3 Improved math and added independency condition.
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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
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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 $\sum x_i \sim \text{Gamma}(n,\lambda)$

$$f(x|\alpha,\beta)=\frac{\beta^α}{\Gamma(\alpha)} \cdot x^{\alpha−1} \cdot e^{−x\beta} $$

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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)