# Tag Info

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$$MSE=variance+Bias^2$$ Variance is the variance of random variable which is your estimator. Your calculation of Bias is correct.

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Bias is zero as $E(\hat p = \frac{1}{n}\sum_i X_i)=\frac{1}{n}E(\sum_i X_i)=\frac{1}{n}\sum_i E(X_i)= \frac{np}{n}=p$ Hence, $MSE=Var(\hat p)+bias^2=Var(\hat p)+0$ $Var(\hat p)=Var(\frac{1}{n}\sum_i X_i)=\frac{1}{n^2}Var(\sum_i X_i)=\frac{1}{n^2}\sum_i Var(X_i)=\frac{np(1-p)}{n^2}=\frac{p(1-p)}{n}$ therefore $MSE(\hat p)=Var(\hat p)=\frac{p(1-p)}{n}$

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I just figured out the answer, it is a good drill for understanding the basic idea about the law of iterated expectations. 1) $E[AB]=E[E[AB|N]]=E[E[A|N]E[B|N]]$, Because $E[A|N]=E[B|N]=N$, so $E[AB]=E[N^2]$. We know $E[N]=\frac{1}{p}$ and $Var(N)=\frac{1-p}{p^2}$. It is easy to get $E[N^2]$. Similarly, $E[AN]=E[E[AN|N]]=E[E[N|N]E[A|N]]$ 2) After getting ...

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2) $${\bf E}[(\hat\Theta _1-\Theta _1)^2] = {\bf E}[(\hat\theta _1-\Theta _1)^2] = var(\Theta)$$ For a normal distribution, we recognize that the variance is determined by the coefficient that comes next to a term of the form $\theta _1^2$. So what we're looking for is the constant that sits next to the $\theta _1^2$ term. We collect all of them, it is ...

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Using Bayes' Rule, the posterior distribution of the variance is given by the product of the likelihood and the prior distribution of the variance divided by the marginal likelihood $$p(\sigma^2|t_1,t_2,\cdots,t_n)=\frac{p(t_1,t_2,\cdots,t_n|\sigma^2)p(\sigma^2)}{ p(t_1,t_2,\cdots,t_n)}=\frac{p(t_1,t_2,\cdots,t_n|\sigma^2)p(\sigma^2)}{\int ... 1 The notation is rather bad, because it implies that we always sum the first c numbers, not a random set. If I understand it right, we have$$\hat S= \frac{n}{c} \sum_{j\in T} \alpha_j where $T \subset \{1,2 \cdots n\}$, $|T|=c$, is a random subset. We could prescribe that the size $c$ is fixed, or that it is itself random. We intend $\hat S$ to be an ...

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