2

The identities $$\mathbb{E}_x\mathbb{E}_\theta \bigl[{\bf 1}_{\{f(x|\theta)>1\}}\log f(x|\theta) \bigr]=\mathbb{E}_\theta\mathbb{E}_x\bigl[{\bf 1}_{\{f(x|\theta)>1\}}\log f(x|\theta) \bigr]\quad \; \; (*)$$ and $$\mathbb{E}_x\mathbb{E}_\theta \bigl[{\bf 1}_{\{f(x|\theta)\le 1\}}\log f(x|\theta) \bigr]=\mathbb{E}_\theta\mathbb{E}_x\bigl[{\bf 1}_{\{f(...


2

$(\lim \sup \frac {S_n} {C_n}>a)= (\lim \sup \frac {X_k+X_{k+1}+...+X_n} {C_n}>a)$ for any $k$ since $\frac {X_1+X_2+..+X_{k-1}} {C_n} \to 0$. From this it is clear that $(\lim \sup \frac {S_n} {C_n}>a)$ belongs to $\sigma (X_k,X_{k+1},...)$ for each $k$ and hence it belongs to $\mathcal T$.


2

First, note that $X_n\xrightarrow{p}X$ iff every subsequence of $\{X_n\}$ has a sub-subsequence converging to $X$ a.s. Take a subsequence $\{\xi_{n_k}\}$. We know that $\xi_{n_k}\xrightarrow{p}\xi$ and, thus, there is a subsequence $\{\xi_{n_{k_j}}\}$ converging a.s. to $\xi$. Since $\{f \text{ is discontinuous at }\xi\}$ is a null set, $f(\xi_{n_{k_j}})\to ...


1

Suppose your estimator for $\mu$ is $\bar Y$. Write $$\begin{split}MSE(\bar Y)&=E(\bar Y^2)-2\mu E(\bar Y)+\mu^2\\&=Var(\bar Y)+[E(\bar Y)]^2-2\mu E(\bar Y)+\mu^2\end{split}$$ If each of the Y's has expected value $\mu$ and variance $\sigma^2$, we have $\frac{\sigma^2}{n}+\mu^2-2\mu^2+\mu^2=\frac{\sigma^2}{n}$ This is in the form of $Var(\bar Y)+Bias(...


1

The mean zero requirement is to ensure the positive definiteness, i.e. if $X$ is in this space and $\operatorname{Cov}(X,X)=0$ then $X$ is the zero random variable. (To have a unique "zero random variable" in this sense, you must first take the quotient by $X \sim Y$ if $X=Y$ a.s., which is the usual procedure used in defining the $L^p$ spaces.) If ...


1

You've made one mistake. There are only $n-1$ possible neighbors of a given vertex. So you should have $$n(n-1)p(1-p)^{n-2}$$


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