Question 1193841: If X is continuous random variable with probability density function
f(x) = {(ax, 0≤x≤1
b[(x-1)]^2, 1
0, otherwise,
find a and b? (make valid assumptions if required and state them)
Answer by parmen(42)   (Show Source):
You can put this solution on YOUR website! **1. Define the Probability Density Function (PDF)**
* **f(x) = { ax, 0 ≤ x ≤ 1
{ b(x-1)², 1 < x ≤ 2
{ 0, otherwise**
**2. Conditions for a Valid PDF**
* **Non-negativity:** f(x) ≥ 0 for all x
* This condition is satisfied since:
* For 0 ≤ x ≤ 1: ax ≥ 0 if a ≥ 0
* For 1 < x ≤ 2: b(x-1)² ≥ 0 for any value of b
* **Normalization:** The total area under the PDF curve must equal 1.
* This means:
* ∫₀¹ ax dx + ∫₁² b(x-1)² dx = 1
**3. Calculate the Integrals**
* ∫₀¹ ax dx = [a(x²/2)]₀¹ = a/2
* ∫₁² b(x-1)² dx = b ∫₀¹ y² dy (where y = x-1) = b[y³/3]₀¹ = b/3
**4. Apply the Normalization Condition**
* a/2 + b/3 = 1
**5. Determine the Relationship between a and b**
* a/2 + b/3 = 1
* 3a + 2b = 6
* **3a = 6 - 2b**
**6. Additional Constraint (Optional)**
* To uniquely determine the values of 'a' and 'b', we would typically need an additional constraint. For example, we could specify the mean or variance of the random variable X.
**7. Solve for a and b**
* Without further constraints, we can express 'a' in terms of 'b' using the equation 3a = 6 - 2b.
* For example:
* If we assume b = 1, then a = (6 - 2)/3 = 4/3
**Therefore:**
* **a = (6 - 2b) / 3**
* **b can be any non-negative value.**
**Note:** To find specific values for 'a' and 'b', you would need additional information about the distribution, such as its mean, variance, or other relevant properties.
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