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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? ...

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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 ... 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 $$\binom{6}{3}\cdot p^{3}\cdot (1-p)^{3}$$ 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 Let$Y$,$X_1$and$X_2$be independent random variables with uniform distribution on$[0,1]$. Now let $$Z=X_1+X_2+Y\mod 1.$$ Then$X$and$Y$are independent conditionally on$X_1$and conditionally on$X_2$but clearly not conditional on$X_1$and$X_2$. So the second condition does not imply the first one. 2 A sequence$X_1,X_2,\ldots$of random variables is said to converge in distribution, or converge weakly, or converge in law to a random variable$X$if $$\lim_{n\to\infty}F_n(x)=F(x)$$ for every number$x\in\mathbb R$at which$F$is continuous, where$F_n(x)=\mathbb P(X_n\le x)$and$F(x)=\mathbb P(X\le x)$. Thus, we need to show that$F(x)$is ... 2 Counterexample Let $$S_n := \sum_{j=1}^n Y_j, \qquad n \in \mathbb{N} \tag{1}$$ a simple random walk on$\mathbb{Z}$, i.e.$Y_j \sim \frac{1}{2} (\delta_1+\delta_{-1})$independent identically distributed random variables. By Stirling's formula, we have $$\mathbb{P}(S_{2n}=0) = 2^{-2n} {2n \choose n} \sim \frac{1}{\sqrt{\pi n}} \qquad \text{and} \qquad ... 2 Hint: We can say :$$\omega \in C \Leftrightarrow (\forall m \in \Bbb N )(\exists p \in \Bbb N)( \forall k \geq p) \quad \left|\sum_{i=k+1}^{+\infty} X_k(\omega)\right| < \frac 1{m+1} $$That gives:$$C=\bigcap_{m \in \Bbb N}\bigcup_{p \in \Bbb N}\bigcap_{k \geq p} Y_{m,p,k}$$Where :$$Y_{m,p,k}=\left\{\omega \in \Omega / \left| \sum_{i=k+1}^{+\infty} ... 2 If$(X_i)$is i.i.d.,$(Y_i)$is i.i.d. and$(X_i)$and$(Y_i)$are independent, this follows from the central limit theorem applied to the i.i.d. sequence$(Z_i)defined by $$Z_i=X_i-Y_i-E(X_1)+E(Y_1).$$ To wit, considering the events $$A_N=[S_2-S_1\geqslant E(S_2)-E(S_1)],$$ one gets $$... 2 The question has been answered, so we give a different derivation. If p\ne 0, our series clearly converges, so the expectation exists. Call it a. On the first trial, either we have a success, in which case the expectation is 0, or we have a failure. In that case, we have wasted a trial, and the expectation is 1+a. It follows that$$a=(1-p)(1+a).$$... 2 Recall that$$\mathbb{E}e^{c B_t} = e^{\frac{1}{2} c^2 t} \tag{1}$$as B_t is Gaussian with mean 0 and variance t. In particular, we see that$$M_t := 4^{B_t} = \exp \bigg( B_t \cdot \log 4 \bigg)$$is not a martingale since$$\mathbb{E}M_t \stackrel{(1)}{=} \exp \left( \frac{1}{2} (\log 4)^2 \cdot t \right)$$is not constant. In fact, by the ... 2 This is saying that the average number of coin flips if you flip n fair coins each with heads probability p is np (you are calculating the mean of a Binomial(n,p) distribution). You can use linearity of expectation - if Y = \sum_i X_i where X_i is Bernoulli(p) and there are n such X_i, then Y is Binomial(n,p) and has that distribution. ... 2 You don't need to use induction or take derivatives; just note that k\binom{n\vphantom{1}}{k}=n\binom{n-1}{k-1}$$ \begin{align} \sum_{k=0}^nk\binom{n}{k}p^k(1-p)^{n-k} &=\sum_{k=0}^nn\binom{n-1}{k-1}p^k(1-p)^{n-k}\\ &=np\sum_{k=0}^n\binom{n-1}{k-1}p^{k-1}(1-p)^{n-k}\\ &=np\,\Big(p+(1-p)\Big)^{n-1}\\[12pt] &=np \end{align} $$2 Exercise: Let H:\mathbb R\to\mathbb R denote a non-decreasing function. Show that H is continuous from the right at a if and only if H(x_n)\to H(a) for at least one sequence (x_n) such that x_n\geqslant a for every n and x_n\to a when n\to\infty, if and only if H(a+1/n)\to H(a) when n\to\infty. Recall that H being continuous from ... 2 I would do it using approximations. First suppose that g \in C_c(\mathbb R^n). Construct a sequence of measures \mu_k that are absolutely continuous w.r.t. Lebesgue measure (or alternatively are finite linear combinations of dirac delta functions) such that \mu_k converges weakly to \mu. Then for each x \in \mathbb R^n, we have \mu_k*g(x) \to ... 2 If X_n\to 0 in distribution, then any \varepsilon does the job. The converse is harder. Here it's the proof of Levy's continuity theorem which will be used. Denoting by \varphi_n the characteristic function of X_n and \mu_n its distribution, we indeed have the equality ... 2 When you know evey moment of a bounded random variable you know its law: a random variable is caracterized by its Fourier transform E\exp iu X . In the case of a bounded X, you have$$ E\exp iu X = E\sum_{n=0}^\infty \frac{(iu X)^n}{n!} =\sum_{n=0}^\infty EX^n\frac{(iu )^n}{n!} $$You can swap E and \sum thanks to the Fubini theorem, because X is ... 2 Let R denote a property each \omega in \Omega may have or not, and A=\{\omega\in\Omega\mid R(\omega)\}, in your case R(\omega) is: \exists n\geqslant1,X_n(\omega)=0. Then, P(R) is simply P(A), that is,$$ P(\exists n\geqslant1,X_n=0)=P(\{\omega\in\Omega\mid \exists n\geqslant1,X_n(\omega)=0\}). $$By the way, this is exactly for the same ... 2 Let (X_t)_{t \geq 0} be a non-negative solution of the SDE$$X_t - x = 3t + 2 \int_0^t \sqrt{X_s} \, dB_s \tag{1}$$for x \geq 0. Applying Itô's formula to f(y) = \frac{1}{\sqrt{y}}, we find$$\frac{1}{\sqrt{X_t}} - \frac{1}{\sqrt{x}} = - \int_0^t \frac{1}{X_s} \, dB_s.$$For \tau_{a,b} := \inf\{t \geq 0; X_t \notin (a,b)\}, 0<a<b, this ... 2 This is a kind of "uniform ergodic theorem" and extends naturally the case X_k=X for each k. Notice that X_k=X_k-X+X and by Birkhoff's ergodic theorem,$$\frac 1n\sum_{k=0}^{n-1}X\circ T^k\to \mathbb E[X\mid\mathcal I]\quad\mbox{ a.s.},$$where \mathcal I denotes the \sigma-algebra of invariant sets, that is, \mathcal I=\{A, T^{-1}(A)=A\}. If ... 2 In order to make the inner product well-defined, we talk about L^2(\Omega,\mathcal F,\mu), where (\Omega,\mathcal F,\mu) is the underlying probability space. But we then extend condition expectation to integrable random variables. We use a projection over the closed subspace L^2(\Omega,\mathcal N,\mu), that is, the vector subspace which consists of ... 1 Rewrite this as z_n=z_{n-1}y_n where z_n=f(x_n) and y_n=1+\alpha/m+\beta\epsilon_n/\sqrt{m}, thus, (y_n) is i.i.d. such that E(y_1)=\gamma with \gamma=1+\alpha/m and \mathrm{var}(y_1)=\beta^2/m. If x_0 is independent of (\epsilon_n), one gets E(z_n)=E(z_{n-1})E(y_n), that is,$$E(f(x_n))=\gamma^nE(f(x_0)). $$Likewise, ... 1 I don't understand the fetish for calculus. There should be a way without it (in fact, personally I prefer avoiding it if necessary): Order the data as follows: x_1 < x_2 \ldots <x_{n}. Let's suppose that x \in [x_i,x_{i+1}]. Then we have the following the$$\|(x_1, \ldots x_n) - (x,x\ldots x) \|_1 = \sum_{j \leq i}(x-x_j) + \sum_{j>i} ... 1 If people vote reflecting their preferences (i.e. voting for their first preference candidate) then somebody who gets over 50% of votes would be the Condorcet candidate. There are other issues: in particular simple plurality systems may discourage some voters from voting for their first preference candidate, and if this happens, then somebody who gets ... 1 Firstly separate the states in communication classes. There are two communication classes which are determined as follows. Start from state0$. Which states can you visit and then return to state$0$? You can make following transitions $$0\xrightarrow{0.5} 2 \xrightarrow{0.9} 0$$ So$0$and$2$communicate, which means that they belong to the same ... 1 There are some common (and very useful) results concerning recurrence and transitivity of states which you might know and/or could use: All states in an irreducible set are either all recurrent or all transient. State$i$is transient if and only if $$\sum_{n=1}^{\infty}p_{ii}^{(n)} < \infty$$ State$i$is recurrent if and ... 1 There is a simple algorithm: let$P$the transition matrix. let$A=I + P$,$j = 1$replace each non$0$entry by$1$replace$A$by$A^2$,$j$by$2j$Go back to$2.$until$j\ge n$Then the graph is connected iff every entry of$A$is$1$. At each iteration, at the end of step$4.$the non zero entries of$A$are the state you can go to with$\le j$... 1 I realize what I did in the other answer was maybe overcomplicated. Define$D_l:=\sup_{k\geqslant l}|X_k-X|$(which is integrable for each$l$), and notice that by Birkhoff's ergodic theorem, $$\frac 1n\sum_{j=0}^{n-1}D_l\circ T^j\to\mathbb E[D_l\mid\mathcal I]\quad \mbox{a.e.}$$ with the same notations as in the other answer. Since$$\frac ... 1 It seems Doob-Dynkin theorem is the answer to your question. Let$X$be a real valued random variable,$\sigma(X):=\{X^{-1}(B), B\in\mathcal B(\mathbb R)\}$. The random variable$Y$is$\sigma(X)$-measurable if and only if there exists$f\colon\mathbb R\to\mathbb R$Borel measurable such that$Y=f(X)$. But there are restrictions on$f\$ in order to ...

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The question, in my opinion, extends to the debate "Bayesian vs classical (or frequentist) statistics". According to Wikipedia there are two major differences in the frequentist and Bayesian approaches to inference: In a frequentist approach to inference, unknown parameters are often, but not always, treated as having fixed but unknown values that are not ...

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