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For simplicity, assume the probability measure can be represented by a density $p(u)$. Then we want to find $g(x)$ to solve: $f(r) = \int_{-\infty}^{\infty} g(\alpha r + u)p(u)du$ (for all $r \in \mathbb{R}$) This reminds me of a convolution. So define: $y(x)=g(-x)$, with Laplace transform $Y(s)=\int_{-\infty}^{\infty} y(x)e^{-sx}dx = G(-s)$. ...

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The question is closely related to Dual space of the space of finite measures, except that one asked "what" is the dual space, and you have a candidate for it. As the comment by t.b. (supported by a quote from Dunford and Schwartz) indicates, there is no satisfactory description of the dual, which sort of implies the negative answer to your question. But ...

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That is a good counterexample. In the exercise there is missing an additional condition on $f \in [0,1]^X$, namely $\sum_{\omega \in X} f(\omega) = 1$. Then everything works fine.

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If you mean the pointwise limit inferior, then in this case it is even simpler than that: $\lim_{n\rightarrow\infty} X_n = 0$ pointwise, which one proves as you did: let $\omega$ be given and pick $N > \omega + 2$. Then, for all $n \geq N$, $\omega < \lfloor N \rfloor$ and so $X_n = 1_{[n,n+1]}(\omega) = 0$. Since this holds for all sufficiently large ...

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If $(X_n)_{n\in\mathbb N}$ is a sequence of random variables with nondegenerate normal distributions, then $P(X_n\in\mathbb Z)=0$ for every $n$, hence $$P(\exists n\in\mathbb N,\,X_n\in\mathbb Z)\leqslant\sum_{n=1}^\infty P(X_n\in\mathbb Z)=0,$$ that is, $$P(\exists n\in\mathbb N,\,X_n\in\mathbb Z)=0.$$ This does not assume independence, only that the ...

2

Note that $$(1-p_n)^{\lfloor x/p_n \rfloor} = (1-p_n)^{x/p_n} (1-p_n)^{\lfloor x/p_n \rfloor - x/p_n}$$ For the first factor, we use the fact that $(1-\epsilon)^{x/\epsilon} \to e^{-x}$ as $\epsilon \to 0$. This is a standard fact from calculus and there are many ways to check it. Note that we can write $(1-\epsilon)^{x/\epsilon} = ... 2 Total variation norm is a norm on measures whereas all the other types of convergence/norms/topologies that you mentioned are on spaces of functions. Nonetheless functions can be measures too. If you have a sequence of measures$\mu_n$and say all of them are absolutely continuous with respect to a third measure$\lambda$, then the measures converge in TV ... 2 For every$s\lt t$,$E(B_s(B_t-B_s))=E(B_s)E(B_t-B_s)=0$because$B_s$and$B_t-B_s$are independent and$B_t-B_s$is centered. A consequence is that$E(B_t(B_t-B_s))=E(B_t(B_t-B_s))-E(B_s(B_t-B_s))=E((B_t-B_s)^2)$. Finally,$B_t-B_s$is centered normal with variance$t-s$hence$E((B_t-B_s)^2)=t-s$. The hypothesis that one starts from$B_0=0$is not ... 2 The core reason is that $$\color{red}{\sup_\color{black}{n}}\color{green}{\Pr}(|X_n-X|\geqslant\varepsilon)\qquad\text{and}\qquad \color{green}{\Pr}(\color{red}{\sup_\color{black}{n}}|X_n-X|\geqslant\varepsilon)$$ have little in common. In general, the latter is (much) larger than the former. For instance, if$\Pr(X_n=1)=1-\Pr(X_n=0)=1/n$for every ... 2 The condition means that, for every$n$, there exists some measurable set$A_n$such that $$\tau\leqslant n\iff(X_1,\ldots,X_n)\in A_n.$$ Thus, every positive stopping time$\tau$can be rewritten as $$\tau=\inf\{n\geqslant1\mid(X_1,\ldots,X_n)\in A_n\},$$ for some well-chosen sequence$(A_n)$. 2 Let$Y_k = X_kX_{k+1}$where the$X_k$are iid standard normal random variables. Then $$E(Y_k) = E(X_k)E(X_{k+1})=0, \\ var(Y_k) = E(Y_k^2) = E(X_k^2)E(X_{k+1}^2)=1.$$ Also $$var\left(n^{-1} \sum_{k=1}^{n-1}Y_k\right) = E\left(n^{-2} \sum_{j=1}^{n-1}\sum_{k=1}^{n-1}X_jX_{j+1}X_kX_{k+1} \right) \\=n^{-2} ... 2 Your definition of \tau is flawed, actually,$$\tau=\int_0^\infty\mathbf 1_{W_t\gt t}\,\mathrm dt,$$hence$$E(\tau)=\int_0^\infty P(W_t\gt t)\,\mathrm dt.$$For every t, W_t is normal centered with variance t hence$$P(W_t\gt t)=P(Z\gt\sqrt{t}),$$where Z is standard normal. By symmetry, P(Z\gt\sqrt{t})=P(Z\lt-\sqrt{t}) hence$$ ... 1 Let$(\delta_k)_{k\geqslant 1}$be a sequence decreasing to$0$. Then$A^{\delta_k}\downarrow \overline{A}$(the closure of$A$for the topology induced by$d$), hence $$\lim_{k\to +\infty}\mathbb P(A^{\delta_k})=\mathbb P(\overline{A}).$$ So the condition we want is$\mathbb P(A)=\mathbb P(\overline{A})$. 1 Whenever we consider a process$(Y_t)_t$of the form$Y_t = f(B_t)$(where$f$is a "nice" function and$(B_t)_t$a Brownian motion), then by Itô's formula $$Y_t - Y_0 = \int_0^t f'(B_s) \, dB_s + \frac{1}{2} \int_0^t f''(B_s) \, ds. \tag{1}$$ Since the stochastic integral is a martingale, this yields in particular that $$\mathbb{E}Y_t = \mathbb{E} \left( ... 1 I agree with your answer. S_3 must be closed regardless. Then, the rest of the cases for which at least one of the paths connects are covered by the following: S_2 is closed; S_0 and S_1 have any other state (probability p) S_0 and S_1 closed, S_2 open (probability p^2q) So the total probability is then p(p + p^2q) = p^2(1+pq), ... 1 The proof is in two steps: it is true for elementary functions: you just have to prove that for every Y_i mesurable with respect to F_{t_{i-1}},$$ E\left[\sum_{i=0}^{N-1} Y_i (B_{t_i} - B_{t_{i-1}})\right] =0 $$(it is quite easy). then, write an element f as an L^2 limit of such elementary processes f_n. Then,$$ E \left[\int_S^T f_n dB - ... 1 You should try Shiryaev's boook. You will find everything you need with a good level of details. Probability - Shiryaev 1 Because, if$f(t)$is a continuous function, then the existence of any$t \in [0,1]$such that$f(t) > \epsilon$implies the existence of a rational$s \in [0,1]$such that$f(s) > \epsilon$. In other words, for any$n \in \mathbb{N}, the two sets \{\omega : \exists t \in [0,1] \text{ s.t. } |f(t,\omega)| > 1/n\} = \{\omega : \exists t \in [0,1] ... 1 For Polish X: Optimal measure M^* exists: Topics in Optimal Transportation (Theorem 1.3 on page 19) and Optimal Transport (Theorem 4.1 on page 32) Equality K_d=W_d holds: Topics in Optimal Transportation (Theorem 1.14 on page 34) and Optimal Transport (Remark 6.5 on page 95). The first of two books is more analysis-oriented, and may be easier to ... 1 Let \mu be a measure and f\in L^{1}(d\mu), f\ge0, and \int fd\mu\neq0 then fd\mu is a bounded measure which can be normalized to be a probability measure. Such an f can be found if the measure is semifinite. For Lebesgue measure we can take the measure ce^{-\lvert x\rvert^{2}}dm where c is a normalizing constant. The idea here was simply to ... 1 Answer: It does not follow. I don't know what your lecturer meant or said, but you cannot draw this conclusion, as Michael's example shows. Let (X_n) be a Markov chain with state space \{0,1\} and transition matrix P=\pmatrix{0&1\cr 1&0}. The unique invariant distribution is \pi=(1/2,1/2), but \mathbb{P}_0(X_n=0)=(-1)^n does not converge ... 1 It can be written as a markov chain to obtain the probabilities. But there are 101 states and couldn't find a simple closed form like in the case of the original gambler's ruin. If the limit was 5 units instead, the markov chain is written as: \begin{align*} A &= \left(\begin{array}{rrrrrr} r & q & 0 & 0 & 0 & p \\ p & r ... 1 Enumerate X as X=\{x_n\mid n\in\mathbb N\}, fix some converging series \sum\limits_nu_n with positive entries, and define g_t on X by g_t(x_n)=tu_n for every n, for some positive t. Then g_t is positive everywhere and \sum\limits_{x\in X}g_t(x)f(x)=t\sum\limits_nu_nf(x_n) converges hence a solution is g=g_t with ... 1 The probability of S[i] and S[j] being equal to 1 is independent of the probability of |i-j|\le K if i\neq j. Such probability is 1/4 if i\neq j and 1/2 if i=j. So let's look at |i-j|. You are taking a pair (i,j) among N^2 possible pairs. There are N pairs with i=j, N-1 with |i-j|=1 and in general there are N-K pairs at ... 1 \displaystyle P\{X > T\} =\exp\left(-\int_0^T h(t)\,\,\mathrm dt\right) where h(t) is the hazard rate. So calculate the area under the hazard rate function from 0 to 6 and apply the formula to get the probability that the machine has not failed in the first 6 years. For part (b), you are asked for \displaystyle P\{X \leq 8\mid X > 6\} = ... 1 Let X \sim {\mathcal U}[-\alpha, \alpha] and let Y be a discrete r.v. taking values \pm\alpha with equal probabilities. Then X+Y \sim {\mathcal U}[-2\alpha, 2\alpha] . I believe that the sum of n \geq 3 independent r.v.'s distributed on [-\alpha, \alpha] cannot be uniform on [-n\alpha, n\alpha] , but I have no proof of this. 1 The proof is based on this standard fact from functional analysis: Theorem. Let H be a Hilbert space and let E \subset H be a linear subspace. Let E^\perp := \{ x \in H : \langle x,y \rangle = 0 \text{ for all } y \in E\}. Then E is dense in H iff E^\perp = \{0\}. So for this proof, H = L^2(\mathcal{F}_T, P) and E is the linear span ... 1 As it turns out, the claim is indeed true. The \sigma-compactness of the space R ensures that R is still \sigma-compact (and thus a Borel-set) in the completion of R. Also (which turns out to be equivalent (for separable metric spaces)) it ensures that every finite measure on R is tight. These two (for separable metric spaces equivalent) ... 1 The (1\{Y_i \leq x\})_{i=1...n} are independent Bernoulli Be(F(x)) random variable. Thus \sum_{i=1}^n 1\{Y_i \leq x\} \sim B(n,F(x)). By De Moivre-Laplace Theorem \left(\sum_{i=1}^n 1\{Y_i \leq x\} - nF(x)\right)/\sqrt{n} converges in distribution to a \mathcal{N}(0,F(x)(1-F(x)) distribution. Rearranging things give you the result. 1 Let u_n be the probability that n Bernoulli trials result in an even number of successes. This occurs if an initial failure is followed by an even number of successes, or an initial success is followed by an odd number of successes. Therefore u_0=1 and for n\geq 1u_n=q u_{n-1}+p(1-u_{n-1}).$Multiplying by$s^n$and adding over$n\$ we see that ...

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