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A martingale is characterized by its characteristic function. apply the property with $f(x) = x$ proves that $W$ is a martingale fix $u$ and apply with $f(x) = \exp(iux)$; define $g(u, t) = \mathrm E(\exp(iuW_t)); M(t) = f(W_t) - \frac 12\int_0^t f''(W_s) ds$ and you get $$1 = M(0) = \mathrm EM(t) = g(u,t) + \frac 12 u^2\int_o^t g(u, s) ds$$ whose ...

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Back for a second try. Running a couple of simulations suggests that the normalized sum $S_n/s_n$ indeed converges, but strangely the histogram of $S_n/s_n$ seems more rectangular than bell-shaped. To understand what the limiting distribution is, calculate the characteristic function of the partial sum $S_n=X_1+\cdots+X_n$: \begin{align} E\exp(itS_n) ... 0 If P(A)>0 then P(B|A)=\frac{P(A \cap B)}{P(A)}, as you saw in elementary probability. So if X=\chi_C then\int_\Omega X(\omega) dP(\omega|A) = \frac{P(C \cap A)}{P(A)} = \frac{1}{P(A)} \int_A X dP.$$Now extend to simple functions and finally random variables (as usual). 1 If you want to show the CLT holds, you'll need a criterion other than the Lindeberg condition. Note that s_n^2< 4^{n+1}. Use this inequality to bound the Lindeberg expression away from zero:$$\sum\limits_{j=1}^{n}E[X_{j}^2\chi_{\{|X_{j}|>\epsilon s_{n}\}}]= \sum_{j=1}^n 4^jP(|X_j|>\epsilon s_n)\ge 4^nP(|X_n|>\epsilon s_n) \ge ...

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Let $$F(x):=x^{13}\cdot(\frac{(-x)^0}{3!}+\frac{(-x)^1}{2!}+\frac{(-x)^2}{1!}+\frac{(-x)^3}{0!})^{13}$$ Then $$4!^{13}\cdot\sum_{k=13}^{52} \frac{1}{k!}[x^k]F(x)=\frac{50972203946555791528902451677555189167087762981}{92024242230271040357108320801872044844750000000000} =0.000553899741\cdots$$ is the required probability.

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As I said in my comment, the prime ($'$) is being used to denote the complement of the set—in this case, $(A \cup B)'$ means not $(A \cup B)$. To provide some intuition behind thanasissdr's answer: Consider that $P(A) = P(B) = 1/2$. In some sense, both $A$ and $B$ cover half of the probability space. But we are told $P((A \cup B)') \equiv P(\text{not }(A ... 1 Some basic identities$P(A\cup B)= P(A) + P(B) - P(A\cap B)P(A-B) = P(A\cap B') = P(A) - P(A\cap B)P(A')= 1-P(A)P(A\cup B)= 1-P\big((A\cup B)'\big) = 1- 0.2 = 0.8$Then,$P(A\cap B) = P(A) + P(B) - P(A\cup B) = 0.2 $Thus,$P(A\cap B')= P(A) - P(A\cap B) = 0.5-0.2 =0.3$2 Let$A$be the event that$\int_0^\infty e^{B_t} \,dt < +\infty$. By the Kolomorgov 0-1 law,$P(A) = 0$or$1$. Now let$B$be the event that$\int_0^\infty e^{-B_t}\,dt < +\infty$. By symmetry,$P(A) = P(B)$. Moreover, on$A \cap B$we have $$\int_0^\infty (e^{B_t} + e^{-B_t}) \,dt < +\infty$$ which is absurd since$e^x + e^{-x} \ge 1$for all ... 1 If$X_i \sim \operatorname{Exponential}(\lambda)$are iid such that $$f_{X_i}(x) = \lambda e^{-\lambda x}, \quad x > 0,$$ then$T = \sum_{i=1}^n X_i \sim \operatorname{Gamma}(n,\lambda)with $$f_T(x) = \frac{\lambda^n x^{n-1} e^{-\lambda x}}{\Gamma(n)}, \quad x > 0.$$ Then it is easy to see that $$\operatorname{E}[n/T] = \int_{x=0}^\infty \frac{n}{x} ... 1 It may help to use the fact that$$\mathbb{E}\left[X\right] = \int_0^{\infty} \mathbb{P}(X \geq x) dx$$for positive continuous variables so that$$\mathbb{E}\left[\frac{n}{X_1+\ldots+X_n}\right] = \int_0^{\infty} \mathbb{P}\left(\frac{n}{X_1+\ldots+X_n} \geq x\right) dx = \int_0^{\infty} \mathbb{P}\left(X_1+\ldots+X_n \leq \frac{n}{x}\right) dx$$Now use ... 1 (Not a proof, but here's some intuition at least) Standard Brownian motion has mean 0 and variance t for 0 \le t \lt \infty Thus, on average the integral becomes...$$\int_0^{\infty} e^0 \ ds=\int_0^{\infty} 1 \ ds$$Which clearly diverges. What you should aim for proving is that Brownian motion crosses between negative and positive values in such a ... 0 The number of partitions of 17 into up to three parts is 33 (see for example A001399), so the number of partitions of 20 into exactly three positive parts is 33, so the number of partitions of 23 into exactly three distinct positive parts is 33, so the number of partitions of 20 into exactly three distinct non-negative positive parts is ... 2 You have three distinct boxes, and want to count ways to put 20 indistinct balls into them. The total count of solutions is, as you have calculated:$$\frac{22!}{2!20!} = 231Now, to generate forbidden solutions you can choose 2 boxes, then put the same number of balls n into each of them, and the remainder into the other. Since the number n can vary ... 0 No, the joint probability distribution does not have to be stationary w.r.t. time. Just imagine a precessing top with no friction. The probability distribution for l^2 is just a delta peaked at some particular value. But the joint probability distribution of (l_1,l_2,l_3) is a delta peak that moves around in the angular momentum space. 1 In calculating F_Y, you did wrong. Note that X^2 and X are not independent. But X^2 \le X iff X \le 1. We have for t \in [0,1]: \begin{align*} \def\P{\mathbf P}\P(Y \le t) &= \P(X \le t)\\ &= \frac 12 t\\ \end{align*} and for t \in [1,4]: \begin{align*} \P(Y \le t) &= \P(Y \le 1) + \P(1 < Y \le t)\\ ... 1 Take k=2, t_1=1 and t_2=2. If (Y_n(1),Y_n(2)) converged in distribution to (X,X), then in particular X_n-X_{2n} would converge to 0 in distribution. But it is not always the case. For example, if X is a symmetric non-degenerated random variable and X_n=X if n is even, -X if n is odd, then taking n odd we would have 2X=0 in ... 2 Finite additivity and countable sub-additivity is equivalent to countable additivity. The proof is below. Let \mu be a measure. It is clear that if \mu is countably additive then it is finitely additive and countably sub-additive. Assume that \mu is countably sub-additive and finitely additive. Consider a collection \{A_n \}_{n=1}^\infty of ... 1 No, not quite that. Conditional Mutual Information is:\begin{align} I(X;Y\mid Z) & =D_{KL} (p(X,Y,Z)\parallel p(X\mid Z)\;p(Y\mid Z)\;p(Z)) \\[2ex] & = \sum_{z\in Z} p_{_Z}(z)\; D_{KL}(p(X,Y\mid Z=z)\parallel p(X\mid Z=z)\;p(Y\mid Z=z)) \\[3ex] & = \sum_{x\in X}\sum_{y\in Y}\sum_{z\in Z} p_{_Z}(z)p_{_{X,Y\mid Z}}(x,y\mid ... 1 Mutual information is often writtenI(X;Y)=D_{KL}(p(X,Y)||p(X)p(Y))$. Your instructor has provided a slightly generalised version which depends on some other variable,$Z$. You are free to take expectations over this. Note that if$Z$is constant, then that leads to the usual definition. 3 It can be shown that nonnegative random variables$X$and$Y$have the same distribution so long as$\mathbb{E}[X^\alpha]=\mathbb{E}[Y^\alpha]$is finite for all$\alpha\in(a,b]$, any$0\le a <b$. Setting$U=\log X$and$V=\log Y, define the functions $$f(\alpha)=\mathbb{E}[1_{\{X > 0\}}e^{\alpha U}],\ g(\alpha)=\mathbb{E}[1_{\{Y > 0\}}e^{\alpha ... 1 Let the RV X take value n^2 with probability \frac{c}{n^2} where c is a norming constant. Then$$E \left( \frac{X}{n^2+X} \right) \geq \sum_{k=n}^{\infty} \frac{c}{k^2} \frac{k^2}{n^2+k^2} \geq \sum_{k=n}^{\infty} \frac{c}{k^2} \frac{n^2}{n^2+n^2}= \frac{1}{2}\sum_{k=n}^{\infty} \frac{c}{k^2} \sim \frac{c}{2n} $$which is not summable. Note that ... 2 Three-dimensional case (n=3) The Wikipedia article on area and volume element states that the area element of the unit sphere is \sin\phi\,\mathrm d\phi\,\mathrm d\theta where I use \phi to denote inclination and \theta to denote azimuth. If you consider your fixed vector as the zenith direction, then my \phi will be the same as yours. To turn ... 1 Hints: Let \arg\min_{{b_0,b_{-0}}}E\left[(X_{n+1}-b_0-b_{-0}'X)^2\right]=(\beta_0,\beta_{-0}')'=\beta\in\mathbb{R}^{n+1}, \quad X=(X_1,\dots,X_n)'. \beta_0=E[X_{n+1}]-\beta_{-0}'E[X], \quad \beta_{-0}=Var(X)^{-1}Cov(X,X_{n+1}). Cov(X, X_{n+1}-\beta_0-\beta_{-0}'X)=0. Use normality of (X_1,\dots,X_{n+1})'. Show that ... 2 The claim holds only true if X is assumed to be real-valued, i.e. \mathbb{P}(|X|<\infty)=1. If this assumption holds true, then the continuity of the measure \mathbb{P} implies$$\mathbb{P}(|X| \geq r) \leq \deltafor r>0 sufficiently large where \delta is chosen as in the assumption for \epsilon := 1. Hence, by assumption, ... 1 It is false. Suppose X is uniformly distributed in the interval [0,1] and Y=1-X. Or that X\sim N(0,1) and Y=-X. Or that X\sim\mathrm{Bernoulli}(1/2) and Y=1-X. Or that X,Y\sim\text{i.i.d. }\mathrm{Exponential}(1). In each of these cases, X has the same distribution, and hence the same expected value as Y, and in each of these cases, ... 0 It is not bounded sorry for the inconvenience. Here is the proof a friend gave me (thanks a lot to him): (It can also be found here) Let consider the set W_1 = \{w = (s,w_1,\dots)| w_1 \leq n\}. Then \mathcal{E}^n_n is bigger than \displaystyle\sum_{w\in W_1}O^n_n(w)*P(w). \begin{align} \sum_{w\in W_1}O^n_n(w)*P(w) &= \sum_{k=0}^n \sum_{i=0}^k ... 1 Let P\{Z>0\}=1 and realize that \{Z>0\}=\bigcup_{n=1}^{\infty}\{Z\geq\frac1{n}\}. That implies that P\{Z\geq\frac1{n}\} converges to P\{Z>0\}=1 so that P\{Z\geq\frac1{n}\}>\frac12 for some n. Consequently \mathbb EZ\geq\frac1{n}\frac12>0. Apply this on Z=X-Y. Edit: Alternatively 1=P\left\{ Z>0\right\} ... 1 Yes. The proof for the general case is a bit messier than the other two, because as you know we usually only retain nonstrict inequalities under a limit process. The idea is that the set X>Y can be broken up into a union of sets X>Y+1/n. If any of these has positive probability, and X<Y has zero probability, then E[X]>E[Y] (why?). The way ... 0 I would say the most interesting thing is that your metric is equal to the L_1 Kantorovich/Wasserstein/transport distance. And there is a lot of research related to this thing (although most of the research is for L_2 versions). Wiki1, Wiki2 Using right version of the distance you can prove path-connectedness of the space of prob. measures with weak ... 1 \begin{align} E[B_t^3|\mathcal{F}_s]&=E[(B_t-B_s+B_s)^3|\mathcal{F}_s]\\ &=E[(B_t-B_s)^3+3(B_t-B_s)^2B_s+3(B_t-B_s)B_s^2+B_s^3|\mathcal{F}_s]\\ &=E[3(B_t-B_s)^2B_s|\mathcal{F}_s]+B_s^3\neq B_s^3\\ E[3tB_t|\mathcal{F}_s]&=E[3t(B_t-B_s+B_s)|\mathcal{F}_s]\\ &=E[3t(B_t-B_s)|\mathcal{F}_s]+E[3tB_s|\mathcal{F}_s]\\ &=3tB_s \neq 3sB_s ... 0 Let \Omega_X be the set of full measure on which X_n \rightarrow X. Define \Omega_Y likewise. Note that \Omega_0:=\Omega_X \cap \Omega_Y has full measure as well. On \Omega_0, X_n \rightarrow X AND X_n \rightarrow Y. Hence X=Y a.s due to uniqueness of limit. 1 Suppose that X_n\to X almost surely as n\to\infty and X_n\to Y almost surely as n\to\infty. Then there exists \Omega'\subset\Omega such that \Pr(\Omega')=1 and for each \omega\in\Omega' |X_n(\omega)-X(\omega)|\to0 $$as n\to\infty. Similarly, there exists \Omega''\subset\Omega such that \Pr(\Omega'')=1 and for each \omega\in\Omega'' ... 1 first you can get b simply be computing the time zero expectation:$$\mathbb{E}(e^{5B_t})= e^{0.5 \times 25 \times t}.$$So b = 12.5. With this value, your final equation$$\exp\{\frac{25(t-s)}{2}+5B_s-bt\} = \exp\{5B_s-bs\}holds and we are done. 2 I believe that \{\tau\leq 1\} is a shorthand for \begin{align*} \{\omega\in\Omega\,|\,\tau(\omega)\leq 1\}. \end{align*} To see that this is consistent with the statement that \{\tau\leq 1\}=\{1\}, note that \begin{align*} \tau(1)=&\,\inf\{t\geq 0\,|\,\max\{t-1,0\}>0\}=\inf\{(1,\infty)\}=1,\\ \tau(2)=&\,\inf\{t\geq ... 3 Hint: Condition on the result of the coin-flip. 0 The formula for a 95% confidence interval with a lower bound will be p - z^*\sqrt{p(1-p)/n}, where p = .45, n = 1060, and z^* = 1.645 cuts 95% from the upper tail of a standard normal distribution. This gives 0.5148181 \approx 51.5\%. The interpretation is we have 95% confidence that more than 51.5% of individuals in the population (from which ... 0 If E(X) is finite, the inequality |e^{ihx}-1|\le |hx| gets you uniform continuity right away:|\varphi(t+h)-\varphi(t)|\le\int|hx|dF_X(x)=|h|E(|X|)\;.$$If X is not integrable, you've already found an upper bound that is free of t, so it suffices to show that$$\lim_{h\to0}\int|e^{ihx}-1|dF_X(x)=0\;,\tag1$$since (1) then implies that for every ... 5 Step I: Pick 6 balls and weigh them 3 on each side. Step II: 1. If the balls balance each other, then weigh the remaining two balls and the heavier one will tilt the balance on its side. 2. If the balls do not balance each other, then the heavier ball should be one of the three balls on the side the balance tilts. From these three balls, ... 0 heropup -> Don't you think we have \hat{\theta}\sim InvGamma(n,n\theta^{-1}) instead of \hat{\theta}\sim InvGamma(n,n\theta) ? 1 The assertion is not true. A counterexample: Pick a random variable Y and define Y_n:=Y and X_n:=Y+1/n. Then X_n converges in distribution to Y, and obviously Y_n converges in distribution to Y. But X_n and Y_n are equal nowhere. 2 Denote by$$p(t,x) := \frac{1}{\sqrt{2\pi t}} \exp \left(- \frac{x^2}{2t} \right), \qquad x \in \mathbb{R},$$the density of the normal distribution with mean 0 and variance t.As you already noted, this function solves the heat kernel equation, i.e.$$\frac{\partial}{\partial t} p(t,x) = \frac{1}{2} \frac{\partial^2}{\partial x^2} p(t,x).$$For f ... 1 For concreteness, let \pi be the random permutation such that X_t = R_{\pi(t)}. Then we have$$ S = \sum_{t=1}^N a_t X_t = \sum_{t=1}^N a_t R_{\pi(t)} = \sum_{t=1}^N a_{\pi^{-1}(t)} R_t. $$It remains to show that for each \pi, the vector (a_{\pi^{-1}(1)},\ldots,a_{\pi^{-1}(n)}) has the same distribution as (a_1,\ldots,a_n). One way to show this is ... 2 I claim that$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;E\big[XE[Y|\mathcal{G}]\big]=E\big[E[X|\mathcal{G}]\cdot E[Y|\mathcal{G}] \big]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(*)$$If we can prove this then we will be finished, because the right-hand side of equation (*) is symmetric in X and Y, and so the ... 0 The probability of exactly 6 dead people at the end of the month is:$$\binom{10}{6}\times(0.6)^6\times(1-0.6)^{10-6}\approx0.25$$More generally, the probability of exactly n dead people at the end of the month is:$$\binom{10}{n}\times(0.6)^n\times(1-0.6)^{10-n}$$1 It follows from the Cauchy-Schwarz inequality that X \cdot Y \in L^1. Therefore, (iii) is a direct consequence of the dominated convergence theorem. 1 It is not certain that 6 people will be dead and 4 alive. That is the most likely outcome, but, you could imagine a very unlucky month where each person died. The likelihood for each number of deaths is plotted below. 0 Hint: if you had ten coins each of which came up heads with probability 60% and you flipped them would you always see exactly 6 heads and 4 tails? If the coins were fair (50-50) would you always see just 5 heads and 5 tails? 0 Assume X and Y are independent. Let's redefine Y and X to \tilde Y and \tilde X so that$$ E( \tilde Y^n )=E(Y ^n g(Y))/E(g(Y))\\ E( \tilde X^n )=E(X ^n f(X))/E(f(X))\\ $$Since g(\ )/E(g(Y)) and f(\ )/E(f(X)) are densities with respect to the distributions of Y and X respectively, this is possible. Then$$ E((\tilde Y-\tilde ... 0 I've found an answer for my question: Theorem 3.4 of Pertti Mattila's Geometry of sets and measures in Euclidean spaces: Fractals and rectifiability. 0 IfA$has an inverse and if$Y=A X$then$X=A^{-1}Y$. So for a (measurable) set$B$in$R^n$$$Q_Y(B)=P(Y\in B)=P(X\in A^{-1}B)=Q_X(A^{-1}B).$$ So, the relationship between the distributions is simple. If the pdf's exist then the special formula can be derived based on the dollowing integral:$\$Q_Y(B)=\iint\cdots ...

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