Question

The estimator Yis a random variable that varies with different random samples; it has a probability distribution function that represents its sampling distribution, and mean and variance. Using the properties on expected values and variances of linear functions of random variables and sum operators, show that:

A. E(Y) = μ
B. Var(Y) σ2/N.

Answer 1

Check Explanation

Explanation:

According to the Central limit theorem, the population mean (μ) is approximately equal to the mean of sampling distribution (μₓ).

And the standard deviation of the sampling distribution (σₓ) is related to the population standard deviation (σ) through

Standard deviation of the sampling distribution = (Population standard deviation)/(√N)

where N = Sample size

σₓ = (σ/√N)

So, population mean (μ) = Mean of sampling distribution (μₓ)

Population Standard deviation = (Standard deviation of the sampling distribution) × √N

= σ × √N

A) The expected value of a given distribution is simply equal to the mean of that distribution.

Hence, the expected value of random variable Y thay varies with different samples is given as

E(Y) = Mean of sampling distribution = μₓ

But μₓ = μ

Hence, E(Y) = μ (Proved)

B) Var (Y) is given as the square of the random distribution's standard deviation.

Var (Y) = (standard deviation of the sampling distribution)²

= (σ/√N)²

= (σ²/N) (Proved)

Hope this Helps!!!