1. **Problem Statement:** Find the PDF, expected value, and variance of a uniform random variable $U \sim \text{Uniform}(a,b)$. 2. **Formula for Uniform Distribution:** - PDF: $$f_U(u) = \frac{1}{b-a} \quad \text{for } a \leq u \leq b$$ - Expected value: $$E[U] = \frac{a+b}{2}$$ - Variance: $$\text{Var}(U) = \frac{(b-a)^2}{12}$$ 3. **Explanation:** - The uniform distribution assumes all values between $a$ and $b$ are equally likely. - The PDF is constant over $[a,b]$ and zero elsewhere. - The mean is the midpoint of the interval. - The variance measures the spread and depends on the length of the interval squared, divided by 12. 4. **Intermediate Work:** - PDF is derived from the total probability being 1 over the interval length: $$\int_a^b f_U(u) du = 1 \Rightarrow f_U(u) = \frac{1}{b-a}$$ - Expected value calculation: $$E[U] = \int_a^b u \cdot \frac{1}{b-a} du = \frac{1}{b-a} \cdot \frac{b^2 - a^2}{2} = \frac{a+b}{2}$$ - Variance calculation: $$E[U^2] = \int_a^b u^2 \cdot \frac{1}{b-a} du = \frac{1}{b-a} \cdot \frac{b^3 - a^3}{3}$$ $$\text{Var}(U) = E[U^2] - (E[U])^2 = \frac{b^3 - a^3}{3(b-a)} - \left(\frac{a+b}{2}\right)^2 = \frac{(b-a)^2}{12}$$ **Final answers:** $$f_U(u) = \frac{1}{b-a}, \quad E[U] = \frac{a+b}{2}, \quad \text{Var}(U) = \frac{(b-a)^2}{12}$$