Ex: Suppose X₁, X₂, ..., X₂ is a random sample from a normal population with variance o². Consider the following estimat
Posted: Thu May 05, 2022 9:16 pm
Or solve the problem with no give an reference like "we know
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I'll give you upvote. I really need to understand.
Ex: Suppose X₁, X₂, ..., X₂ is a random sample from a normal population with variance o². Consider the following estimator for o², n Σ(x − x) - i=1 S*2 n a Show that S*2 is a biased estimator of o². b Estimate the amount of bias when S*2 is used to estimate o².
Solution: a From Chapter 8, we know that statistic (n − 1)S²/σ² has a x² distribution with (n − 1) degrees of freedom, where n Σ(x − x) - $²=i=1 n - 1 From Chapter 6, we know that a x² distribution has a mean equal to its degrees of freedom. Therefore, we get E[(n − 1)S²/o²] = (n − 1), and S*² = (n − 1)S²/n. From this information, we can find the mean of S*2 as - - 0² E(S²³²) = E( - 8( (" - 198²) - = ² 8( (n = 115²) - DS² - E = = -(n − 1) n n n In other words, in repeated sampling, S*2 values tend to center around a value slightly lower than o², namely, o²(n − 1)/n. Therefore, we conclude that S*2 is a biased estimator of o².
Solution: b The amount of bias when S*2 is used to estimate o² is (n − 1) - Bias = E(S*²) — σ² = 0² 0² 0² n n Because o² > 0, the negative bias indicates that S*2 systematically underes- timates o². Note that S*2 defined with (n − 1) instead of n in the denominator is an unbiased estimator of o² and therefore is more commonly used to estimate o² than S*2. It is also commonly referred to as the sample variance.
that statistic..." .
I'll give you upvote. I really need to understand.
Ex: Suppose X₁, X₂, ..., X₂ is a random sample from a normal population with variance o². Consider the following estimator for o², n Σ(x − x) - i=1 S*2 n a Show that S*2 is a biased estimator of o². b Estimate the amount of bias when S*2 is used to estimate o².
Solution: a From Chapter 8, we know that statistic (n − 1)S²/σ² has a x² distribution with (n − 1) degrees of freedom, where n Σ(x − x) - $²=i=1 n - 1 From Chapter 6, we know that a x² distribution has a mean equal to its degrees of freedom. Therefore, we get E[(n − 1)S²/o²] = (n − 1), and S*² = (n − 1)S²/n. From this information, we can find the mean of S*2 as - - 0² E(S²³²) = E( - 8( (" - 198²) - = ² 8( (n = 115²) - DS² - E = = -(n − 1) n n n In other words, in repeated sampling, S*2 values tend to center around a value slightly lower than o², namely, o²(n − 1)/n. Therefore, we conclude that S*2 is a biased estimator of o².
Solution: b The amount of bias when S*2 is used to estimate o² is (n − 1) - Bias = E(S*²) — σ² = 0² 0² 0² n n Because o² > 0, the negative bias indicates that S*2 systematically underes- timates o². Note that S*2 defined with (n − 1) instead of n in the denominator is an unbiased estimator of o² and therefore is more commonly used to estimate o² than S*2. It is also commonly referred to as the sample variance.