}\\
}