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Given: $(X,Y)$ is Uniformly distributed on a disc of unit radius. Then, the joint distribution of $n$ such points, $((X_1,Y_1), ..., (X_n,Y_n))$, each drawn independently, will have joint pdf: $$f( (x_1,y_1), ..., (x_n,y_n) ) = \begin{cases}\pi^{-n}& (x_1^2 + y_1^2 < 1) & \text{ & } \cdots \text{ & }&(x_n^2 + y_n^2 < 1) \\ 0 ... 13 Here are some asymptotics for the first part: Claim For every fixed m, the random variable$$ S_n^m = T_{n}^m - \log n - (m-1) \log\log n+\log((m-1)!) $$converges in distribution to the "extreme value" distribution independent of m given by$$ F_{S_\infty}(s) = \exp\left(-\mathrm e^{-s}\right). $$Proof For frog j and t\ge 0, let ... 12 The range of an n-dimensional vector measure is always closed. See the paper (and references therein) P. R. Halmos (1948), The range of a vector measure, Bull. Amer. Math. Soc. 54, 416–421. So the set will always be closed. 11 To talk about probabilities, you first need to define a probability space, that is a measure space \Omega such that \mu(\Omega)=1. The measurable subsets of \Omega are your events. For instance, if you are to choose a number in \{1,2\}, you could set \Omega:=\{1,2\} and put the counting measure on it, divided by 2. There are four events: ... 10 Let \phi(t) = E(e^{itX}) and \psi(t) = E(e^{itY}) be the characteristic functions of X and Y. By hypothesis, we have$$\forall t \in \Bbb R,\qquad \phi(t)(1-\psi(t))=0$$The function \phi being continuous with \phi(0)=1, we can find an open neighborhood U of 0 where \phi does not vanish. Therefore \psi(t)=1 for every t \in U, which ... 9 As the other respondents have already noted, probability is already defined as a ratio. But here is another thing to think about. Normally when you take the ratio of two quantities, the result is measured in the units of the ratio of the units. For example, if you travel 10 meters in 5 seconds, your average velocity is 10/5 meters per second. But ... 9 I think the situation is similar to that in algebra. In elementary school, you learned that 1+1=2. It was kinda obvious, right? In rigorous advanced algebra, however, you first have to define “1”, “2”, “+” and then you must prove that 1+1=2. Similarly, probability theory at an undergraduate level uses some informal but intuitively sound notions ... 8 No, because you're not in a situation where you always remember two other wrong answers at the moment you're about to mark your choice. The analysis in the Monty Hall problem that leads to "it is an advantage to switch" depends on the fundamental assumption that you will be shown a non-prize door that you haven't chosen no matter which choice you made ... 8 There is a very elegant theorem published by Beardwood, Halton and Hammersley in 1959 that gives an alsmost-sure convergence for the length of the shortest path divided by \sqrt{N}. See this poster for a statement in a general setting. Also, the convergence holds for expectations. In 1989, Rhee and Talagrand showed that the length of the shortest path is ... 8 Thanks for merico's crucial hint "4 coulmns of arrows going to the right", following is my answer. By collapsing some of the vertices and edges, the honey comb is equivalent to following ladder graph: Forget about the matrix expression in above figure first. Some of the edges in this ladder graph has been labeled by a number. It means that edge in the ... 8 Theorem. \lim_{n\to\infty} K(n,n)/\binom {2n}n^2=16/9. This theorem is a corollary of the following results. For every integers 0\le a,b\le n put c_{a,b}(n)=\binom {2n-a-b}{n-a}/\binom {2n}n. Lemma 1. For every integers a,b\ge 0 there exists a limit \lim_{n\to\infty} c_{a,b}(n)=2^{-(a+b)}. Proof. It follows from the equality c_{a,b}(n)=\frac ... 7 Watch the phrase "almost surely". Here is a written-out statement of the problem: Suppose that for each n, there exists an event A_n with P(A_n) = 1 such that for every \omega \in A_n, we have |X_n(\omega)| \le Y(\omega). Show that there exists an event A with P(A) = 1 such that for every \omega \in A, we have \sup_n |X_n(\omega)| \le ... 7 The question is what does it mean to be tangible. The sets that Tsitsiklis mentions are "non measurable sets" and they exist as a consequence of the axiom of choice. However it is consistent with the failure of the axiom of choice that no such set exists, and that we can - in fact - assign probabilities to all the subsets of the real line/plane/etc. ... 6 By definition (essentially), a density function for a random variable X is precisely the Radon-Nikodym derivative of the probability measure on \mathbb{R} induced by X. So, this fact amounts entirely to the following claim: Lemma: Let \mu be a Borel probability measure on \mathbb{R} which is absolutely continuous with respect to Lebesgue measure ... 6 Let (\Omega,\mathcal{F},P) be a probability space, i.e. \Omega is a non-empty set, \mathcal{F} is a sigma-algebra of subsets of \Omega and P:\mathcal{F}\to [0,1] is a probability measure on \mathcal{F}. Now, suppose we have a function X:\Omega\to\mathbb{R} and we want to "measure" the probability of X belonging to some subset of \mathbb{R}. ... 6 Let us formulate your problem mathematically. Let X_n be the outcome of each coin toss. P(X_n = 1) = 1- p and P(X_n = -1) = p Let H_{n} be your bet size on n-th toss and this depends on outcomes of previous coin tosses X_1,... X_n. Let S_n = \sum_{i=1}^n X_n Then your final wealth is W_n = W_0+\sum_{i=1}^n H_n X_n = W_0 + \sum_{i=1}^n H_n ... 6 Measure theory gives a unified mathematical and conceptual framework for general probability theory. The two classic scenarios in probability theory are the discrete and continuous measures, which are treated quite separately: In the discrete case, we have a finite or countable space \Omega and assign a probability p(x) to each point x \in \Omega, ... 6 There is a Borel set E in \mathbb R^2 such that F := \{x-y\colon (x,y) \in E\} is not a Borel set. Let A := \{f \in \mathbf{C}\colon (f(1), f(0)) \in E\}. Then A \in \mathcal{B}_{\left[0,\infty\right)}. How about T(A)? In fact$$ T(A) = \{g \in \mathbf{C}\colon g(0)=0, g(1) \in F\} and is not Borel. added Mar 10 Why is T(A) not Borel? ... 6 You want to look at the version of this question on MO. In his answer, Stefan Geschke indicates there that the axiom of choice is needed to exhibit examples, since any such finitely additive measure gives us sets of reals without the Baire property. (In a comment, Clinton Conley provides details.) The point is that there are models of set theory without the ... 6 I've a proof which relies on Birkhoff ergodic theorem, so I'm not sure it fits the term "elementary". Anyway, we don't need to know the fact that ergodic measures are the extreme points of the convex set of invariant measures. Assume that \lambda:=\frac{\mu+\nu}2 is ergodic. Then by von Neumann's or Birkhoff's ergodic theorem, we get that given ... 6 You are correct that this is not always possible. It depends upon p. For N flips there are 2^N sequences, each with a certain probability that you can figure out if you know the probability the coin shows heads using the binomial distribution. To make a fair game, you need to be able to express \frac 12 as the sum of some set of these ... 6 Your answer depends, as you guessed on the process u \mapsto \sigma_u. You can amplify your approach using the Ito-isometry with the BDG-inequality: \begin{align} \mathbb{E}[|\int_0^t\sigma_udW_u - & \int_0^s\sigma_u \,dW_u|^{2p}] \stackrel{\text{BDG}}{\leq} c(p) \mathbb{E}[(\int_s^t |\sigma_u|^2 \, du)^p] \\ & \leq c(p) \mathbb{E}[(t-s)^{p-1} ... 6 The length of the longest initial run of tails is a random variable that can take on values 0,1,2,\ldots with probability 1/2, 1/4, 1/8,\ldots. The probability of winning is generally P_{\text{win}}(\mathbf{a}) = \sum_{i=0}^{\infty}\frac{a_i}{2^i}, $$where a_i=1 if a run of i tails followed by a head is a win and a_i=0 if it's a loss. You can ... 6 There is a HUGE difference between being true for an arbitrarily large n, and being true an infinite set. For a completely trivial example, you can prove that every subset of the natural numbers is bounded using the same logic, which is clearly nonsense. To see a basic non-trivial example, consider closed sets in \Bbb R. By induction finite unions of ... 6 Let f : \mathbb{R} \to \mathbb{R} be convex. This means that at every point a \in \mathbb{R}, there is an affine linear function l_a : \mathbb{R} \to \mathbb{R} which is dominated by f, i.e.$$ l_a(x) \leq f(x) $$and l_a(a) = f(a). When f is differentiable, for example, then l_a is the tangent to f at a. When f is strictly convex, we ... 6 You may assume the first point A at (1,0) and the second point B=(\cos\phi,\sin\phi) being uniformly distributed on the circle. The probability measure is then given by {1\over2\pi}{\rm d}\phi. The distance D:=|AB| computes to 2\left|\sin{\phi\over2}\right|, and we obtain$${\mathbb E}(D)={1\over 2\pi}\int_{-\pi}^\pi ...

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Integrate the LHS and the RHS of the pointwise identity $$X=\int_0^X\mathrm dx=\int_0^\infty\mathbf 1_{X\geqslant x}\,\mathrm dx.$$ This shows that the desired formula for $E[X]$ holds irrespectively of the hypothesis that $X$ is discrete or continuous or neither discrete nor continuous, as soon as $X\geqslant0$ almost surely, and that two formulas are ...

5

Let $Y_i$ be independent RVs with $P(Y_i=2^i)=1/i^2$ and then $P(Y_i=-1) = 1-(1/i^2)$. Note that $E[Y_i] >0,$ and $E[Y_i]$ is increasing in $i$, thus $S_n := \sum_{i=1}^n Y_i$ is a submartingale relative to $\mathcal{F}_n:= \sigma(Y_i,1\leq i \leq n)$. (It is an extremely biased random walk.) Clearly $E[S_n] > n E[Y_1] \to \infty$ since ...

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There is a natural notion of picking a number from the uniform distribution on $(0,1)$ a $\kappa$ number of times where $\kappa$ is an uncountable cardinal number. It is the product of the probability measures. The problem is that in the usual formulation of probability theory, not every set denotes an event that one can assign a probability too. Using the ...

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If I didn't make a mistake, such a random variable must always exist. Let $[-\infty,+\infty]$ be the set of real numbers along with the two infinities $\pm\infty$ in the order topology and with the Borel $\sigma$-algebra. Throughout this proof, random variables will be allowed to take values in $[\infty,+\infty]$. Terminology: Suppose that for each \$n\ge ...

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