We have talked about the fact that the sample mean estimator X = X; is an unbiased estimator of the mean u for identically distributed X1, X2, ..., Xn: E[X] = variance estimator, on the other hand, is not an unbiased estimate of the true variance o²: for µ. The straightforward Vý = E1(X; – X)², we get that E[V½] = (1 – )o². Instead, the following bias-corrected sample variance estimator is unbiased: V = „4 E (X; – X)². This unbiased estimator is typically what is called the sample variance. Use the fact that E[V] = o² to show that E[V] = (1 – )0². Hint: The proof is short, it can be done in a few lines. (b) that Var[X] = 02, for iid variables with variance o?. We can similarly ask what the variance is We also discussed the variance of the sample mean estimator, and concluded

TF.5 A small hotel has 5 rooms. Let X be the number of occupied rooms on any given day. The pmf p(x) of X is given in the table. x 0 1 2 3 4 5 p(x) .05 .05 .20 .25 .30 .15 Find the means E(X) and E(X2) , the variance V(X) and the standard deviation SD(X) of X.

1. Suppose a X value is produced by a random process that has a mean of 1 and a variance of 9, while a Y value is produced by a random process that has a mean of 9 and a variance of 1. Let Z = 3X - 2Y. Compute: a. the mean and variance of Z assuming X and X are independent. b. the mean and variance of Z assuming X and X are not independent and their correlation coefficient is 0.50.

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