Math Problem Statement
Solution
This problem involves a piecewise probability density function (PDF) defined as:
\begin{cases} \frac{\alpha - 1}{x^\alpha}, & x \geq 1 \\ 0, & x < 1 \end{cases}$$ where $$\alpha > 2$$. ### Task 1: Show that $$f(x)$$ is a valid probability density function. A function is a PDF if: 1. $$f(x) \geq 0$$ for all $$x$$, and 2. The integral of $$f(x)$$ over all $$x$$ equals 1: $$\int_{-\infty}^\infty f(x) \, dx = 1.$$ Since $$f(x) = 0$$ for $$x < 1$$, we only need to check $$x \geq 1$$: $$\int_1^\infty \frac{\alpha - 1}{x^\alpha} dx.$$ Let’s compute this integral: $$\int_1^\infty \frac{\alpha - 1}{x^\alpha} dx = (\alpha - 1) \int_1^\infty x^{-\alpha} dx.$$ For $$\alpha > 1$$, the integral $$\int x^{-\alpha} dx$$ converges: $$\int x^{-\alpha} dx = \frac{x^{1-\alpha}}{1-\alpha}.$$ Evaluate at the bounds $$1$$ to $$\infty$$: $$\int_1^\infty x^{-\alpha} dx = \left[ \frac{x^{1-\alpha}}{1-\alpha} \right]_1^\infty = \frac{1}{\alpha - 1}.$$ Thus: $$\int_1^\infty f(x) dx = (\alpha - 1) \cdot \frac{1}{\alpha - 1} = 1.$$ Hence, $$f(x)$$ is a valid PDF. --- ### Task 2: Determine the cumulative distribution function $$F_X(x)$$. The cumulative distribution function $$F_X(x)$$ is defined as: $$F_X(x) = \int_{-\infty}^x f(t) dt.$$ For $$x < 1$$: $$F_X(x) = 0 \quad \text{(since $$f(t) = 0$$ for $$t < 1$$)}.$$ For $$x \geq 1$$: $$F_X(x) = \int_1^x \frac{\alpha - 1}{t^\alpha} dt.$$ Compute the integral: $$\int_1^x t^{-\alpha} dt = \left[ \frac{t^{1-\alpha}}{1-\alpha} \right]_1^x = \frac{x^{1-\alpha} - 1}{1-\alpha}.$$ Thus: $$F_X(x) = (\alpha - 1) \cdot \frac{x^{1-\alpha} - 1}{1-\alpha} = 1 - x^{1-\alpha}, \quad x \geq 1.$$ So: $$F_X(x) = \begin{cases} 0, & x < 1, \\ 1 - x^{1-\alpha}, & x \geq 1. \end{cases}$$ --- ### Task 3: Calculate the expectation $$\mathbb{E}[X]$$. The expectation is given by: $$\mathbb{E}[X] = \int_{-\infty}^\infty x f(x) dx.$$ Since $$f(x) = 0$$ for $$x < 1$$, we only integrate over $$x \geq 1$$: $$\mathbb{E}[X] = \int_1^\infty x \cdot \frac{\alpha - 1}{x^\alpha} dx = (\alpha - 1) \int_1^\infty x^{1-\alpha} dx.$$ Using the earlier result for $$\int x^{-\alpha} dx$$: $$\int_1^\infty x^{1-\alpha} dx = \frac{x^{2-\alpha}}{2-\alpha} \Big|_1^\infty = \frac{1}{\alpha - 2}.$$ Thus: $$\mathbb{E}[X] = (\alpha - 1) \cdot \frac{1}{\alpha - 2} = \frac{\alpha - 1}{\alpha - 2}.$$ --- ### Final Results: 1. $$f(x)$$ is a valid probability density function. 2. The cumulative distribution function $$F_X(x)$$ is: $$F_X(x) = \begin{cases} 0, & x < 1, \\ 1 - x^{1-\alpha}, & x \geq 1. \end{cases}$$ 3. The expectation $$\mathbb{E}[X]$$ is: $$\mathbb{E}[X] = \frac{\alpha - 1}{\alpha - 2}.$$ --- Do you have any questions or want detailed steps for a specific part? ### Related Questions: 1. What is the variance $$\text{Var}(X)$$ of this distribution? 2. How does $$F_X(x)$$ change as $$\alpha \to \infty$$? 3. What happens to $$\mathbb{E}[X]$$ if $$\alpha$$ approaches its lower bound (e.g., $$\alpha \to 2^+$$)? 4. Can we derive higher moments of $$X$$, like $$\mathbb{E}[X^2]$$? 5. How does the behavior of $$f(x)$$ compare to a uniform distribution for large $$\alpha$$? ### Tip: When verifying if a function is a PDF, always confirm the normalization condition using proper integration techniques.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability Theory
Integration
Cumulative Distribution Functions
Expected Value
Formulas
Normalization condition for PDF: \( \int_{-\infty}^{\infty} f(x) \, dx = 1 \)
CDF: \( F_X(x) = \int_{-\infty}^x f(t) \, dt \)
Expected value: \( \mathbb{E}[X] = \int_{-\infty}^{\infty} x f(x) \, dx \)
Theorems
Properties of Probability Density Functions
Fundamental Theorem of Calculus
Suitable Grade Level
University Level
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