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If $X\sim \exp(\lambda_1)$, $Y\sim\exp(\lambda_2)$ and $\lambda_1\neq\lambda_2$, the sum $Z=X+Y$ has pdf given by the convolution \begin{align} f_Z(z) &=\int_{-\infty}^\infty f_Y(z-x)f_X(x)dx \\&=\lambda_1\lambda_2\int_0^z e^{-\lambda_2(z-x)}e^{-\lambda_1x}dx \\&=\lambda_1\lambda_2 e^{-\lambda_2z}\int_0^z e^{-(\lambda_1-\lambda_2)x}dx \\&=\frac{\lambda_1\lambda_2}{\lambda_1-\lambda_2} e^{-\lambda_2z}(1- e^{-(\lambda_1-\lambda_2)z}) \\&=\frac{\lambda_1\lambda_2}{\lambda_1-\lambda_2} (e^{-\lambda_2z} - e^{-\lambda_1z}) \end{align} which is the two-parameter hyperexponential distribution.

If $X\sim \exp(\lambda_1)$, $Y\sim\exp(\lambda_2)$ and $\lambda_1\neq\lambda_2$, the sum $Z=X+Y$ has pdf given by the convolution \begin{align} f_Z(z) &=\int_{-\infty}^\infty f_Y(z-x)f_X(x)dx \\&=\lambda_1\lambda_2\int_0^z e^{-\lambda_2(z-x)}e^{-\lambda_1x}dx \\&=\lambda_1\lambda_2 e^{-\lambda_2z}\int_0^z e^{-(\lambda_1-\lambda_2)x}dx \\&=\frac{\lambda_1\lambda_2}{\lambda_1-\lambda_2} e^{-\lambda_2z}(1- e^{-(\lambda_1-\lambda_2)z}) \\&=\frac{\lambda_1\lambda_2}{\lambda_1-\lambda_2} (e^{-\lambda_2z} - e^{-\lambda_1z}) \end{align} which is the two-parameter hypoexponential distribution.

If $X\sim \exp(\lambda_1)$, $Y\sim\exp(\lambda_2)$ and $\lambda_1>\lambda_2$, the sum $Z=X+Y$ has pdf given by the convolution \begin{align} f_Z(z) &=\int_{-\infty}^\infty f_Y(z-x)f_X(x)dx \\&=\lambda_1\lambda_2\int_0^z e^{-\lambda_2(z-x)}e^{-\lambda_1x}dx \\&=\lambda_1\lambda_2 e^{-\lambda_2z}\int_0^z e^{-(\lambda_1-\lambda_2)x}dx \\&=\frac{\lambda_1\lambda_2}{\lambda_1-\lambda_2} e^{-\lambda_2z}(1- e^{-(\lambda_1-\lambda_2)z}) \\&=\frac{\lambda_1\lambda_2}{\lambda_1-\lambda_2} (e^{-\lambda_2z} - e^{-\lambda_1z}) \end{align} which is the two-parameter hyperexponential distribution.

If $X\sim \exp(\lambda_1)$, $Y\sim\exp(\lambda_2)$ and $\lambda_1\neq\lambda_2$, the sum $Z=X+Y$ has pdf given by the convolution \begin{align} f_Z(z) &=\int_{-\infty}^\infty f_Y(z-x)f_X(x)dx \\&=\lambda_1\lambda_2\int_0^z e^{-\lambda_2(z-x)}e^{-\lambda_1x}dx \\&=\lambda_1\lambda_2 e^{-\lambda_2z}\int_0^z e^{-(\lambda_1-\lambda_2)x}dx \\&=\frac{\lambda_1\lambda_2}{\lambda_1-\lambda_2} e^{-\lambda_2z}(1- e^{-(\lambda_1-\lambda_2)z}) \\&=\frac{\lambda_1\lambda_2}{\lambda_1-\lambda_2} (e^{-\lambda_2z} - e^{-\lambda_1z}) \end{align} which is the two-parameter hyperexponential distribution.

If $X\sim \exp(\lambda_1)$, $Y\sim\exp(\lambda_2)$ and $\lambda_1>\lambda_2$, the sum $Z=X+Y$ has pdf given by the convolution \begin{align} f_Z(z) &=\int_{-\infty}^\infty f_Y(z-x)f_X(x)dx \\&=\lambda_1\lambda_2\int_0^z e^{-\lambda_2(z-x)}e^{-\lambda_1x}dx \\&=\lambda_1\lambda_2 e^{-\lambda_2z}\int_0^z e^{-(\lambda_1-\lambda_2)x}dx \\&=\frac{\lambda_1\lambda_2}{\lambda_1-\lambda_2} e^{-\lambda_2z}(1- e^{-(\lambda_1-\lambda_2)z}) \\&=\frac{\lambda_1\lambda_2}{\lambda_1-\lambda_2} (e^{-\lambda_2z} - e^{-\lambda_1z}) \end{align} which is not a Gamma distribution.

If $X\sim \exp(\lambda_1)$, $Y\sim\exp(\lambda_2)$ and $\lambda_1>\lambda_2$, the sum $Z=X+Y$ has pdf given by the convolution \begin{align} f_Z(z) &=\int_{-\infty}^\infty f_Y(z-x)f_X(x)dx \\&=\lambda_1\lambda_2\int_0^z e^{-\lambda_2(z-x)}e^{-\lambda_1x}dx \\&=\lambda_1\lambda_2 e^{-\lambda_2z}\int_0^z e^{-(\lambda_1-\lambda_2)x}dx \\&=\frac{\lambda_1\lambda_2}{\lambda_1-\lambda_2} e^{-\lambda_2z}(1- e^{-(\lambda_1-\lambda_2)z}) \\&=\frac{\lambda_1\lambda_2}{\lambda_1-\lambda_2} (e^{-\lambda_2z} - e^{-\lambda_1z}) \end{align} which is the two-parameter hyperexponential distribution.