![Given a random variable X with E[X] = 0 and Var(X) = 1, find the expected value and variance of Z = X2.](/_next/image?url=https%3A%2F%2Fohijasnjvprzodcmxltt.supabase.co%2Fstorage%2Fv1%2Fobject%2Fpublic%2Fwhatsapp-images%2F1777400452184-0d7db896f034e4a3.png&w=3840&q=75)
This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Step 1: Calculate the expected value of Z, E[Z]. We are given Z = X^2. The variance of a random variable X is defined as Var(X) = E[X^2] - (E[X])^2. We are given E[X] = 0 and Var(X) = 1. Substitute these values into the variance formula: 1 = E[X^2] - (0)^2 1 = E[X^2] - 0 E[X^2] = 1 Since Z = X^2, then E[Z] = E[X^2]. Therefore, E[Z] = 1. Step 2: Calculate the variance of Z, Var(Z). The variance of Z is defined as Var(Z) = E[Z^2] - (E[Z])^2. We know Z = X^2, so Z^2 = (X^2)^2 = X^4. Thus, Var(Z) = E[X^4] - (E[Z])^2. We already found E[Z] = 1, so Var(Z) = E[X^4] - (1)^2 = E[X^4] - 1. To determine E[X^4], we need more information about the specific distribution of X. Consider two possible distributions for X that satisfy E[X]=0 and Var(X)=1: Case 1: X takes values -1 and 1 with probabilities P(X=-1) = 0.5 and P(X=1) = 0.5. • E[X] = (-1)(0.5) + (1)(0.5) = 0. • E[X^2] = (-1)^2(0.5) + (1)^2(0.5) = 1(0.5) + 1(0.5) = 1. • Var(X) = E[X^2] - (E[X])^2 = 1 - 0^2 = 1. In this case, Z = X^2 will always be 1^2 = 1 or (-1)^2 = 1. So Z is always 1. If Z is always 1, then E[Z] = 1 and Var(Z) = 0. Case 2: X takes values -sqrt(3), 0, sqrt(3) with probabilities P(X=-sqrt(3)) = 1/6, P(X=0) = 4/6, P(X=sqrt(3)) = 1/6. • E[X] = (-sqrt(3))(1/6) + (0)(4/6) + (sqrt(3))(1/6) = 0. • E[X^2] = (-sqrt(3))^2(1/6) + (0)^2(4/6) + (sqrt(3))^2(1/6) = 3(1/6) + 0 + 3(1/6) = 1. • Var(X) = E[X^2] - (E[X])^2 = 1 - 0^2 = 1. In this case, Z = X^2 can take values: If X = -sqrt(3), Z = (-sqrt(3))^2 = 3. If X = 0, Z = 0^2 = 0. If X = sqrt(3), Z = (sqrt(3))^2 = 3. So Z can be 0 or 3. P(Z=0) = P(X=0) = 4/6 = 2/3. P(Z=3) = P(X=-sqrt(3)) + P(X=sqrt(3)) = 1/6 + 1/6 = 2/6 = 1/3. Now calculate E[Z] and Var(Z) for this distribution: E[Z] = (0)(2/3) + (3)(1/3) = 0 + 1 = 1. (Matches Step 1) E[Z^2] = (0)^2(2/3) + (3)^2(1/3) = 0 + 9(1/3) = 3. Var(Z) = E[Z^2] - (E[Z])^2 = 3 - (1)^2 = 3 - 1 = 2. In this case, Var(Z) = 2, which is positive. Since Var(Z) can be 0 or 2 (or other positive values depending on the distribution of X), Var(Z) can be positive, and its exact value depends on the distribution of X. Step 3: Evaluate the given options. A. E[Z] = 1 but Var(Z) can be positive, depending on the distribution of X. This matches our findings. B. E[Z] = 1 and Var(Z) = 0 because squaring removes variability when variance is 1. This is incorrect, as Var(Z) can be positive. C. Neither E[Z] nor Var(Z) can be determined from the given information. This is incorrect, as E[Z] can be determined. D. E[Z] = 0 because E[X] = 0 and Var(Z) = 1 by definition. This is incorrect, as E[Z] = 1. The correct statement is A. A. E[Z] = 1 but Var(Z) can be positive, depending on the distribution of X. 3 done, 2 left today. You're making progress.