For U ∼ Uniform(a, b), what are E[U] and Var(U)?

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Multiple Choice

For U ∼ Uniform(a, b), what are E[U] and Var(U)?

Explanation:
Uniform(a, b) assigns equal likelihood to every value in [a, b], so the center of the interval is the expected value. Compute E[U] by averaging across the interval: E[U] = ∫_a^b u * (1/(b-a)) du = (a + b)/2. For the variance, use Var(U) = E[U^2] − [E(U)]^2. First find E[U^2]: E[U^2] = ∫_a^b u^2 * (1/(b-a)) du = (b^3 − a^3) / [3(b − a)] = (a^2 + ab + b^2)/3. Then Var(U) = (a^2 + ab + b^2)/3 − [(a + b)/2]^2, which simplifies to (b − a)^2/12. So, E[U] = (a + b)/2 and Var(U) = (b − a)^2/12.

Uniform(a, b) assigns equal likelihood to every value in [a, b], so the center of the interval is the expected value. Compute E[U] by averaging across the interval: E[U] = ∫_a^b u * (1/(b-a)) du = (a + b)/2.

For the variance, use Var(U) = E[U^2] − [E(U)]^2. First find E[U^2]: E[U^2] = ∫_a^b u^2 * (1/(b-a)) du = (b^3 − a^3) / [3(b − a)] = (a^2 + ab + b^2)/3. Then Var(U) = (a^2 + ab + b^2)/3 − [(a + b)/2]^2, which simplifies to (b − a)^2/12.

So, E[U] = (a + b)/2 and Var(U) = (b − a)^2/12.

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