PROOF: Reproductive Property of the Gamma Distribution / Probability Theory

In this page, we show that the sum of two gamma-distributed random variables, Y:=X1+X2, also follows the gamma distribution: i.e. the gamma distribution has the reproductive property.

Reproductive Property of the Gamma Distribution

If the two random variables X_1 and X_2 follow the gamma distributions Gamma(k_1,\theta) and Gamma(k_2,\theta):

(1)   \begin{equation*} \text{PDF : } \; f_{X_i}(x_i) = \frac{ x_i^{k_i-1} e^{-x_i/\theta} }{ \Gamma(k_i)\; \theta^{k_i} }  \qquad (i = 1,2), \end{equation*}

then the sum of the two random variables Y:= X_1 + X_2 follows the gamma distribution Gamma(k_1+k_2,\theta):

(2)   \begin{equation*} \text{PDF : } \; f_Y(y) = \frac{ y^{k_1+k_2-1} e^{-y/\theta} }{ \Gamma(k_1+k_2)\; \theta^{k_1+k_2} }. \end{equation*}



Preparation for the Proof

The following equation holds:

(3)   \begin{equation*} \int_0^y x^{k_1-1} (y-x)^{k_2-1} dx \quad = \quad \frac{\Gamma(k_1)\; \Gamma(k_2)}{\Gamma(k_1+k_2)}y^{k_1+k_2-1}. \end{equation*}

If we transform the variable x in the left side of the above equation (3) into z:=x/y, then we obtain x=yz, dx=ydz, and z\in[0,1), thus

(4)   \begin{eqnarray*} &&\int_0^y x^{k_1-1} (y-x)^{k_2-1} dx \\ &=& \int_0^1 (yz)^{k_1-1} (y-yz)^{k_2-1} ydz\\ &=& y^{k_1+k_2-1} \int_0^1 z^{k_1-1} (1-z)^{k_2-1} dz. \end{eqnarray*}

The integration on the last part of the above equation (4) is the beta function, which is written as

(5)   \begin{equation*}  \int_0^1 z^{k_1-1} (1-z)^{k_2-1} dz \quad = \quad \frac{\Gamma(k_1)\; \Gamma(k_2)}{\Gamma(k_1+k_2)}, \end{equation*}

using the gamma functions.
Therefore we obtain

(6)   \begin{equation*} \int_0^y x^{k_1-1} (y-x)^{k_2-1} dx \quad = \quad \frac{\Gamma(k_1)\; \Gamma(k_2)}{\Gamma(k_1+k_2)}y^{k_1+k_2-1}. \end{equation*}



Generally the PDF of the sum of two random variables Y:=X_1+X_2, f_Y, is derived from the convolution of the PDFs, f_{X_1} and f_{X_2}:

(7)   \begin{equation*} f_Y(y) = \int_0^y f_{X_1}(x)f_{X_2}(y-x)dx. \end{equation*}

If we substitute the PDFs of the gamma distributions (1) into the above convolution (7), we obtain

(8)   \begin{eqnarray*} f_Y(y)  &=& \int_0^y f_{X_1}(x)f_{X_2}(y-x)dx \\ &=& \int_0^y \frac{x^{k_1-1} e^{-x/\theta} }{\Gamma(k_1)\;\theta^{k_1}} \cdot \frac{(y-x)^{k_2-1} e^{-(y-x)/\theta} }{\Gamma(k_2)\; \theta^{k_2}} dx \\ &=& \frac{e^{-y/\theta} }{\Gamma(k_1)\;\Gamma(k_2)\;\theta^{k_1+k_2}} \int_0^y x^{k_1-1} (y-x)^{k_2-1} dx. \end{eqnarray*}

The integration on the last part of the above equation (8) equals to the equation (3) which is shown in the Preparation, thus we obtain the following equation:

(9)   \begin{eqnarray*} f_Y(y)  &=& \frac{e^{-y/\theta} }{\Gamma(k_1)\;\Gamma(k_2)\;\theta^{k_1+k_2}} \int_0^y x^{k_1-1} (y-x)^{k_2-1} dx \\ &=& \frac{e^{-y/\theta} }{\Gamma(k_1)\;\Gamma(k_2)\;\theta^{k_1+k_2}} \cdot \frac{\Gamma(k_1)\; \Gamma(k_2)}{\Gamma(k_1+k_2)}y^{k_1+k_2-1} \\ &=& \frac{y^{k_1+k_2-1} \; e^{-y/\theta} }{\Gamma(k_1+k_2)\;\theta^{k_1+k_2}}. \end{eqnarray*}



Remark 1

The gamma distribution Gamma(k,\theta) with the integer k and the parameter \theta = 1/\lambda implies the Erlang distribution Erlang(k,\lambda). The proof in this page also holds in the Erlang distribution. [Reproductive Property of the Erlang Distribution (in JPN) ]


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