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3 years ago

12

SOMEONE HELP PLEASEEE !!!

1 answer:

denis23 [38]3 years ago

5 0

Answer:

if im right its 13.09 miles since perimiter is just the total length of the outside lines.

Step-by-step explanation:

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Suppose that E(θˆ1) = E(θˆ2) = θ, V(θˆ 1) = σ2 1 , and V(θˆ2) = σ2 2 . Consider the estimator θˆ 3 = aθˆ 1 + (1 − a)θˆ 2. a Show

katen-ka-za [31]

Answer:

Step-by-step explanation:

Given that:

$E( \hat \theta _1) = \theta \ \ \ \ E( \hat \theta _2) = \theta \ \ \ \ V( \hat \theta _1) = \sigma_1^2 \ \ \ \ V(\hat \theta_2) = \sigma_2^2$

If we are to consider the estimator $\hat \theta _3 = a \hat \theta_1 + (1-a) \hat \theta_2$

a. Then, for $\hat \theta_3$ to be an unbiased estimator ; Then:

$E ( \hat \theta_3) = E ( a \hat \theta_1+ (1-a) \hat \theta_2)$

$E ( \hat \theta_3) = aE ( \theta_1) + (1-a) E ( \hat \theta_2)$

$E ( \hat \theta_3) = a \theta + (1-a) \theta = \theta$

b) If $\hat \theta _1 \ \ and \ \ \hat \theta_2$ are independent

$V(\hat \theta _3) = V (a \hat \theta_1+ (1-a) \hat \theta_2)$

$V(\hat \theta _3) = a ^2 V ( \hat \theta_1) + (1-a)^2 V ( \hat \theta_2)$

Thus; in order to minimize the variance of $\hat \theta_3$ ; then constant a can be determined as :

$V( \hat \theta_3) = a^2 \sigma_1^2 + (1-a)^2 \sigma^2_2$

Using differentiation:

$\dfrac{d}{da}(V \ \hat \theta_3) = 0 \implies 2a \ \sigma_1^2 + 2(1-a)(-1) \sigma_2^2 = 0$

⇒

$a (\sigma_1^2 + \sigma_2^2) = \sigma^2_2$

$\hat a = \dfrac{\sigma^2_2}{\sigma^2_1+\sigma^2_2}$

This implies that

$\dfrac{d}{da}(V \ \hat \theta_3)|_{a = \hat a} = 2 \ \sigma_1^2 + 2 \ \sigma_2^2 > 0$

So, $V( \hat \theta_3)$ is minimum when $\hat a = \dfrac{\sigma_2^2}{\sigma_1^2+\sigma_2^2}$

As such; $a = \dfrac{1}{2}$ if $\sigma_1^2 \ \ = \ \ \sigma_2^2$

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gregori [183]

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tamaranim1 [39]

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(y¹-y¹/x¹-x¹)

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Step-by-step explanation:

1/3 or one-third is your answer!

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meriva

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