Special Continuous Distributions I – Gamma and Chi-Square
Example 1 (Gamma Distribution):
Problem Let X ∼ Gamma(α = 2, θ = 3), where α is the shape parameter and θ
the scale parameter. Find P (X ≤ 4).
Solution
Step 1: Recall the Gamma CDF (Erlang form) For a Gamma random variable
with integer shape parameter α = n:
n−1
X (x/θ)k
FX (x) = P (X ≤ x) = 1 − e−x/θ .
k!
k=0
Step 2: Substitute α = 2 and θ = 3
x
FX (x) = 1 − e−x/3 1 + .
3
Let
x 4
y= = ≈ 1.333.
θ 3
Then
4
FX (4) = 1 − e−4/3 1 + .
3
Sam Maseno University August 19, 2025
, Special Continuous Distributions I – Gamma and Chi-Square
Step 4: Numerical computation
e−4/3 ≈ 0.2636, 1+ 4
3 = 7
3 ≈ 2.3333.
e−4/3 1 + 4
3 ≈ 0.2636 × 2.3333 ≈ 0.6151.
FX (4) = 1 − 0.6151 ≈ 0.3849.
Final Answer
P (X ≤ 4) ≈ 0.385
Remark (Gamma Tables) Using incomplete gamma tables, we compute
4
P (2, y), y= ≈ 1.33.
3
Interpolating between P (2, 1.30) ≈ 0.3732 and P (2, 1.40) ≈ 0.4082 yields
P (2, 1.33) ≈ 0.3837,
which agrees with the direct calculation.
Sam Maseno University August 19, 2025
, Special Continuous Distributions I – Gamma and Chi-Square
Example 2
Let X ∼ Gamma(α = 3, θ = 2). That is, the shape parameter is 3 and the scale
parameter is 2.
1. Probability Density Function (PDF):
1
f (x; α, θ) = xα−1 e−x/θ , x>0
Γ(α)θα
Substituting α = 3, θ = 2:
1
f (x) = x3−1 e−x/2 , x>0
Γ(3) 23
Since Γ(3) = (3 − 1)! = 2:
1 2 −x/2
f (x) = x e , x>0
16
2. Mean and Variance:
E[X] = αθ = 3 × 2 = 6
Var(X) = αθ2 = 3 × 22 = 12
Sam Maseno University August 19, 2025
, Special Continuous Distributions I – Gamma and Chi-Square
The Chi-Square Distribution
The chi-square distribution is a special case of the gamma distribution.
Specifically, if a random variable X follows a chi-square distribution with k
degrees of freedom, then:
2 k
X ∼ χ (k) ⇐⇒ X ∼ Gamma ,2
2
Where:
k ∈ N is the number of degrees of freedom (a positive integer).
The gamma distribution here has shape α = k2 and scale θ = 2.
Probability Density Function (PDF):
1
f (x; k) = k/2 x(k/2)−1 e−x/2 , x > 0
2 Γ(k/2)
Key Properties:
Mean: µ = k and Variance: σ 2 = 2k
Additivity Property of the Chi-Square Distribution If X1 ∼ χ2 (k1 ),
X2 ∼ χ2 (k2 ), and X1 and X2 are independent, then:
X1 + X2 ∼ χ2 (k1 + k2 )
Sam Maseno University August 19, 2025
Example 1 (Gamma Distribution):
Problem Let X ∼ Gamma(α = 2, θ = 3), where α is the shape parameter and θ
the scale parameter. Find P (X ≤ 4).
Solution
Step 1: Recall the Gamma CDF (Erlang form) For a Gamma random variable
with integer shape parameter α = n:
n−1
X (x/θ)k
FX (x) = P (X ≤ x) = 1 − e−x/θ .
k!
k=0
Step 2: Substitute α = 2 and θ = 3
x
FX (x) = 1 − e−x/3 1 + .
3
Let
x 4
y= = ≈ 1.333.
θ 3
Then
4
FX (4) = 1 − e−4/3 1 + .
3
Sam Maseno University August 19, 2025
, Special Continuous Distributions I – Gamma and Chi-Square
Step 4: Numerical computation
e−4/3 ≈ 0.2636, 1+ 4
3 = 7
3 ≈ 2.3333.
e−4/3 1 + 4
3 ≈ 0.2636 × 2.3333 ≈ 0.6151.
FX (4) = 1 − 0.6151 ≈ 0.3849.
Final Answer
P (X ≤ 4) ≈ 0.385
Remark (Gamma Tables) Using incomplete gamma tables, we compute
4
P (2, y), y= ≈ 1.33.
3
Interpolating between P (2, 1.30) ≈ 0.3732 and P (2, 1.40) ≈ 0.4082 yields
P (2, 1.33) ≈ 0.3837,
which agrees with the direct calculation.
Sam Maseno University August 19, 2025
, Special Continuous Distributions I – Gamma and Chi-Square
Example 2
Let X ∼ Gamma(α = 3, θ = 2). That is, the shape parameter is 3 and the scale
parameter is 2.
1. Probability Density Function (PDF):
1
f (x; α, θ) = xα−1 e−x/θ , x>0
Γ(α)θα
Substituting α = 3, θ = 2:
1
f (x) = x3−1 e−x/2 , x>0
Γ(3) 23
Since Γ(3) = (3 − 1)! = 2:
1 2 −x/2
f (x) = x e , x>0
16
2. Mean and Variance:
E[X] = αθ = 3 × 2 = 6
Var(X) = αθ2 = 3 × 22 = 12
Sam Maseno University August 19, 2025
, Special Continuous Distributions I – Gamma and Chi-Square
The Chi-Square Distribution
The chi-square distribution is a special case of the gamma distribution.
Specifically, if a random variable X follows a chi-square distribution with k
degrees of freedom, then:
2 k
X ∼ χ (k) ⇐⇒ X ∼ Gamma ,2
2
Where:
k ∈ N is the number of degrees of freedom (a positive integer).
The gamma distribution here has shape α = k2 and scale θ = 2.
Probability Density Function (PDF):
1
f (x; k) = k/2 x(k/2)−1 e−x/2 , x > 0
2 Γ(k/2)
Key Properties:
Mean: µ = k and Variance: σ 2 = 2k
Additivity Property of the Chi-Square Distribution If X1 ∼ χ2 (k1 ),
X2 ∼ χ2 (k2 ), and X1 and X2 are independent, then:
X1 + X2 ∼ χ2 (k1 + k2 )
Sam Maseno University August 19, 2025
