Problem 1 - Thomas Calculus

Question

Determine which are probability density functions and justify your answer. \(f(x)=\frac{1}{18} x\) over \([4,8]\)

Step-by-Step Solution

Answer

The function is not a PDF because its integral over \([4, 8]\) is not 1.

Step 1: Understand PDF Conditions

For a function to be a probability density function (PDF), it must satisfy two conditions: 1) The function must be non-negative over its defined interval, and 2) The integral of the function over that interval must equal 1.

Step 2: Check Non-negativity

Examine the given function \(f(x) = \frac{1}{18}x\) over the interval \([4, 8]\). Since \(x\) is between 4 and 8, and \(\frac{1}{18}\) is a positive constant, \(f(x)\) is non-negative across this interval.

Step 3: Compute the Integral

Calculate the integral of \(f(x) = \frac{1}{18}x\) from 4 to 8.\[\int_{4}^{8} \frac{1}{18}x \,dx = \frac{1}{18} \int_{4}^{8} x \,dx = \frac{1}{18} \left[ \frac{1}{2}x^2 \right]_{4}^{8} = \frac{1}{18} \left( \frac{1}{2} (8^2) - \frac{1}{2} (4^2) \right)\]

Step 4: Simplify the Integral

Continue simplifying the result:\[= \frac{1}{18} \left( \frac{1}{2} (64) - \frac{1}{2} (16) \right) = \frac{1}{18} \left( 32 - 8 \right) = \frac{1}{18} \times 24 = \frac{24}{18} = \frac{4}{3}\]Since \(\frac{4}{3} eq 1\), the integral does not equal 1.

Step 5: Conclusion

The function \(f(x) = \frac{1}{18}x\) is non-negative on the interval \([4, 8]\), but the integral over this interval does not equal 1. Therefore, it does not satisfy both conditions to be a PDF.

Key Concepts

PDF Conditions

To determine whether a function is a probability density function (PDF), we need to ensure it satisfies two essential conditions. First, the function should be non-negative over its defined interval. This means that if we look at any part of the interval specified, the function value should not dip below zero. Non-negative functions ensure that probabilities, which cannot be negative, stay valid.

Second, the integral of the function over this interval must add up to exactly 1. The integral essentially represents the total area under the curve of the function over its interval. Since this area corresponds to all possible outcomes within the range, it must sum to 1, reflecting 100% probability distributed across the entire interval. If both these conditions are satisfied, the function can be considered a valid PDF.

Non-Negativity of Functions

Non-negativity is a crucial property for probability density functions. It guarantees that probabilities assigned to events are sensible since probabilities can't be negative. In the example given, we have the function \( f(x) = \frac{1}{18}x \) evaluated over the interval \([4,8]\).

• The term \( \frac{1}{18} \) is positive, scaling the input \( x \).
• Since \( x \) within our interval is always between 4 and 8, each value sampled will inherently be positive.

Therefore, since all components in this function are non-negative over the entire specified interval, the function itself remains non-negative. However, non-negativity alone is not enough to establish the function as a PDF, as we'll see in the next section.

Integral of Functions

The integral of a function helps us understand the total area under the curve over a specified interval. For probability density functions, this integral must exactly equal 1. This ensures the total probability (representing the complete interval range) sums up to 1, indicating certainty in one of the possible outcomes occurring.

For our function \( f(x) = \frac{1}{18}x \), we compute the integral over the interval from 4 to 8:\[ \int_{4}^{8} \frac{1}{18}x \,dx \].

Carrying out the integral, it evaluates to \( \frac{4}{3} \), which is not 1. This outcome means that, although the function is non-negative, it doesn't fulfill the second condition required for a PDF, rendering it invalid as a probability density function.