Does the sum of two independent exponentially distributed random variables with different rate parameters follow a gamma distribution?

Question

Suppose we have two independent exponentially distributed random variables with means $400$ and $200$, so that their respective rate parameters are $1/400$ and $1/200$. Do these random variables then follow a gamma distribution with shape parameter equal to $2$ and rate parameter equal to $1/300$?

I know that two independent exponentially distributed random variables with the same rate parameter follow a gamma distribution with shape parameter equal to the amount of exponential r.v.'s involved and rate parameter equal to the rate parameter of those exponential r.v.'s, but what about exponentially distributed r.v.'s with different rate parameters?

I looked online but could not find the answer, so I suppose that the answer is no. Is there a simple way to get the convoluted distribution of two exponentially distributed r.v.'s with different rate parameters?

Answer 1

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.

Answer 2

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}) , \ \ z\geq 0 \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=\lambda $, 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^2 \int_0^z e^{-\lambda(z-x)}e^{-\lambda x}dx \\&=\lambda^2 e^{-\lambda z}\int_0^z 1 dx \\&=\lambda^2 z e^{-\lambda z}, z\geq 0 \end{align} which is the two-parameter Gamma(2, $\lambda$) distribution.

Also, similar results can be calculated with LST.