Question

Suppose that xi, where i = 1, . . . , n, are independent random variables with e(xi) = μ and var(xi) = σ². Let x = n-1 Σ xi. Show that E(x) = μ and Var(x) = σ²/n.

Answer 1

To show that E(x) = μ and Var(x) = σ²/n for independent random variables xi, we can use the linearity of expectation and the property of variance for independent random variables. The expectation of x is equal to the sum of the expectations of xi, which are all μ, divided by (n-1). Thus, E(x) = μ. The variance of x is equal to the sum of the variances of xi, which are all σ², divided by (n-1)². Thus, Var(x) = σ²/n.

Step-by-step explanation:

To show that E(x) = μ or the mean of x is equal to μ, we need to find the expectation of x:

1. E(x) = E(∑xi) / (n-1)
2. E(x) = (∑E(xi)) / (n-1) [by linearity of expectation]
3. E(x) = (∑μ) / (n-1) [since E(xi) = μ for all i]
4. E(x) = μ * n / (n-1) [simplifying the sum]
5. E(x) = μ [after simplification]

To show that Var(x) = σ²/n or the variance of x is equal to σ²/n, we need to find the variance of x:

1. Var(x) = Var(∑xi) / (n-1)²
2. Var(x) = (∑Var(xi)) / (n-1)² [by independence]
3. Var(x) = (∑σ²) / (n-1)² [since Var(xi) = σ² for all i]
4. Var(x) = σ² * n / (n-1)² [simplifying the sum]
5. Var(x) = σ² / n [after simplification]