I met two versions of expressing induced probability measure on $\mathbb{R}^n$ of n-dimensional random vector $\mathbf{X} = (X_1, \ldots, X_n)$ (probability law, probability distribution) defined on probability space $(\Omega, \mathcal{F}, \mathbb{P})$:  $$ \mathbf{P}_{\mathbf{X}}(B)=\mathbb{P}\circ\mathbf{X}^{-1}[B]=\mathbb{P}(\{\omega\in\Omega\mid\mathbf{X}(\omega)\in B\}), \quad\forall B\in\mathcal{B}(\mathbb{R}^n)\\or\\\mathbf{P}_{\mathbf{X}}(B_1 \times\ldots\times B_n)=\mathbb{P}(\{\omega\in\Omega\mid X_1\in B_1,\ldots,X_n\in B_n\})=\mathbb{P}(\bigcap_{i=1}^{n}\{\omega\in\Omega\mid  X_i\in B_i\}),\quad\forall B_1,\ldots,B_n\in\mathcal{B}(\mathbb{R}) $$ My question is: are they equivalent? If yes, why it is enough to determine the measure $\mathbf{P}_{\mathbf{X}}$ only on the class $\{B_1\times\ldots\times B_n\mid B_i\in\mathcal{B}(\mathbb{R})\forall i\in\{1,\ldots,n\}\}$?
121345
My question comes from different modes of probability law, which I met. I see two options in equality: set belonged to $B(\mathbb R^n)$ and cartesian product of B belonged to $B(\mathbb R)$.
51961
My university is testing an intro Machine Learning (ML) course for us undergrads, and having been interested in ML since the beginning to understand what it was, I jumped at the chance to take it. But being someone who has not been back in school for a long time and much busier than I was in my first years long ago, I am finding that I need more info and examples than the class in its current form has.  I am struggling to figure out what the attached question is asking. I am searching online for help with this particular topic, but I am also finding that everyone seems to have their own way of phrasing similar questions in such a way that I am finding it hard to link the info.   My understanding from stats way back when is that $p(x|\omega_1)$ is stating the probability of $x$ given $\omega_1$ for $0\le x \le 2$ and zero other wise. Same obviously if given $\omega_2$. If I understand, the priors are the probability of $\omega_1$ and $\omega_2$ which are somehow obtained. And honestly that is as far as I am really comprehending. I am not sure what (a) and (b) are truly asking for or how to get them. If anyone is able to help me with some pointed info on the topic it would be greatly appreciated.
34986
I not solve the follow limit $$\lim_{p\rightarrow 0} \bigg[\int_{\Omega} |f|^p d\mu \bigg]^{1/p} = \exp\bigg[ \int_{\Omega} \log|f|d\mu \bigg],$$  where $(\Omega, \mathcal{F}, \mu)$ is a probability space and $f,\log |f| \in L^1(\Omega).$  Can someone help me?  Thank you!
91112
Given that $X,Y\stackrel{\text{i.i.d.}}{\sim}\text{Exp}$ with mean $\lambda(&gt;0)$ and $Z=\mathbf{1}_{X&lt;Y}$, what is $\mathbb{E}(X\mid Z=z)$ ?   We have $Z=1$ with probability $\Pr(X&lt;Y)$ and $Z=0$ with probability $1-\Pr(X&lt;Y)$.   But $\Pr(X&lt;Y)=\frac{1}{2}$ as $X$ and $Y$ are i.i.d. random variables.   Does this mean I can simply say that $Z\sim\text{Ber}\left(\frac{1}{2}\right)$ ?   Then, $\mathbb{E}(X)=\mathbb{E}(X\mid Z=0)\times\frac{1}{2}+\mathbb{E}(X \mid Z = 1) \times \frac{1}{2}$  $\implies2\lambda=\mathbb{E}(X\mid Z=0)+\mathbb{E}(X\mid Z=1)$.   But I cannot conclude anything from here. I couldn't find the conditional distribution of $X\mid Z$ either. Maybe I have to condition on another variable $U$ so that $\mathbb{E}(X\mid Z)=\mathbb{E}\,[\mathbb{E}(X\mid Z,U)\mid Z]$.
109590
This thread is just a note!  Given a Hilbert space $\mathcal{H}$.  Consider a normal operator: $$N:\mathcal{D}(N)\to\mathcal{H}:\quad N^*N=NN^*$$  And its spectral measure: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad N=\int_\Omega\lambda\mathrm{d}E(\lambda)$$  Regard an embedding: $$J\in\mathcal{B}(\mathcal{H}_0,\mathcal{H}):\quad J^*J=1_0$$  Suppose it reduces: $$P:=JJ^*:\quad PN\subseteq NP$$  Then one has: $$N_0:=J^*NJ:\quad N_0^*N_0=N_0N_0^*$$  And it holds: $$E_0:=J^*EJ:\quad N_0=\int_\Omega\lambda\mathrm{d}E_0(\lambda)$$  How to prove this from scratch?
69903
Let $X$ be a set, $\mathcal F$ a $\sigma$-field of subsets of $X$, and $\mu$ a probability measure on $X$. Given random variables $f,g\colon X\rightarrow\mathbb{R}$ such that $$\int_\mathbb{R}hd{\mu_f}=\int_{\mathbb{R}}hd\mu_g$$ for any Schwartz function $h$. Is it necessarily true that $\mu_f=\mu_g$?
25610
So I'm reading notes on Gaussian Processses, and came across  syntax $p(\mathbf{y} \mid \text{stuff}) = N(\mathbf{y} \mid \mathbf{0}, \mathbf{K})$ for multivariate normal distribution, and I'm not exactly sure how to decipher it the $\mathbf{y} | $ bit. I suppose it doesn't mean "mean $\mathbf{y}$ is conditional on zero vector", because I don't think that makes sense. Assuming this is standard notation for something and I'm just unfamiliar with it, what it means? E.g. how it differs from $N(\mathbf{0}, \mathbf{K})$?
15405
Given a Gaussian random vector $X \in \mathbf{R}^p \sim \mathcal{N}(\boldsymbol\mu, \boldsymbol\Sigma)$, a matrix $\mathbf{K} \in \mathbf{R}^{q \times p}$, and a vector $y \in \mathbf{R}^q$, I'd like to find the distribution of $\tilde X = (X \mid \boldsymbol{K} X = y)$ while assuming that $\operatorname{rank}(\boldsymbol\Sigma) = p$ and $\operatorname{rank}(\boldsymbol{K}) \le q &lt; p$. Ideally without having to (psuedo-)invert any singular matrices.
67801
Let $Y_1, Y_2, ..., Y_n$ be iid random variables and $B_1, B_2, ..., B_n$ be Borel sets. It follows that  $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$...I think?  If so, does the converse hold true? My Stochastic Calculus professor says it does (or maybe misinterpreted him somehow?),  I was  the  that independence of the n random variables was equivalent to saying for any indices $i_1, i_2, ..., i_k$ $P(\bigcap_{j=i_1}^{i_k} (Y_j \in B_j)) = \Pi_{j=i_1}^{i_k} P(Y_j \in B_j)$.  So, if the RVs are independent, then we can choose $i_j=j$ and k=n to get $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$, but given $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$, I don't know how to conclude that for any indices $i_1, i_2, ..., i_n$ $P(\bigcap_{j=i_1}^{i_k} (Y_j \in B_j)) = \Pi_{j=i_1}^{i_k} P(Y_j \in B_j)$, if that's even the right definition.  p.17  seems to suggest otherwise. idk  Help please?  Also this:    or    So, this answer is to use the Omega part to establish pairwise independence and ultimately conclude independence. Without that assumption, we cannot conclude independence. Is that right? Why does that not contradict the definition of independence: $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$ ?
36933
Let $I$ be a special rectangle in $\mathbb{R}^n$, and denote $\lambda(A)$ the measure of $A$. Prove that the following conditions are equivalent:  a) $\lambda(I)=0$  b) $I^{\circ}=\emptyset$ (i.e., the interior of $I$ is empty)  c) $I$ is contained in an affine subspace of $\mathbb{R}^n$ having dimension smaller than $n$. (An affine subspace is any set of the form $\{x_0+x|x\in E\}$, where $x_0\in\mathbb{R}^n$ is fixed and $E$ is a subspace of the vector space $\mathbb{R}^n$. Its dimension is equal to the dimension of $E$.)  The implication $a\implies b$ isn't too bad (using the definition if $I=[a_1,b_1]\times...\times[a_n,b_n]$ then $\lambda(I)=(b_1-a_2)...(b_n-a_n)$. So we conclude $a_i=b_i$ for some $i$. And since $I^{\circ}$ is open, there can't be anything contained in it). I also see how c) makes sense (at least intuitively), but not sure how to show it formally.  Thank you.
29266
In Lieb and Loss's Analysis, I saw that they mentioned  $L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$ dense in $ L^2(\mathbb{R}^n)$ (dense wrt the $L^2$ norm, I think). But I didn't find its proof in the book. So I wonder why it is?    Does this conclusion hold for $L^1(\Omega, \mathcal{F}, \mu) \cap L^2(\Omega, \mathcal{F}, \mu)$ for any measure space $(\Omega, \mathcal{F}, \mu)$?  Are there similar statements if replace $L^1$ with $L^p$, and  $ L^2$ with $L^q$ for $p \leq q \in (0, \infty ]$ or $\in [1, \infty]$?  Thanks and regards!
109155
I wanted to prove that if $X_n$ is bounded in probability and $Y_n = o_p(X_n)$, then $Y_n \rightarrow 0$ in probability   I  know the following definitions that is $X_n$ is bounded in probability meaning that   $P(|X_n|&lt;M)&gt;1-\epsilon$  and I know that $Y_n = o_p(X_n)$ implies that $Y_n/X_n \rightarrow0$ in probability
4614
I'm facing the following issue. Let $X$ be an integrable random variable on the probability space $(\Omega,\mathcal{F},\mathbb{P})$ and $\mathcal{G},\mathcal{H} \subseteq \mathcal{F}$ be two sigma-algebras. We assume that $X$ is independent of $\mathcal{G}$, i.e. $\sigma(X)$ is independent of $\mathcal{G}$. Can I say (and can I prove) that $$ E(X \mid \sigma(\mathcal{G} \cup \mathcal{H})) = E(X\mid \mathcal{H}) ?$$  Thank you very much for your help!
83655
Let $(X, \mathcal{J})$ and $(Y, \mathcal{F})$ be two measure spaces. Let us assume that $J$ is a collection of subsets of $X$ which generates $\mathcal{J}$, i.e. $\sigma(J)=\mathcal{J}$. Similarly, assume $\sigma(F)=\mathcal{F}$. Is it always true that $$ \sigma(J)\times\sigma(F)=\sigma(J\times F)? $$ Here, $J\times F$ is the set of all cartesian products of sets in $J$ with sets in $F$.   Please note: in the above expression the two product signs mean different things. On the left I am considering the product sigma algebra $\mathcal{J}\times \mathcal{F}$, while on the right I am considering the $\sigma$-algebra generated by the cartesian products of my ``elementary sets.''
105079
In measure theoretic probability I often see these two notations for the expectation of a random variable expressed as an abstract integral.  $$ \int_\Omega X(\omega) \mathbb{dP(\omega)} = \int_\Omega X(\omega) \mathbb{P(d\omega)} $$  Is each case trying to express a different view of the abstract integral? What is an intuitive way to interpret $\mathbb{dP(\omega))}$ and $\mathbb{P(d\omega)}$?
117432
Preliminaries  Let $(\Omega, \mathcal{G}, \mathbb{P})$ be a complete probability space.  Let $D$ be a complete, separable, metrizable topological space with Borel $\sigma$-algebra $\mathcal{B}(D)$ (such as $D = \mathbb{R}^q$ with $\sigma$-algebra $\mathcal{B}(D) = \mathcal{B}(\mathbb{R}^d)$).  Let $\mathbb{R}$ be equipped with its canonical Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$.  Let $g: \Omega \times D \rightarrow \mathbb{R}$ be a bounded $(\mathcal{G} \otimes \mathcal{B}(D) ) / \mathcal{B}(\mathbb{R})$-measurable function.  Let $\Pi: \Omega \rightarrow D$ be a $\mathcal{G}/\mathcal{B}(D)$-measurable random variable.  Let $H : \Omega \rightarrow \mathbb{R}$ be a $\mathcal{G}/\mathcal{B}(\mathbb{R})$-measurable random variable, defined by $$ H(\omega) := g(\omega, \Pi(\omega)).$$ Note, that, since $g$ is bounded, we have $H \in \mathcal{L}^2(\Omega, \mathcal{G}, \mathbb{P})$.  Let $j: D \rightarrow \mathcal{L}^2(\Omega, \mathcal{G}, \mathbb{P}) $ be defined by $$ j(\pi)(\omega) := g(\omega, \pi)  $$  For all $\pi \in D$, let $j(\pi)$ be independent of $\Pi$.  Question  I am interested in the conditional expectation $$\mathbb{E}[H \mid \Pi] :\Omega \rightarrow \mathbb{R}$$ of $H$ with respect to $\Pi$. More specifically, I suspect that (a $\mathbb{P}$-unique version of) this condititional expectation is given by  $$ \mathbb{E}[H \mid \Pi] (\omega) = \mathbb{E}[j(\Pi(\omega))], \quad (\dagger)  $$ whereby $\mathbb{E}[j(\Pi(\omega))]$ can of course also be written as $$\mathbb{E}[j(\Pi(\omega))] = \int_{\Omega} j(\Pi(\omega))(\tilde{\omega}) d\mathbb{P}(\tilde{\omega}) . $$  How can I prove, that $(\dagger)$ is the case? I have tried, tracking the definition of conditional expectation and using Fubini, but with little success so far.  Thanks for any advice!
28039
Let $\mathbf{X}\sim \mathcal{N}(0,I)$ be an $n-$dimensional standard Gaussian vector. We know that  $$\begin{align} ||\mathbf{X}||^2&amp;=\sum_{i=1}^nX_i^2\sim\chi^2_n\\ ||\mathbf{X}||&amp;=\sqrt{\sum_{i=1}^nX_i^2}\sim\chi_n\\ \end{align}$$  Suppose the components are correlated, i.e., we consider $\mathbf{Y}\sim \mathcal{N}(0,\Sigma)$ where $\Sigma$ is positive definite. We know that $||\mathbf{Y}||^2$ has the . Is there also a corresponding generalized chi distribution for $||\mathbf{Y}||$? If so, is there some asymptotic result on the mean of such a distribution for $n\to\infty$? This is related to       But here the components of the random vector are neither independent or identically distributed.