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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

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 $${\mathbb E}(D)={1\over 2\pi}\int_{-\pi}^\pi ... 5 Let n\ge 3. Suppose that the permutation \alpha takes k to a_k (k=1 to n). Let \alpha' be the permutation that takes k to (n+1)-a_k. Then the number of local maxima (minima) of \alpha is the same as the number of local minima (maxima) of \alpha'. Thus by symmetry the expected number of local maxima and the expected number of local ... 5 First, there are things that are much easier given the abstract formultion of measure theory. For example, let X,Y be independent random variables and let f:\mathbb{R}\to\mathbb{R} be a continuous function. Are f\circ X and f\circ Y independent random variables. The answer is utterly trivial in the measure theoretic formulation of probability, but ... 4 The statement is not correct: Let X,Y be independent Gaussian random variables with mean 0 and variance 1. Then$$\mathbb{E}(X^2)=1$$but$$\mathbb{E}(X \cdot Y) = \mathbb{E}(X) \cdot \mathbb{E}(Y)=0.$$Concerning your edit: Using a similar argumentation, it is not difficult to see that$$\mathbb{E}(X \cdot Y) = \mathbb{E}(X) \cdot \mathbb{E}(Y) ...

4

Note that the series lasts $7$ games if and only if each team wins $3$ times in the first $6$ games; so the probability is $$\binom{6}{3}\cdot{\left (\frac{1}{2}\right )}^{6}$$ If the probabilities of winning for A were $p$ then the result is $$\binom{6}{3}\cdot p^{3}\cdot (1-p)^{3}$$

4

For a Markov process $(X_t)_{t \geq 0}$ we define the generator $A$ by $$Af(x) := \lim_{t \downarrow 0} \frac{\mathbb{E}^x(f(X_t))-f(x)}{t} = \lim_{t \downarrow 0} \frac{P_tf(x)-f(x)}{t}$$ whenever the limit exists in $(C_{\infty},\|\cdot\|_{\infty})$. Here $P_tf(x) := \mathbb{E}^xf(X_t)$ denotes the semigroup of $(X_t)_{t \geq 0}$. By Taylor's formula ...

4

Start with: $x=P\left[A\cap B\right]$, $y=P\left[A^{c}\cap B^{c}\right]$, $a=P\left[A\cap B^{c}\right]$ and $b=P\left[A^{c}\cap B\right]$ where $a,b,x,y\geq0$ and $a+b+x+y=1$. Then: $$\left|P\left(A\cap B\right)-P\left(A\right)P\left(B\right)\right|=\left|x-\left(a+x\right)\left(b+x\right)\right|=\left|xy-ab\right|\leq\max\{xy,ab\}$$ Here ...

4

Maybe this can help you to understand the concept of conditional expectation, behind your question. Suppose you have a probability space $(\Omega, \mathcal P (\Omega), \mathbb{P})$, where $\mathcal P (\Omega)$ denotes the set of all possible subsets of $\Omega$ (evidently, a $\sigma$-algebra), and $\mathbb{P}$ is a probability measure (in this case, a ...

4

Your computation is valid. Probably the easiest way to show it is to note that, for every $t$, $$E(\mathrm e^{\mathrm iuW_\tau}\mid\tau=t)=E(\mathrm e^{\mathrm iuW_t}\mid\tau=t)=E(\mathrm e^{\mathrm iuW_t}),$$ where the second identity holds thanks to the independence of $W_t$ and $\tau$. One knows that, for every $t$, $E(\mathrm e^{\mathrm iuW_t})=\mathrm ... 3 Note that$E[X \mid \sigma(Y)]$is$\sigma(Y)$-measurable. We will prove the existience of a measurable$h$for any$\sigma(Y)$-measurable random variable$Z$. First let$Z = 1_A$be a characteristic function of some$A \in \sigma(Y)$. Then$A = Y^{-1}[B]$for some Borel set$B$. Let$h = 1_B$, then for any$\omega \in \Omega$: $$1_B\circ Y(\omega) = 1 \iff ... 3 X_n\overset{p}{\rightarrow} X\Rightarrow X_{n}\overset{d}{\rightarrow }X\Rightarrow \Phi_{n}(x)\rightarrow \Phi(x) pointwise where \Phi_n(\cdot),\ \Phi(\cdot) are the characteristic functions of X_n,X respectively. Now, Since X_n are Gaussian, \Phi_n(x)=e^{jx^T\mu_n-x^TC_nx/2} where \mu_n:=EX_n,\ C_n:=\mbox{Cov}(X_n) Then, since \exp(\cdot) is ... 3 Adding the hypothesis that the random variables are independent, if every X_k is normal \mathcal N(\mu_k,\sigma_k^2) and if the series \sum\limits_k\mu_k and \sum\limits_k\sigma_k^2 both converge then yes, the series \sum\limits_kX_k converges in distribution. Easiest approach: characteristic functions and Lévy's convergence theorem. 3 Try X \sim N(0,1) and Y=X when |X|\lt k and Y=-X when |X|\ge k for some non-negative k. Then X and Y have standard normal distributions and cov(X,Y) is a continuous increasing function of k, negative when k is close to 0 and positive when k is large. So for some k you will have cov(X,Y)=0 3 \sigma-fields, also known as \sigma-algebras, are used to model "events we can assign a probability to". We want some properties from these sort of events: If we can assign a probability to two events, then we can assign a probability that both events happen. If we can assign a probability to two events, then we can assign a probability that at least ... 3 The answer cannot be \frac 14. If there are a huge number of students, put the first into the first group (change the names if necessary) The second friend goes in this group with probability \frac 14 (here is where the huge number comes in-we imagine there are not fewer slots left in the first group). The third also goes in the first group with ... 3 I'd sooner post this as a comment — this is a question I've long meant to get better acquainted with but simply haven't had the time — but it won't fit there. First, if you (or, more likely, a computer) always pick the word with the greatest probability, then it's not a random process. But even if the process is going about in some random ... 3 Durrett's probability book appears to still be free (on author's page). Your subject is embedded in Chapter 3 Central Limit Theorems (Weak Convergence, Characteristic Functions etc., leading to Continuity Theorem 3.3.6). Rick Durrett's Probability: Theory and Examples book Theorem 3.3.6. Levy's continuity theorem: Let \mu_n, 1\leq n \leq \infty be ... 3 Hi this is not always possible and it is the subject of one famous theorem in financial mathematics known as the Fundamental Theorem of Asset Pricing that claims that under some conditions there exist such measure change that turns (morally) semimartingales into local martingales. In a series of articles by Delbean and Schachermayer (available on their ... 3 Note that this is a "small"problem. Since everything is a multiple of 25, we might as well assume that we start with 4 gold coins, and we bet until we either have 6 gold coins or none. Let random variable X be the number of bets until the game is over. It looks as if we are only being asked about E(X). Let e_1 be the expected further length ... 3 The first one certainly comes nowhere near implying the second. For example, suppose X is uniformly distributed on (0,1) and Y is 1 or 0 with probabilities X and 1-X, and Z is 1 or 0 with those same probabilities, and they're conditionally independent given X. Then \mathbb E(Y\mid X)=\mathbb E(Z\mid X)=X, so those are certainly not ... 3 Take a look at this family of sets:$$\mathcal A=\left\{\left(\frac1n, 1-\frac1n\right), n\in\mathbb N\right\}$$If you take any two elements A,B from \mathcal A, then A\cup B is in \mathcal A. However, the union of all elements of \mathcal A is not an element of \mathcal A. Let A_n = \left(\frac1n, 1-\frac1n\right). In a way, you can ... 3 A random variable just a fancy name for a measurable function on a probability space (\Omega, \cal{F}, P). When we write \{X \in S\}, this is just notational shorthand for \{ \omega \in \Omega : X(\omega) \in S\}. If you don't see this, I recommend you read a basic measure theoretic probability book, like Durrett's Probability: Theory and Examples, ... 2 You can indeed prove by induction, that in an algebra \mathcal A you have$$ \bigcup_{i=1}^n A_i \in \mathcal A $$for all n\in\mathbb N and A_i\in\mathcal A. So induction gives you that every finite union, no matter how many sets involved, is again an element of the algebra. Still \bigcup_{i=1}^\infty A_i is not a union of this kind! 2 That Y={\rm E}[X\mid Y] and X={\rm E}[Y\mid X] means in particular that$$ {\rm E}[Y;A]={\rm E}[X;A], \quad A\in\sigma(Y),\quad\text{and}\quad{\rm E}[Y;A]={\rm E}[X;A], \quad A\in\sigma(X). $$Since \mathcal{E}=\{C\subseteq\Omega\mid {\rm E}[Y;C]={\rm E}[X;C]\} is a sigma-algebra (why?) containing \sigma(X) and \sigma(Y) we conclude that$$ ... 2 The usual answer is that of course measure theory is not only provides the right language for rigorous statements, but allows achieve progress not possible without it. The only place I found a different point of view is a remarkable book by Edwin Jaynes, Probability Theory: The Logic of Science, which is real pleasure to read. Here is an exctract from ... 2 The converse is not true. The Cauchy random variable is finite a.s. but it does not have defined its mean. Same with the Pareto distribution for$\alpha \leq 1$where the mean is infinite. The property you mention comes from measure theory. If a function$f > 0$is not finite a.e. then the integral of$f$is already infinite over$\{f=\infty\}$. For ... 2 I would do it using approximations. First suppose that$g \in C_c(\mathbb R^n)$. Construct a sequence of measures$\mu_k$that are absolutely continuous w.r.t. Lebesgue measure (or alternatively are finite linear combinations of dirac delta functions) such that$\mu_k$converges weakly to$\mu$. Then for each$x \in \mathbb R^n$, we have$\mu_k*g(x) \to ...

2

There are $\binom{52}{13}$ bridge hands. We assume they are equally likely. (Since real shuffling is not an reliable randomizer, this is not necessarily a realistic assumption.) Now we count the "favourables," the hands that have a King and Ace of the same kind. Consider first the hands that have both the $\heartsuit$ King and the $\heartsuit$ Ace. These ...

2

(Assuming you have 8 identical balls and 6 boxes): Before checking the below answer check the link I gave in the comments above. Also this is the same thing I tried to point in your last question:Two probability questions. A:= B1 is not empty. B:= B1 and B2 are not both empty. Size of sample space = $\binom{13}{5}$ \$P(B1\ is\ empty) = ...

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