Question
3. Let be a random sample from a population with mean
and variance
. Consider the estimator
(i) Find the bias of the estimator .
(ii) Find the variance of .
(iii) Check whether it is more efficient than sample mean.
Answer :
Word Count : 221
The estimator is given by [ \hat{\mu} = \frac{1}{n}\sum_{i=1}^{n} X_i + 2 ] (i) Bias of the estimator The bias of an estimator (\hat{\mu}) is defined as [ \text{Bias}(\hat{\mu}) = E[\hat{\mu}] - \mu ] Now, compute (E[\hat{\mu}]): [ E[\hat{\mu}] = E\left[\frac{1}{n}\sum_{i=1}^{n} X_i + 2\right] = E\left[\frac{1}{n}\sum_{i=1}^{n} X_i\right] + 2 ] [ E\left[\frac{1}{n}\sum_{i=1}^{n} X_i\right] = \frac{1}{n}\sum_{i=1}^{n} _____ __________ ___ _____ _______ ________ _______ _________ ________ ______.
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The estimator is given by [ \hat{\mu} = \frac{1}{n}\sum_{i=1}^{n} X_i + 2 ] (i) Bias of the estimator The bias of an estimator (\hat{\mu}) is defined as [ \text{Bias}(\hat{\mu}) = E[\hat{\mu}] - \mu ] Now, compute (E[\hat{\mu}]): [ E[\hat{\mu}] = E\left[\frac{1}{n}\sum_{i=1}^{n} X_i + 2\right] = E\left[\frac{1}{n}\sum_{i=1}^{n} X_i\right] + 2 ] [ E\left[\frac{1}{n}\sum_{i=1}^{n} X_i\right] = \frac{1}{n}\sum_{i=1}^{n} _____ __________ ___ _____ _______ ________ _______ _________ ________ ______.
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Get Full Answer on WhatsApp
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