How Do Mean and Variance Relate in a Gamma Distribution?

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SUMMARY

The discussion confirms the relationships for the mean (μ) and variance (σ²) of a Gamma distribution, specifically stating that μ = αθ and σ² = αθ². The derivation of these formulas utilizes the properties of the Gamma function, including the recurrence relation and integration techniques. The calculations for E(X) and E(X²) are presented in detail, leading to the final variance expression. The consensus among participants validates the correctness of these mathematical expressions.

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  • Understanding of Gamma distribution and its parameters (α and θ)
  • Familiarity with the Gamma function and its properties
  • Knowledge of integration techniques, particularly improper integrals
  • Basic statistics concepts, including mean and variance
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prove that μ=αθ and σ^2=αθ^2




μ = E(X)
...= ∫(x = 0 to ∞) x *(1/(θ^α Γ(α))) x^(α-1) e^(-x/θ) dx
...= (1/(θ^α Γ(α))) ∫(x = 0 to ∞) x^α e^(-x/θ) dx
...= (1/(θ^α Γ(α))) ∫(t = 0 to ∞) (tθ)^α e^(-t) * (θ dt), letting t = x/θ
...= (θ/Γ(α)) ∫(t = 0 to ∞) t^α e^(-t) dt
...= (θ/Γ(α)) Γ(α+1), by definition of Gamma function
...= (θ/Γ(α)) * α Γ(α), by Gamma recurrence
...= αθ.

E(X^2) = ∫(x = 0 to ∞) x^2 * (1/(θ^α Γ(α))) x^(α-1) e^(-x/θ) dx
...= (1/(θ^α Γ(α))) ∫(x = 0 to ∞) x^(α+1) e^(-x/θ) dx
...= (1/(θ^α Γ(α))) ∫(t = 0 to ∞) (tθ)^(α+1) e^(-t)* (θ dt), letting t = x/θ
...= (θ^2/Γ(α)) ∫(t = 0 to ∞) t^(α+1) e^(-t) dt
...= (θ^2/Γ(α)) Γ(α+2), by definition of Gamma function
...= (θ^2/Γ(α)) * α(α+1) Γ(α), by Gamma recurrence (used twice)
...= α(α+1)θ^2.

So, σ^2 = E(X^2) - (E(X))^2 = α(α+1)θ^2 - (αθ)^2 = αθ^2




i would like to know if it is correct and tnx :biggrin:
 
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mcguiry03 said:
prove that μ=αθ and σ^2=αθ^2




μ = E(X)
...= ∫(x = 0 to ∞) x *(1/(θ^α Γ(α))) x^(α-1) e^(-x/θ) dx
...= (1/(θ^α Γ(α))) ∫(x = 0 to ∞) x^α e^(-x/θ) dx
...= (1/(θ^α Γ(α))) ∫(t = 0 to ∞) (tθ)^α e^(-t) * (θ dt), letting t = x/θ
...= (θ/Γ(α)) ∫(t = 0 to ∞) t^α e^(-t) dt
...= (θ/Γ(α)) Γ(α+1), by definition of Gamma function
...= (θ/Γ(α)) * α Γ(α), by Gamma recurrence
...= αθ.

E(X^2) = ∫(x = 0 to ∞) x^2 * (1/(θ^α Γ(α))) x^(α-1) e^(-x/θ) dx
...= (1/(θ^α Γ(α))) ∫(x = 0 to ∞) x^(α+1) e^(-x/θ) dx
...= (1/(θ^α Γ(α))) ∫(t = 0 to ∞) (tθ)^(α+1) e^(-t)* (θ dt), letting t = x/θ
...= (θ^2/Γ(α)) ∫(t = 0 to ∞) t^(α+1) e^(-t) dt
...= (θ^2/Γ(α)) Γ(α+2), by definition of Gamma function
...= (θ^2/Γ(α)) * α(α+1) Γ(α), by Gamma recurrence (used twice)
...= α(α+1)θ^2.

So, σ^2 = E(X^2) - (E(X))^2 = α(α+1)θ^2 - (αθ)^2 = αθ^2




i would like to know if it is correct and tnx :biggrin:

It looks OK.

RGV
 

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