2. Let X1, X2, ? , X9 denote a random sample from a...

2. Let X1, X2, ? , X9 denote a random sample from a population with mean ? and variance ?^2. Consider the following two estimators of ?: ??1 = (X1 + X2 + ? + X9) / 9 ??2 = (3X1 - X6 + 2X4) / 4 (a) (4 pts) Is either of the estimators unbiased? (b) (4 pts) Which estimator is better? Explain why.

Transcript

00:01 So in the problem, we have been given two estimators of the mean mu of this random sample x1 through x9.

00:11 So x1 to x9 is a population.

00:13 It has a mean mu, sigma square variance.

00:17 And we've been given two estimators of mu, theta 1 hat and theta 2 hat.

00:22 And in part a, we've been asked to determine is either of these estimators unbiased? so in order to figure out which of these estimators is unbiased or biased, you compare the expected value of the estimator to the mean.

00:40 So the expected value of theta 1 hat is just the expected value of this.

00:49 And since we know expected value is distributive, so we can say that this is 1 over 9 and this is the expected value of.

00:58 X1 all the way through expected value of x9 right and each of these is mu so this becomes me in part sorry it's not a new part the next part or something else altogether so the expected value of theta 2 is the same thing you carry out the expected value operation on each one of these terms and so you take the one fourth out which is a constant and you distribute the expected value of onto each one of them.

01:35 So this would be expected value of 3x1 minus expected value of x6 plus 2 expected value, sorry, i'll write that part out later, but expected value of 2x4.

01:52 And if there's a constant within the expected value, obviously take the term out.

01:57 So 3 of expected value of x1, and this would be 2 times expected value.

02:01 Value of x4 and each of these are also mu so this ends up being um three mu minus mu two mu plus two mu which is four mu right i think so i think that's the answer since this is four so yeah it's uh x1 minus xxx plus 2x4 over 4 that is the one i have over here so i think i might have not sure let me just compute it out myself so this is also mu so since both are mu that is both are equal to the actual population mean so therefore theta 1 hat and theta 2 hat are unbiased estimators all right and in part b i believe we've been asked to find which estimator is better and why so in order to find which estimator is better we have to check the variation within the data right because if the data is to spread apart then the estimated mean can be really far from the actual population mean but if the data is clumped together that would not be the case and the variability of data is is expressed via variance.

03:39 So let's calculate the variances of the estimators that are given to us.

03:44 So this is basically the same thing where the variance gets carried over.

03:49 And you just end up squaring the term inside.

03:53 So it's 1 over 9 square times variance of x1 all the way through variance of x9...

Question

2. Let X1, X2, ? , X9 denote a random sample from a population with mean ? and variance ?^2. Consider the following two estimators of ?: ??1 = (X1 + X2 + ? + X9) / 9 ??2 = (3X1 - X6 + 2X4) / 4 (a) (4 pts) Is either of the estimators unbiased? (b) (4 pts) Which estimator is better? Explain why.

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