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$\newcommand\th\theta\newcommand\R{\mathbb R}$Indeed, the densities $f_\th:=d\mu_\th/d\nu$ are defined only $\nu$-a.e., and hence it makes no sense to define a maximum likelihood estimate (MLE) as a maximizer of $f_\th(x)$ in $\th$ for arbitrary versions of the densities. Usually, though, it is assumed that versions $f_\th$ of the densities can be chosen so ...

2

Let me calculate the expectation value of $\alpha$. The probability distribution of $X$ is invariant under orthogonal transformations, so without loss of generality I can orient the unit vector $w$ along one of the axes, $w_i=\delta_{ip}$, $p\in\{1,2,\ldots d\}$. Then $$\mathbb{E}[\alpha]=\mathbb{E}\left(X^T(XX^T+\lambda I)^{-1}XX^T(XX^T+\lambda I)^{-1}X\... 2 To find the dependence of s_{nm} on t=a\cdot b, we take a=(t,\sqrt{1-t^2},0,0,\ldots 0), b=(1,0,0,0,\ldots 0), so that$$s_{nm} = \mathbb E[H_n(X^\top a)H_m(X^\top b)]=\mathbb E[H_n(X_1 t+X_2\sqrt{1-t^2})H_m(X_1)].$$The marginal distribution P(X_1,X_2) of two elements from a vector that is uniformly distributed on the d-dimensional unit sphere ... 1 In general, a maximum likelihood estimator (MLE) does not have to be measurable. For instance, suppose that f(x|\theta)=g(x-\theta), where g(x)=1(0<x<1). Then, for any (x_1,\dots,x_n)\in\mathbb R^n, any number \hat\theta(x_1,\dots,x_n)\in(\max_i x_i-1,\min_i x_i) is a maximizer of the likelihood f(x_1|\theta)\cdots f(x_n|\theta) (in real \... 1 The correct version of this formula is$$P(\max_j |\epsilon_j| \ge x) = 1-(1-P(|\epsilon_1| \ge x))^p \underset{x \to\infty}\sim p \,P(|\epsilon_1| \ge x)$$for each real p>0, which follows because (1-u)^p=1-(p+o(1))u as u\to0. (The reproduction quality of the preview of the book is indeed terrible. It is also clear that 1-(1-P(|\epsilon_1| \ge x))... 1 Without loss of generality, let the final time to be t=1 (if it is not, we can make it so by rescaling time as \alpha'=t\alpha and \beta'=t\alpha). Then, consider a single trajectory of the process that starts and ends on state A and makes x jumps (x has to be even). We can represent any such trajectory in terms of the jump times,$$\vec{t}=(t_{0}...

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The coefficients $a_i$ of the random variables $X_i$ are not any prior probabilities at all -- because prior probabilities are coefficients, not of random variables, but of probability distributions. The choice $a_i\propto 1/\sigma_i$ in your setting equalizes the variances of the random variables $a_iX_i$, and that is all it does. Even though priors have ...

1

$\newcommand{\der}{\mathrm{der}}\newcommand{\tdert}{\mathrm{dert}}\newcommand{\erf}{\operatorname{erf}}\newcommand{\eqs}{\overset{\text{sign}}=}\newcommand{\tder}{\widetilde\der}$The answer is no. Indeed, let $x:=\xi_1$ and $y:=\xi_2$, so that $\xi_3=1-x-y$, $0\le x\le y\le1-x-y$, whence $x\in[0,1/3]$. Consider further the case $y=x\in(0,1/3]$, so that \...

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The answer to your first question is positive. For brevity, I will use the notation $a\pm b$ to indicate the interval $(a-b,a+b)$. Let $U_n:=f(X_n,Y_n)$. By the assumption, for every $\varepsilon>0$, \begin{align} \mathbb{P}\big(\mathbb{E}[U_n\,|\,X_n]\notin\alpha\pm\varepsilon\big)&\to 0 \qquad \text{as $n\to\infty$,} \tag{A1} \\ \mathbb{P}\...

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