Question

Let X be a random variable with mean μ and variance σ². Given two independent random samples of sizes n₁ = 9 and n₂ = 7, with sample means X₁-bar and X₂-bar, if

X-bar = k X₁-bar + (1 – k) X₂-bar, 0 < k < 1, is an unbiased estimator for μ. If X₁-bar and X₂-bar are independent, find the value of k that minimizes the standard error of X-bar.

Answer 1

Solution :

Given data,

n1 = 9 , n2 = 7

E(k x-bar 1 + (1-k) x-bar 2) = k E(x-bar1) + (1-k) E(x-bar 2) = k mu + (1-k) mu =

k mu + mu - k mu = mu

Thus, the estimator is unbiased.

var(x-bar 1) = sigma^2/n1 and var(x-bar 2) = sigma^2/n2

var(k x-bar 1 + (1-k) x-bar 2) = k^2 var(x-bar 1) + (1-k)^2 var(x-bar 2) + 2 covar(x-bar 1, x-bar 2) =

k^2 sigma^2/n1 + (1-k)^2 sigma^2/n2 + 0 =

sigma^2 (k^2/n1 + (1-k)^2/n2)

The standard error is minimized when the variance is minimized.

Taking the derivative with respect to a, we get

sigma^2(2k/n1 + (2k-2)/n2)

This equals 0 when 2k/n1 + (2k-2)/n2 = 0, or

2k/n1 = -(2k-2)/n2 or, cross-multiplying,

2k n2 = -(2kn1 - 2 n1), or

2k n1 + 2k n2 = 2 n1, or

2k(n1+n2) = 2n1, or

k = 2n1/2(n1+n2), or

k = n1/(n1+n2)

sub n1 and n2 values in above equation.

k = 9/(9+7)

k = 9/16

k = 0.5625

There for the k value is 0.5625

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