Problem 5 - Thomas Calculus

Question

In Exercises \(1-8,\) determine which are probability density functions and justify your answer. $$ f(x)=\left\\{\begin{array}{ll}{\frac{1}{x^{2}}} & {x \geq 1} \\ {0} & {x<1}\end{array}\right. $$

Step-by-Step Solution

Answer

The function is a probability density function because it is non-negative and integrates to 1.

Step 1: Define the Criteria for a Probability Density Function

Recall that a function \( f(x) \) is a probability density function (PDF) if it satisfies two conditions: First, \( f(x) \geq 0 \) for all \( x \), meaning the function must be non-negative everywhere. Second, the integral over the entire space must equal 1, meaning the total probability must be 1.

Step 2: Check Non-negativity

Examine the function \( f(x) = \begin{cases} \frac{1}{x^2} & x \geq 1 \ 0 & x < 1 \end{cases} \). In this case, \( f(x) = \frac{1}{x^2} \) is always positive for \( x \geq 1 \) and zero for \( x < 1 \). Therefore, \( f(x) \geq 0 \) for all \( x \), satisfying the first condition.

Step 3: Calculate the Integral of the Function

Calculate the integral of \( f(x) \) over its entire range, which is from \( x = 1 \) to \( x = \infty \). This integral is given by: \[\int_{1}^{\infty} \frac{1}{x^2} \, dx\]

Step 4: Evaluate the Improper Integral

To evaluate the integral \[\int_{1}^{\infty} \frac{1}{x^2} \, dx\]we compute: \[= \lim_{b \to \infty} \int_{1}^{b} \frac{1}{x^2} \, dx = \lim_{b \to \infty} \left[-\frac{1}{x} \right]_{1}^{b} \] \[= \lim_{b \to \infty} \left(-\frac{1}{b} + 1 \right) = 1\].The integral evaluates to 1, meeting the second condition for a PDF.

Step 5: Conclusion

Since the function \( f(x) \) is non-negative for all \( x \) and the integral over its entire range is equal to 1, we conclude that \( f(x) \) is a probability density function.

Key Concepts

Non-negativity Condition

In probability theory, the non-negativity condition is an essential criterion for determining if a function can serve as a probability density function (PDF). A PDF is used to describe the likelihood of a random variable taking on a particular value within a continuum. To ensure a valid PDF, the function must be non-negative across its entire domain, meaning it cannot produce a negative probability, which would defy logical interpretation.

Let's consider the example function given:

• For values of \( x \) greater than or equal to 1, \( f(x) = \frac{1}{x^2} \), which is always positive because the square of a real number is positive.
• For values of \( x \) less than 1, \( f(x) = 0 \). In this case, the function remains non-negative, precisely zero.

Thus, the function satisfies the non-negativity condition across all values of \( x \). This condition is crucial to ensure that the computed probabilities make sense, as probabilities must always be non-negative.

Integral Evaluation

The second condition for a function to qualify as a probability density function is that its integral over the entire domain must equal 1. This total integral represents the sum of probabilities for all possible outcomes, which must logically add up to one, reflecting the certainty that some outcome occurs.

For the function in question, we evaluate the integral over its entire range from \( x = 1 \) to \( x = \infty \): \[\int_{1}^{\infty} \frac{1}{x^2} \, dx \]The integral calculates the area under the curve \( f(x) \), representing the accumulated probability over all valid values. If this integral result equals 1, it indicates that the function meets the summation condition for probabilities, confirming it as a legitimate PDF.

Improper Integral

In mathematical terms, an improper integral is key when dealing with infinite limits of integration, such as in our example where \( x \) ranges from 1 to infinity. Calculating such integrals can initially seem daunting, as they involve handling infinite behavior.

To solve an improper integral, like:\[\int_{1}^{\infty} \frac{1}{x^2} \, dx\]We use the limiting process:

• Replace the infinity part of the integral with a variable, say \( b \), approaching infinity.
• Evaluate the integral with this finite upper limit.
• Take the limit as \( b \) approaches infinity.

The integral for our function is:\[\lim_{b \to \infty} \left( -\frac{1}{x} \right)_{1}^{b} = \lim_{b \to \infty} \left(-\frac{1}{b} + 1 \right) = 1\]The final value being 1 indicates that the improper integral condition is met, concretely verifying that the total probability sums to 1, which corroborates the function's status as a PDF.