3 Improved math and added independency condition. edit approved Sep 25 '15 at 20:30 Waldir Leoncio 1,49255 gold badges2424 silver badges3535 bronze badges gamma distribution is made of exponential distribution that is exponential distribution is base for gamma distribution. then if $$f(x|\lambda)=\lambda e^{−\lambda x}$$ we have $$\sum x_i \sim \text{Gamma}(n,\lambda)$$$$\sum_n x_i \sim \text{Gamma}(n,\lambda)$$, as long as all $$X_i$$ are independent. $$f(x|\alpha,\beta)=\frac{\beta^α}{\Gamma(\alpha)} \cdot x^{\alpha−1} \cdot e^{−x\beta}$$ gamma distribution is made of exponential distribution that is exponential distribution is base for gamma distribution. then if $$f(x|\lambda)=\lambda e^{−\lambda x}$$ we have $$\sum x_i \sim \text{Gamma}(n,\lambda)$$ $$f(x|\alpha,\beta)=\frac{\beta^α}{\Gamma(\alpha)} \cdot x^{\alpha−1} \cdot e^{−x\beta}$$ gamma distribution is made of exponential distribution that is exponential distribution is base for gamma distribution. then if $$f(x|\lambda)=\lambda e^{−\lambda x}$$ we have $$\sum_n x_i \sim \text{Gamma}(n,\lambda)$$, as long as all $$X_i$$ are independent. $$f(x|\alpha,\beta)=\frac{\beta^α}{\Gamma(\alpha)} \cdot x^{\alpha−1} \cdot e^{−x\beta}$$ 2 added 95 characters in body edited Aug 3 '15 at 8:42 Andy 15.5k1111 gold badges6565 silver badges8989 bronze badges gamma distribution is made of exponential distribution that is exponential distribution is base for gamma distribution. then if f(x|λ)=λe^( −λx)$$f(x|\lambda)=\lambda e^{−\lambda x}$$ we have ∑xi~gamma(n,λ) $$\sum x_i \sim \text{Gamma}(n,\lambda)$$ f(x|α,β)=β^α/Γ(α)* x^(α−1) * e^(−xβ) $$f(x|\alpha,\beta)=\frac{\beta^α}{\Gamma(\alpha)} \cdot x^{\alpha−1} \cdot e^{−x\beta}$$ gamma distribution is made of exponential distribution that is exponential distribution is base for gamma distribution. then if f(x|λ)=λe^( −λx) we have ∑xi~gamma(n,λ) f(x|α,β)=β^α/Γ(α)* x^(α−1) * e^(−xβ) gamma distribution is made of exponential distribution that is exponential distribution is base for gamma distribution. then if $$f(x|\lambda)=\lambda e^{−\lambda x}$$ we have $$\sum x_i \sim \text{Gamma}(n,\lambda)$$ $$f(x|\alpha,\beta)=\frac{\beta^α}{\Gamma(\alpha)} \cdot x^{\alpha−1} \cdot e^{−x\beta}$$ 1 answered Aug 3 '15 at 8:22 hasanmisaii 1111 bronze badge gamma distribution is made of exponential distribution that is exponential distribution is base for gamma distribution. then if f(x|λ)=λe^( −λx) we have ∑xi~gamma(n,λ) f(x|α,β)=β^α/Γ(α)* x^(α−1) * e^(−xβ)