For the function f(x) = a + bx, 0 ≤ x ≤ 1, to be a valid probability density function, which one of the following statements is correct?
a = 0.5, b = 1
To determine if a function is a valid probability density function (PDF), it must satisfy two fundamental conditions over its defined interval. For the given function, \(f(x) = a + bx\) on the interval \(0 \le x \le 1\), these conditions are crucial.
Probability Density Function (PDF) Conditions
A function \(f(x)\) is considered a valid probability density function over an interval \([A, B]\) if it meets the following two criteria:
- Non-Negativity: The function's values must be non-negative for all \(x\) within the interval. That is, \(f(x) \ge 0\) for all \(x \in [A, B]\).
- Normalization: The total area under the curve of the function over its entire defined interval must be equal to 1. This means the integral of \(f(x)\) over the interval must be 1. Mathematically, \(\int_{A}^{B} f(x) \, dx = 1\).
Applying Conditions to \(f(x) = a + bx\)
1. Non-Negativity Condition (\(f(x) \ge 0\))
For \(f(x) = a + bx\) to be non-negative over \(0 \le x \le 1\), the function must be greater than or equal to zero at all points in this interval. Since \(f(x)\) is a linear function, its minimum value will occur at one of the endpoints, \(x=0\) or \(x=1\).
- If \(b \ge 0\), the function is non-decreasing. The minimum value will be at \(x=0\). Therefore, \(f(0) = a \ge 0\) must hold.
- If \(b < 0\), the function is decreasing. The minimum value will be at \(x=1\). Therefore, \(f(1) = a + b \ge 0\) must hold.
2. Normalization Condition (\(\int_{0}^{1} f(x) \, dx = 1\))
We need to integrate \(f(x) = a + bx\) from \(x=0\) to \(x=1\) and set the result equal to 1:
\(\int_{0}^{1} (a + bx) \, dx = 1\)
Let's perform the integration:
\(\left[ ax + \frac{bx^2}{2} \right]_{0}^{1} = 1\)
Now, substitute the limits of integration:
\(\left( a(1) + \frac{b(1)^2}{2} \right) - \left( a(0) + \frac{b(0)^2}{2} \right) = 1\)
\(a + \frac{b}{2} - 0 = 1\)
\(a + \frac{b}{2} = 1\)
This is the essential equation that \(a\) and \(b\) must satisfy for the function to be a valid probability density function.
Evaluating the Options
Now, let's examine each given option using the derived conditions, especially \(a + \frac{b}{2} = 1\) and the non-negativity condition.
| Option | Values (a, b) | Normalization Check (\(a + \frac{b}{2}\)) | Non-Negativity Check (\(f(x) \ge 0\)) | Valid PDF? |
|---|---|---|---|---|
| 1 | \(a = 1, b = 4\) | \(1 + \frac{4}{2} = 1 + 2 = 3 \ne 1\) | \(f(x) = 1+4x\). Minimum at \(x=0\) is \(f(0)=1 \ge 0\). (Satisfied) | No (Fails Normalization) |
| 2 | \(a = 0.5, b = 1\) | \(0.5 + \frac{1}{2} = 0.5 + 0.5 = 1\). (Satisfied) | \(f(x) = 0.5+x\). Minimum at \(x=0\) is \(f(0)=0.5 \ge 0\). (Satisfied) | Yes (Satisfies both) |
| 3 | \(a = 0, b = 1\) | \(0 + \frac{1}{2} = 0.5 \ne 1\) | \(f(x) = x\). Minimum at \(x=0\) is \(f(0)=0 \ge 0\). (Satisfied) | No (Fails Normalization) |
| 4 | \(a = 1, b = -1\) | \(1 + \frac{-1}{2} = 1 - 0.5 = 0.5 \ne 1\) | \(f(x) = 1-x\). Minimum at \(x=1\) is \(f(1)=0 \ge 0\). (Satisfied) | No (Fails Normalization) |
Conclusion for Valid Probability Density Function
Based on our evaluation, only the pair of values \(a = 0.5\) and \(b = 1\) satisfies both the non-negativity and normalization conditions required for a valid probability density function over the interval \(0 \le x \le 1\). For this option, \(f(x) = 0.5 + x\), which is always positive or zero on the interval, and its integral from 0 to 1 is exactly 1.
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Which one of the following statements is true?
