Final answer:
In this case, if X ~ Γ(α, 1), then βX ~ Γ(α, 1/β).
ᶠ_ˣ⁽ˣ⁾ ⁼ ⁽¹/Γ⁽α⁾⁾ * β^α * ˣ^⁽α⁻¹⁾ * ᵉ^⁽⁻ˣ/β⁾
Step-by-step explanation:
To compute the moment generating function (MGF) for X ~ Γ(α, β), we need to follow these steps:
- 1. Recall that the moment generating function (MGF) of a random variable X is defined as
, where t is a real number. - 2. Substitute the given density function fX(x) = (1/Γ(α)) * β^α * x⁽α⁻¹⁾* e⁽⁻ˣ/β⁾ into the MGF formula: M(t) = E(e⁽ᵗˣ⁾ ) = ∫(e⁽ᵗˣ⁾ ) * fX(x)) dx from 0 to ∞.
- 3. Simplify the expression inside the integral by substituting the given values:
- f_X(x) = (1/Γ(α)) * β^α * x⁽α⁻¹⁾ * e⁽⁻ˣ/β⁾.
- 4. Apply the integral operation to calculate the MGF:
- M(t) = ∫(e⁽ᵗˣ⁾ * (1/Γ(α)) * β^α * x⁽α⁻¹⁾ * e⁽⁻ˣ/β⁾) dx from 0 to ∞.
- 5. Combine the exponential terms:
- M(t) = (1/Γ(α)) * β^α * ∫(x⁽α⁻¹⁾ * e^((t-1)x/β)) dx from 0 to ∞.
- 6. Apply the gamma function property:
- ∫(x⁽α⁻¹⁾ * e⁽⁽ᵗ⁻¹⁾ˣ/β⁾⁾ dx = β^α * Γ(α) * (t-1)⁽⁻α⁾
- 7. Substitute this result back into the MGF expression:
- M(t) = (1/Γ(α)) * β^α * β^α * Γ(α) * (t-1)⁽⁻α⁾
- 8. Simplify the expression:
- M(t) = (β(t-1))⁽⁻α⁾
To compute the mean and variance of X:
1. Recall that for X ~ Γ(α, β), the mean (μ) and variance (σ²) can be expressed as follows:
μ = α * β
σ² = α * β²
If X ~ Γ(α, 1) and we want to find the distribution of βX:
1. Recall that if X ~ Γ(α, β), then βX ~ Γ(α, 1/β).
2. Therefore, if X ~ Γ(α, 1), then βX ~ Γ(α, 1/β).