John Smith
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 Oct 1 awarded Popular Question 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