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0

You may prove by induction that $\delta(\mathcal{G}_1),\dots,\delta(\mathcal{G}_k),\mathcal{G}_{k+1},\dots,\mathcal{G}_n$ are independent (for all $k\le n$)... For $k=0$ this is the assumption ($\mathcal{G}_{1},\dots,\mathcal{G}_n$ are independent). For $k>1$ assume that ...


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Confirming Brian Tung's answer. We have 16 chips; these are 6 red, 7 white, 3 blue There are $\dbinom{16}{4}$ ways to choose any 4 chips. From these we need to subtract those which are not at-least one each kind. There are $\dbinom{10}{4}$ ways to choose 4 not-red chips. There are $\dbinom{9}{4}$ ways to choose 4 not-white chips. There are ...


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The interpretation here is that the balls are distinguishable as are the boxes. This gives the combinatorial species $$\mathfrak{S}_{=M}(\mathcal{E} + \mathcal{U}\mathcal{Z} + \mathfrak{P}_{\ge 2}(\mathcal{Z})).$$ This translates into the bivariate generating function with $z$ representing the number of balls and $u$ the number of boxes with a ...


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Extended Comment, not Answer. I do not believe all of the speculations in the Question and Comments are exactly right. Here are some things I believe are correct, assuming you are dealing with an M/M/1 queue. The formula $W = \lambda/(\mu + \lambda) = \rho/(1 - \rho),$ where $\rho = \lambda/\mu$ is for the average number of customers in an M/M/1 queueing ...


1

Counterexample: Let $n=2$ and $\theta=1$; $\epsilon_1,\epsilon_2\sim_{i.i.d}\text{Bernoulli}(0.5)$. Take $x_1=\epsilon_2-\epsilon_1$, and $x_2=\epsilon_1-\epsilon_2$. Then $$y_1=\epsilon_2\text{; }y_2=\epsilon_1$$ and $$\frac{1}{2}=P\{y_1=0,y_2=0|x_1=0,x_2=0\}$$ $$\ne P\{y_1=0|x_1=0,x_2=0\}P\{y_2=0|x_1=0,x_2=0\}=\frac{1}{2}\times\frac{1}{2}$$ So, $y$-s ...


1

I think barak manos is doubly counting some selections. Following Marconius's tip, we have \begin{align} \binom{6}{2}(7)(3) + (6)\binom{7}{2}(3) + (6)(7)\binom{3}{2} & = (15)(7)(3) + (6)(21)(3) + (6)(7)(3) \\ & = 315 + 378 + 126 = 819 \end{align} different ways to select chips to satisfy the condition. Note that there are $$ \binom{6}{4} ...


1

Using Hoeffding's inequality $$\text{Pr}\left(\left|\bar X\right|>\epsilon\right)\le 2\exp\left(-\frac{1}{2}\epsilon^2n\right)$$


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You can easily get this from the usual proof of WLLN for random variables with finite variance. We have $\Bbb E X_j^2=1/2$. Hence $$\Bbb E\frac{|X_1+\dots+X_n|^2}{n^2}=\frac1{2n}.$$So Chebyschev's inequality says the probabilty you ask about is no larger than...


1

Define $\tilde\tau_x=\inf\{t\ge 0:\tilde B(t)=x\}$ where $\tilde B(0)=0$. Using the translation invariance, symmetry, and the reflection principle for BM $$P\{\tau_0>T\}=P\{\tilde\tau_1>T\}=\frac{2}{\sqrt{\pi}}\int_{0}^{\frac{1}{\sqrt{2T}}} e^{-u^2}du$$ ...


2

An example of a filtered probability space is $([0,1), \mathbf{B}_{[0,1)}, \{\mathcal{F}_t\}_{t\geq 0} , \mathbb{P})$, where, for each $t\in \mathbb{R}, t\geq 0$, define $$ \mathcal{F}_t = \mathbf{B}_{\left[0\,,\,1-\frac{1}{t+1}\right]}=\textrm{ the Borel $\sigma$-algebra defined in } \left [0\,,\,1-\frac{1}{t+1} \right] $$ Note that, by definition of ...


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The condition to ensure the tightness is that $\displaystyle \frac{f(x)}{x} \to \infty$ as $x \to \infty$. In fact if e.g. $f(x)=|x|$ then $$\sup_n \int f \, dμ_n<\infty$$ is only a necessary condition for the tightness.


1

You do not need any Algebraic Number Theory to start; of course there are some problems where you would need some but at the start you rather will not encounter them. Some Analytic Number Theory seems at least very useful. As the type of questions is quite related. A book I would recommend is "Introduction to Analytic and Probabilistic Number Theory: ...


1

Since $Y^i = \theta^T X^i + E^i$, by the substitution principle $$(Y^i \mid X^i =x^i) = \theta^T x^i + E^i,$$ where $x^i$ is a constant. Since the $E^i$ are independent, the $Y^i$ are conditionally independent. More explicitly, conditional pairwise independence holds because \begin{align} P(Y^i \in{A_i}, Y^j \in{A_j}\mid X^i=x^i,X^j=x^j) &= P( E^i ...


9

It would generally not be true even if they were independent. For example if $X,Y,Z$ were identically and independently continuously distributed then they can come in any order with equal probability so $ P(X \leq Y \leq Z) = \frac16$ but similarly $P(X \leq Y)P(Y \leq Z) = \frac12 \times \frac12 = \frac14$.


7

Let $X=Z$, and $Y$ such that $P(X \le Y) = \frac{1}{2}$ and $P(X=Y)=0$, then in order for the equality to hold we must have \begin{eqnarray} 0 &=& P(X=Y) \\ &=&P(X \le Y \le Z) \\ &=& P(X \le Y) P(Y \le X)\\ &=& P(X \le Y) \left( 1- P(X<Y) \right) \\ &=& \frac{1}{4} \end{eqnarray} So the equality does not hold in ...


1

Is your question just about $\sigma$-algebras or about filtration?? If you are just looking for an example of an uncountable increasing family of $\sigma$-algebras, there are many examples. One of the simplest examples is: for each $t\in \mathbb{R}, t\geq 0$, define $ \mathcal{F}_t = \mathbf{B}_{[0,t+1]}\textrm{ the Borel $\sigma$-algebra defined in } ...


0

Your intuition for the measure $d\mu = f d \lambda$ is very reasonable and it is interesting to note how the function is distributed. How much area it has acumulated between $-1$ and $1$ at which point $f(x) = e^{-x^2/2}/{\sqrt{2\pi}}$ goes from concave to convex (f''(x)=0). Since $$f''(x) = \frac{1}{\sqrt{2\pi}} \bigg(-e^{-x^2/2} + x^2e^{-x^2/2} = 0 $$ ...


2

You can look at $$\mathcal B_r=\Big\{A\subseteq [0,1]\times[0,1]\ \bigg|\ A\cap [0,r]\times[0,1]\text{ is Borel, }\land (A\cap(r,1]\times[0,1]\in\{(r,1]\times[0,1],\varnothing\})\Big\}$$ For $r\in[0,1]$. Where Borel means the standard Borel sets of $[0,1]\times[0,1]$. Then for $r\leq s$ we have $\cal B_r\subseteq B_s$, since whenever $A\in\cal B_r$, we ...


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$a)$ $P(\text{2 2-year-old cats live to 3 years old})=0.82666^2=0.68$ $b)$ $P(\text{ 8 2-year-old cats live to 3 years old}) = 0.82666^8 = 0.22$


4

Perhaps the most commonly used example is the natural filtration of Brownian motion, i.e. $\mathcal{F}_t=\sigma(W_s:s \in [0,t])$. This sort of thing is usually how filtrations are created.


0

You can do the same thing as you do in the discrete-time case by just thinking of the $X_n$ as being indexed by the rationals. Explicitly, choose a bijection $f:\mathbb{N}\to\mathbb{Q}$, and for each $r\in\mathbb{R}$, let $\mathcal{F}_r$ be the $\sigma$-algebra generated by the $X_n$ such that $f(n)<r$.


5

Let $A_n$ be independent events with $\mathbb{P}(A_n)=1/n$, and define $X_n=1_{A_n}$. Then $X_n\to0$ in probability, but $X_n$ does not converge almost everywhere. Apply the second Borel-Cantelli lemma twice; once to the sequence $A_n$ and also to the sequence $A_n^c$, to conclude that $$P([X_n=1\mbox{ infinitely often}] \cap [X_n=0\mbox{ infinitely ...


1

No $m$ need not be equal to $g(n)$. In fact $g(n) = m + f(n)$. The construction is concerned with the sequences of lenght $n$ that follow $x_1 \ldots x_m 1\ldots 1$ we are not appending $n - g(n)$ digits (as far as I understand). For example $m = 2$ $n = 4$ $g(n)=2$ $2^{n-g(n)}-1=2$. The sequences we have are $$x_1 x_2 1 1\\ x_1x_2 10\\ x_1x_2 01\\ $$ note ...


0

We know that $$\Bbb{P}(|x_n-L_x|>\epsilon) \xrightarrow[n \to \infty]{}0 \\ \Bbb{P}(|y_n-L_y|>\epsilon) \xrightarrow[n \to \infty]{}0 $$ This implies that for every $k$ there is a $N_k$ such that $n\geq N_k$ $$\Bbb{P}\bigg(|x_n-L_x|>\frac{1}{k}\bigg) \leq \frac{1}{2^k} \\ \Bbb{P}\bigg(|y_n-L_y|>\frac{1}{k}\bigg) \leq \frac{1}{2^k} $$ Let ...


0

In short, $E(X|\mathcal{A})(\omega)$ is the value of the random variable $E(X|\mathcal{A})$ when outcome $\omega$ occurs. It sounds like you are specifically looking at sub-$\sigma$-algebras that are generated by partitions of the entire collection of outcomes: $\Omega=\cup A_i$ where the $A_i$ are pair-wise disjoint and $\mathcal{A}=\sigma(A_1,A_2,...)$. ...


1

First, assuming $A \in \sigma(X)$ and $B \in \sigma(\tilde{X})$, your statement $$ \tilde{P}(A) = \int_A \frac{e^{-\frac{(X+c)^2}{2}}}{e^{-\frac{X^2}{2}}} dP $$ should be $$ \tilde{P}(A) = \int_A e^{-Xc - \frac{1}{2}c^2} dP. \qquad (1) $$ Then $$ \tilde{P}(A) = \frac{1}{\sqrt{2\pi}}\int_A e^{-xc - \frac{1}{2}c^2} e^{-\frac{1}{2}x^2} dx = ...


1

$\phi$ holds almost surely on $S$ means that there is a null set $N$ such that $$\omega \in S \setminus N\Rightarrow \phi(\omega) \text{ holds} $$ This is your Option $3$. Note that this notion is only interesting when $P(S)>0$. That is, on a null set, every property holds almost surely. The idea to consider regular conditional probabilities is not ...


1

The correct approach is always to consider the common sample space. If the events given are independent then the common sample space can be constructed. If they are not independent then either the dependence structure is given on a common sample space or the common sample space is not defined. In the case of independent dice rolling the sample space is ...


1

Given a measureable $f:(X,\mathcal{M},\mu)\to(Y,\mathcal{N})$ (a measure space to a measurable space) one can define a measure $\nu=f_{*}\mu$ on $(Y,\mathcal{N})$ by $$ \nu(B)=\mu(f^{-1}(B)). $$ Both $f$ and $\mu$ are necessary to define it, so $f_{*}\mu$ seems like good notation to me.


3

Just so that everyone knows what we are talking about here, let me rephrase in more familiar notation. Suppose $(\Omega, \mathcal{F}, P)$ is a probability space, and $(M, \mathcal{M})$ is a measurable space. If $X : \Omega \to M$ is a random variable (i.e. a $(\mathcal{F}, \mathcal{M})$-measurable function), it induces a pushforward measure on $(M, ...


1

It seems like you may be referring to the Law of Large Numbers. The Wikipedia page linked to gives a good explanation, from what I can see. See also: the relevant part of this SE answer for more information on the theorem, and why it is important, this SE question for another good, layman oriented explanation of the theorem, the top-rated answer to this SE ...


2

Possibly you're thinking of the Poisson distribution. Suppose you have a one-in-$1{,}000{,}000$ chance of success on each trial, and there are $3{,}600{,}000$ trials. The expected number of successes is then $3.6$. If we ask for the probability that there are exactly $5$ successes, we get $$ \frac{3.6^5 e^{-3.6}}{5!} = \frac{3.6^5 e^{-3.6}}{120} \approx ...


2

This answer is first adapted from comments in: does convergence in $L^p$ imply convergence almost everywhere. Consider a probability space $[0, 1]$ with uniform measure. Define the random variable $X_{2^i + k}$ by $$X_{2^i + k} = 2^{i/2}\chi_{\left[\frac{k}{2^i}, \frac{k+1/2}{2^i}\right]} - 2^{i/2}\chi_{\left[\frac{k + 1/2}{2^i}, \frac{k+1}{2^i}\right]}$$ ...


0

No, independence does not follow merely from this observations. This fact gives you only pairwise independence and this is not sufficient to assure independence of events. (See https://notesofastatisticswatcher.wordpress.com/2012/01/02/pairwise-independence-does-not-imply-mutual-independence/) We might think that \begin{align} \Delta_1 &\perp X_s \\ ...


1

This is the Borel sigma algebra on the Baire space which is homeomorphic to irrationals. So there can be at most continuum many such sets. Here's the wiki article about Baire space: https://en.m.wikipedia.org/wiki/Baire_space_(set_theory)


1

The author of that proof skips a lot of steps! It seems that the author is secretly making use of the Borel-Cantelli lemma. Equation (*) shows that $\sum_{n=1}^{\infty} Pr[|b_n|\leq |X_n|] < \infty$, which means (by Borel-Cantelli) that, with prob 1, the event $|b_n|\leq |X_n|$ only happens for a finite number of integers $n$. This also means that ...


1

If $a$ and $b$ are independent there is a very simple solution. First convince yourself that $$a + b \leq C \Leftrightarrow \mathrm{max}(a,b) \leq \frac{C}{2}$$ Then due to independence, $P\{\mathrm{max}(a,b) \leq \frac{C}{2}\} = P\{a \leq \frac{C}{2}\}P\{b \leq \frac{C}{2}\}$ I am assuming both $a$ and $b$ have uniform distributions. Then, $P\{a \leq ...


0

Can be done in $\min(O(A\log B),O(B\log A))$ using binary serach.


2

I'll take the statement "All players are equally skilled." to mean that every player has a $50\%$ chance to win against every other player. $7$ players have been eliminated by $p_5$; the probability that $p_1$ was not among them is $24/31$. If $p_1$ has not been eliminated, the identity of the winner of $p_5$'s branch of the tournament is irrelevant to her; ...


2

Hint: $$1_{\{|X|>a\}} \leq \left| \frac{X}{a} \right|^4 1_{\{|X|>a\}} \leq \left| \frac{X}{a} \right|^4. $$ Now integrate both sides.


0

I'll give a concrete example. Say you throw two dice and get a pair of numbers: $$ \begin{array}{cccccccccc} (1,1) & (1,2) & (1,3) & \cdots & (1,6) \\ (2,1) & (2,2) & (2,3) & \cdots & (2,6) \\ (3,1) & (3,2) & (3,3) & \cdots & (3,6) \\ \vdots & \vdots & \vdots & & \vdots \\ (6,1) & (6,2) ...


3

Let $Y = E(X \mid G)$. Note that $$\int X^+ - X^-\, dP = \int X\, dP = \int Y\, dP = \int_{Y>0} Y\, dP + \int_{Y<0} Y \, dP = \int_{Y>0} X\, dP + \int_{Y<0} X \, dP $$ More over since $X$ and $Y$ have the same distribution $E [|X|] = E[|Y|]$. This implies $$\int X^+ + X^-\, dP = \int |X|\, dP = \int |Y|\, dP = \int_{Y>0} Y\, dP - ...


0

Since $X,Y$ are random variables in a probility space $(\Omega,\sigma,P)$, the sets $\{\omega:X(\omega)=x\}$, $\{\omega:Y(\omega)=y\}$ are measurable and so it makes sense to talk about the probability of their intersection. The probability distribution of $X,Y$ play a role here, but it doesn't seem to be what you're thinking of. The probability $P$ is one ...


0

The distribution of $Y$ given $X$ is neither the distribution of $Y$ nor the distribution of $X$. The distribution of $Y$ given $X$ can be interpreted as the best guess you would give for $Y$ given the outcome of $X$. So it refers to values of $Y$ but it has available information from $X$.


0

Here is some detail you might find useful Consider $$\mathcal D_A:=\{ E \in \mathcal A \ | \ A \text{ and } E \text{ are independent} \}$$ Let $E_1,E_2 \in \mathcal D_A$ therefore $$P(A \cap E_1) = P(A)P(E_1)\\ P(A \cap E_2) = P(A)P(E_2) $$ We would like to see that $$P(A \cap E_1 \cap E_2) = P(A)P(E_1)P(E_2) $$ This follows for any pair of sets ...


1

The crucial point is the level of significance of the test, if you consider $\epsilon = 2^{-1}$ then you are rejecting the null conjecture (It is a random array) in many cases. For some reason you are bound to think that many $0$ may occur. So you will reject randomness at the event the first $m$ digits are zero. You could also think that it is not very ...


1

A class $\mathcal{C}$ of sets is $\cap$-stable if it is closed under $\cap$. It means that, if $E,F\in\mathcal{C}$ then $E\cap F\in\mathcal{C}$. In the proof of the theorem it is used that the class $\mathcal{B}$ of all sets of type $\cap_{j\in J} M_j$, where $J\subset \{ 1, \ldots, n \}$, is $\cap$-stable. It is easy to prove such statement. The ...


0

\begin{equation}P \lbrace Y_{n+1}=j | Y_{n}=i \rbrace =P_{ij}=Q_{ji}..........(i) \end{equation} (because the algorithm given which is the Hastings-Metropolis Algorithm generates a time reversible markov chain.) Now we are told that $Q$ is defined on the Markov chain $ \lbrace X_{n}, n=1,2,3,.. \rbrace$ it follows that equation $(i)$ can be written as ...


3

Let me give you first a non-rigorous "proof" of the statement; afterwards I'll explain how to justify it. If we define $$f_n(x) := n x 1_{[0,1/n]}(x) + 1_{[1/n,\infty)}(x),$$ then $f_n \uparrow 1_{(0,\infty)}$. Obviously, $f_n$ is not differentiable at $x=0$ at $x=1/n$, but for the moment we forget about this (that's the non-rigorous part) and apply ...


0

As it turned out, my previous logic was wrong. Here is how it should be done. Marginal probability of choosing nest $k$ is $$P_{nB_k} = P\left[\max_{j\in B_k} U_{njk} \geq \max_{j\in B_s} U_{njs}, \forall s\neq k \right]\\ = P\left[W_{nk}+\epsilon_{nk}+\max_{j\in B_k}(Y_{nj}+e_{nj}) \geq W_{ns}+\epsilon_{ns}+\max_{j\in B_s}(Y_{nj}+e_{nj}), \forall s\neq k ...



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