Problem 22 - Calculus Concepts

Question

Indicate whether the function could be a probability density function. Explain. \(h(x)=\left\\{\begin{array}{ll}6\left(x-x^{2}\right) & \text { when } 0 \leq x \leq 1 \\ 0 & \text { elsewhere }\end{array}\right.\)

Step-by-Step Solution

Answer

Yes, it is a probability density function; it is non-negative and integrates to 1 over its domain.

Step 1: Confirm Non-negativity

For the function to be a probability density function (pdf), it must be non-negative for all values in its domain. Given \( h(x) = 6(x-x^2) \), rewrite it as \( h(x) = 6x(1-x) \). Since \( x \) ranges from 0 to 1, both \( x \) and \( (1-x) \) are non-negative. Hence, \( h(x) \) is non-negative within this range.

Step 2: Calculate the Total Integral

Another condition for a function to be a probability density function is that the integral over its entire domain equals 1. We compute the integral of \( h(x) = 6(x-x^2) \) over the interval from 0 to 1. \[ \int_0^1 6(x-x^2) \, dx = \int_0^1 6x - 6x^2 \, dx \]

Step 3: Solve the Integral

Evaluate the integral:\[ \int_0^1 6x - 6x^2 \, dx = \left[ 6 \frac{x^2}{2} - 6 \frac{x^3}{3} \right]_0^1 = \left[ 3x^2 - 2x^3 \right]_0^1 \]Substitute the limits:\[ =(3(1)^2 - 2(1)^3) - (3(0)^2 - 2(0)^3) = 3 - 2 = 1 \]The integral evaluates to 1.

Step 4: Conclusion

Since \( h(x) \) is non-negative over its domain and the integral of \( h(x) \) over the entire domain equals 1, \( h(x) \) satisfies both conditions to be a probability density function.

Key Concepts

Non-negative Function

A non-negative function is crucial for determining if a function can be a probability density function (PDF). Simply put, the function must not produce negative values for any input within its domain. This is because probabilities cannot be negative—negative probabilities don't make logical sense in probability theory. For our function, given as:- \( h(x) = 6(x-x^2) \) within the interval from 0 to 1- \( h(x) = 0 \) elsewhereWe need to ensure \( h(x) \) is non-negative over its defined range. To inspect its non-negativity:* Rewrite \( h(x) \) as \( 6x(1-x) \).* Note that in the interval between 0 and 1, both \( x \) and \( 1-x \) are non-negative since \( 0 \leq x \leq 1 \).Therefore, the product of \( x \) and \( 1-x \)—which is the expression for \( h(x) \)—results in non-negative values for this interval, confirming the non-negativity of the function in its domain.

Integral of Function

Another pivotal criterion for a function to qualify as a probability density function is having an integral over its domain that equals 1. This condition signifies that the total "probability mass" across the function's range sums to exactly 1, which is a foundational aspect of probability. To achieve this, we calculate the integral of the given function.The integral is defined over the range where the function is positive:\[\int_0^1 6(x-x^2) \, dx = \int_0^1 6x - 6x^2 \, dx\]To solve this integral, compute separately:* Calculate \( \int 6x \, dx \) and \( \int 6x^2 \, dx \)* Evaluate using the antiderivative: - \( \left[ 3x^2 - 2x^3 \right]_0^1 \)Substituting the limits into the evaluated integral:* When \( x = 1 \): \( 3(1)^2 - 2(1)^3 = 3 - 2 = 1 \)* When \( x = 0 \): \( 3(0)^2 - 2(0)^3 = 0 \)Thus, the total integral computes to 1, fulfilling the requirement for a probability density function.

Domain of Function

Understanding the concept of a function's domain is key when examining whether it can be a probability density function. The domain of a function denotes the set of all possible input values (\( x \)) for which the function is defined and non-zero. For our example function \( h(x) \), the domain is specifically given as:- \( 0 \leq x \leq 1 \)- Elsewhere, \( h(x) = 0 \)Therefore, within this interval, the function is active and must demonstrate all the necessary properties of a probability density function:* Non-negativity for each \( x \)* An integral of 1 across this rangeBeyond this interval, the function naturally returns zero, as there are no contributions to the probability outside of the defined range. This supports the notion that the function operates strictly within its defined domain, matching the requirements for being a valid PDF. In summary, correctly interpreting the domain ensures that the properties of the probability density function are effectively assessed within the appropriate range. This concept is fundamental to probability theory.