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seen Feb 29 '12 at 9:04

Feb
29
awarded  Supporter
Feb
29
awarded  Scholar
Feb
29
comment Deriving posterior of Beta distribution
thanks for your help
Feb
29
accepted Deriving posterior of Beta distribution
Feb
29
revised Deriving posterior of Beta distribution
made some progress
Feb
29
asked Deriving posterior of Beta distribution
Feb
23
awarded  Editor
Feb
23
revised Convergence in distribution, probability, and 2nd mean
added 151 characters in body
Feb
23
asked Convergence in distribution, probability, and 2nd mean
Feb
19
comment Expected value and variance of arithmetic mean of random variables
Yea, I saw that and thought the covariance term was 0 because the random variables were drawn iid.
Feb
18
comment Expected value and variance of arithmetic mean of random variables
So I got $\mathbb{E}(\bar{X}) = \frac{1}{n}\sum^n_{i=1}{\mathbb{E}(X_i)} = \frac{1}{n}n\frac{0.1}{0.1+0.5} = \frac{1}{6} $ and $\mathbb{V}(\bar{X}) = \frac{1}{n^2}\sum^n_{i=1}{\mathbb{V}(X_i)} = \frac{1}{n^2}n\frac{0.1*0.5}{(0.1+0.5)^2(0.1+0.5+1)} = \frac{1}{n}\frac{0.05}{(0.36)(1.6)}$
Feb
18
awarded  Student
Feb
18
comment Expected value and variance of arithmetic mean of random variables
ok thanks. this was very helpful.
Feb
18
comment Expected value and variance of arithmetic mean of random variables
So I can change $\mathbb{E}(\bar{X})$ to $\sum^n_{i=1}\mathbb{E}(\bar{X_i})$, which I can calculate since I know the expected value of a single random variable from the Beta distribution. Similarly for the variance, I can do: $\mathbb{V}(\bar{X}) = \frac{1}{n^2}\sum^n_{i=1}{\mathbb{V}(X_i)}$
Feb
18
asked Expected value and variance of arithmetic mean of random variables