The probability density function (PDF) of the Pareto distribution is given by:

$f(x) = \frac{B a^B}{x^{B+1}}, \quad \text{for } x \geq a, \ B > 0, \ a > 0$

A. Finding the Mean of $X$:

The mean of a continuous random variable $X$ is given by:

$\mathbb{E}(X) = \int_{a}^{\infty} x f(x) dx$

Substitute the PDF of the Pareto distribution into the formula:

$\mathbb{E}(X) = \int_{a}^{\infty} x \frac{B a^B}{x^{B+1}} dx$

Simplifying the expression:

= B a^B \int_{a}^{\infty} \frac{1}{x^B} dx = B a^B \int_{a}^{\infty} x^{-B} dx$$ The integral of $$x^{-B}$$ is given by: $$\int x^{-B} dx = \frac{x^{1-B}}{1-B}$$ Now applying the limits: $$\mathbb{E}(X) = B a^B \left[ \frac{x^{1-B}}{1-B} \right]_{x=a}^{\infty}$$ At $$x = \infty$$, the term $$\frac{1}{x^{B-1}} \to 0$$ for $$B > 1$$: $$\mathbb{E}(X) = B a^B \left( 0 - \frac{a^{1-B}}{1-B} \right) = B a^B \left( \frac{a^{1-B}}{B-1} \right) = \frac{a B a^{1-B}}{B-1} = \frac{a}{B-1}$$ Thus, the mean exists only when $$B > 1$$, and it is: $$\mathbb{E}(X) = \frac{a}{B-1}$$ ### B. Finding the Variance of $$X$$: The variance is given by: $$\text{Var}(X) = \mathbb{E}(X^2) - (\mathbb{E}(X))^2$$ First, find $$\mathbb{E}(X^2)$$: $$\mathbb{E}(X^2) = \int_{a}^{\infty} x^2 f(x) dx = \int_{a}^{\infty} x^2 \frac{B a^B}{x^{B+1}} dx = B a^B \int_{a}^{\infty} x^{2-B} dx$$ The integral of $$x^{2-B}$$ is: $$\int x^{2-B} dx = \frac{x^{3-B}}{3-B}$$ Applying the limits: $$\mathbb{E}(X^2) = B a^B \left[ \frac{x^{3-B}}{3-B} \right]_{x=a}^{\infty}$$ At $$x = \infty$$, the term $$\frac{1}{x^{B-3}} \to 0$$ for $$B > 2$$: $$\mathbb{E}(X^2) = B a^B \left( 0 - \frac{a^{3-B}}{3-B} \right) = B a^B \left( \frac{a^{3-B}}{B-3} \right) = \frac{a^2}{B-2}$$ Thus, $$\mathbb{E}(X^2)$$ exists only when $$B > 2$$, and it is: $$\mathbb{E}(X^2) = \frac{a^2}{B-2}$$ Now, using the formula for variance: $$\text{Var}(X) = \mathbb{E}(X^2) - (\mathbb{E}(X))^2 = \frac{a^2}{B-2} - \left( \frac{a}{B-1} \right)^2$$ Simplify the expression: $$\text{Var}(X) = \frac{a^2}{B-2} - \frac{a^2}{(B-1)^2} = a^2 \left( \frac{1}{B-2} - \frac{1}{(B-1)^2} \right)$$ Thus, the variance of $$X$$ is: $$\text{Var}(X) = a^2 \left( \frac{B-1 - (B-2)}{(B-2)(B-1)^2} \right) = a^2 \left( \frac{1}{(B-2)(B-1)^2} \right)$$ So the variance is: $$\text{Var}(X) = \frac{a^2}{(B-2)(B-1)^2}$$ This variance exists only when $$B > 2$$. ### Summary: 1. **Mean**: $$\mathbb{E}(X) = \frac{a}{B-1}$$, exists for $$B > 1$$. 2. **Variance**: $$\text{Var}(X) = \frac{a^2}{(B-2)(B-1)^2}$$, exists for $$B > 2$$. Would you like further clarification or details? --- Here are some related questions: 1. What happens to the mean and variance as $$B \to \infty$$? 2. Can we interpret the Pareto distribution's tail behavior for small values of $$B$$? 3. How is the Pareto distribution related to the power law distribution? 4. How can the Pareto distribution be applied in economics or risk management? 5. Can you derive the cumulative distribution function (CDF) of the Pareto distribution? **Tip:** The Pareto distribution is often used to model wealth distributions, where a small percentage of the population holds a large proportion of the wealth!