# Tag Info

You want to prove the statement: $$\lim_{n\to\infty}\sum_{i=1}^{n}c=c \implies c=0$$ Instead, you can prove the equivalent statement: $$c\neq0 \implies \lim_{n\to\infty}\sum_{i=1}^{n}c \neq c$$ And this is rather simple, as you can use the exact trick that you were trying to avoid: $$c\neq0 \implies ... 9 Define X_n to be such that X_n is 0 with probability 1-\frac{1}{n} and n^2 with probability \frac{1}{n}. It is the case that E[X_n]=n \to \infty. But for any positive k we have \mathbb{P}(X_n > k) = \frac{1}{n} \to 0. Thus showing a counter example. 8 In the context of probability theory, you can write$$ \Pr(\varnothing) = \Pr(\varnothing \cup \varnothing) = \Pr(\varnothing)+\Pr(\varnothing). $$Then if you subtract \Pr(\varnothing) from both sides, you get$$ 0 = \Pr(\varnothing). $$Thus you don't need infinite series. That \infty\cdot0=0 does not make sense in certain broad contexts, since for ... 6 "\Rightarrow": Use$$|\mathbb{E}(X_n 1_A)-\mathbb{E}(X 1_A)| \leq \mathbb{E}(|X_n-X|).$$"\Leftarrow": Show$$\mathbb{E}(|X_n-X|) = \mathbb{E}[(X_n-X) 1_{\{X_n-X \geq 0\}})]+ \mathbb{E}[(X-X_n) 1_{\{X_n-X<0\}}],$$and conclude that$$\mathbb{E}(|X_n-X|) \leq 2 \sup_{A \in \mathcal{F}} |\mathbb{E}[(X_n-X) 1_A]|.$$6 From the definition of a (probability) measure, you know that$$P\left(\bigcup_{_{n\in\mathbb{N}}}E_n\right) = \sum_{n\in\mathbb{N}}P(E_n)\qquad (1)$$if the E_i are pairwise dijoint (E_i\cap E_j =\emptyset,\ i\neq j). Now, define$$E_1 = A_1,\ E_2 = A_2\backslash A_1,\ E_3=A_3\backslash (A_1\cup A_2), .. $$or, more formally,$$ E_n = ...
Yes, it's true. By the very definition of $\liminf$, there exists a subsequence such that $$\liminf_{n \to \infty} \mathbb{E}(X_n ) = \lim_{k \to \infty} \mathbb{E}(X_{n_k}). \tag{1}$$ Since, by assumption, $X_{n} \to X$ in probability (hence in particular $X_{n_k} \to X$ in probability), we can choose a further subsequence of $(X_{n_k})_{k \in ... 5 Neither of them holds, in general, for stochastic integrals. The trouble already starts if you consider measures which need not to be non-negative, i.e. signed measures. For a signed measure$\mu: (\Omega,\mathcal{A}) \to \mathbb{R}$we cannot expect that the triangle inequality $$\left| \int f(x) \, \mu(dx) \right| \leq \int |f(x)| \, \mu(dx) \tag{1}$$ ... 5 If you can argue that$\Pr(A \cap B) \le \Pr(A)$, then you can also argue that$\Pr(A \cap B) \le \Pr(B)$by symmetry, giving the desired result. If you need a formal argument to show that$\Pr(A \cup B) \ge \Pr(A)$, consider writing$A \cup B$as a union of the disjoint events$A$and$B \setminus A$. 5 Your reasoning is basically right but your notation is objectionable. What does "$X=x$is a constant" mean? If$x$is a constant, then it would came outside the outer expectation - and the law of total expectation would not apply. You should simply write $$E (X\, Y) =E (E(X \, Y \mid X))=E (X \, E(Y\mid X))$$ This is indeed true for any$X,Y$. And, by the ... 5 Note that$E[X+Y]=E[X]+E[Y]$holds in full generality, even if$X$and$Y$are not mutually independent. Proofs of linearity of expectation do not assume independence of$X$and$Y$. Here's one for example. In other words you do not need to impose that restriction. 4 To simplify the subscripts, I'm going to consider just the first few letters the monkey types. Add a variable$k$to all my subscripts if you want to apply the reasoning at an arbitrary point in the string. It is true that$E(X_1) = 26^{-5}$and also that$E(X_2) = 26^{-5}$. We just have to cycle through the$26^6$equally-likely possibilities for the first ... 4 Community wiki answer so the question can be marked as answered: As Alex remarked: yes, this is correct. 4 Note that $$\hat\theta_n\sim \frac 1 {1+\sum_{i=1}^{n-1}U_i}$$ for i.i.d. standard uniforms$U_i$. Now see this and this 4 Hint: Consider each of the$10$possible values for$A$For example$P(A=2)=P(X=1 \text{ and } Y=1)=0.1^2$while$P(A=3)=P(X=1 \text{ and } Y=2)+P(X=2 \text{ and } Y=1)=0.1\times 0.4+0.4 \times 0.1$4 The complement of the middle-thirds Cantor set is a countable union of (open) intervals. However, the Cantor set itself has uncountably many elements, and any two of them are separated by a point not in the Cantor set. So no countable union of intervals can produce it. Thus$B$is not closed under complement and therefore it is not a$\sigma$-algebra. 4 Let$X_L$be$L$times a Bernoulli random variable (that is, it takes values$0$and$L$with equal probability). Its expectation is$L/2$which tends to infinity with$L$. However,$P(X_L>0)=1/2$for all$L$. 4 If you require a sigma algebra to be closed under uncountable union, then the only measure on the real numbers would be the zero measure, because every subset of$\mathbb{R}$would have to be in the sigma algebra (at least assuming that the intervals were.) 4 The variable$X$is the sample mean and the variable$Y$is the sample variance times$(n-1)$. So Basu's theorem implies that they are independent. The distribution of$Y$is$\chi^2_{n-1}$as the sum of the squares of the$n$iid normal random variables$X_i-X$, (where$X$is used and so there are$n-1$degrees of freedom instead of$n). 4 Since \begin{align*} \left\| \sum_{k=1}^n (X_k-\mathbb{E}(X_k)) - \sum_{k=1}^m (X_k-\mathbb{E}(X_k)) \right\|_{L^2}^2 &= \left\| \sum_{k=m+1}^n (X_k-\mathbb{E}(X_k)) \right\|_{L^2}^2 \\ &= \sum_{k=m+1}^n \sum_{\ell=m+1}^n \text{cov}(X_k,X_{\ell}) \end{align*} for alln \geq m, we find by the Cauchy Schwarz inequality \begin{align*} \left\| ... 3 Write f(X) = f(X)^+ - f(X)^-. Then, observe that the bound f \geq -c implies f(X)^- \leq c a.s., so that \mathbb{E}[f(X)] = \mathbb{E}[f(X)^+ - f(X)^-] = \mathbb{E}[f(X)^+] - \mathbb{E}[f(X)^-] \geq \mathbb{E}[f(X)^+] - c $$and reorganizing the terms give what you want: \mathbb{E}[f(X)^+] \leq \mathbb{E}[f(X)] + c. Note that this does not ... 3 Without any additional assumptions on the Lévy process (X_t)_{t \geq 0}, a central limit theorem does not hold true. Let (X_t)_{t \geq 0} be a (one-dimensional) Lévy process with Lévy triplet (b,\sigma^2,\nu). Define$$T(x) := \nu((x,\infty)) + \nu((-\infty,-x))$$and$$U(x) := \sigma^2+2 \int_0^x y T(y) \, dyfor x>0. There is the ... 3 Because the normal curve has a bell shape, there is more area under it near the average. It appears that we are dealing with a standard normal distribution, so you want to solve for little x Using the CDF, \begin{align*} .95 &= P(X\leq x) -P(X\leq -x) \\ &=P(X\leq x)-[1-P(X\leq x)]\tag{1}\\ &=2P(X\leq x)-1\\ &=2\Phi(x)-1 \end{align*} where ... 3 Observe that\max_{1\leq m \leq n} X_m \geq \lambda \Leftrightarrow \max_{1\leq m \leq n} X_m + c \geq \lambda + c \implies \max_{1\leq m \leq n} \lvert X_m + c \rvert \geq \lambda + c\max_{1\leq m \leq n} \lvert X_m + c \rvert \geq \lambda + c \Leftrightarrow \max_{1\leq m \leq n} \lvert X_m + c \rvert^2 \geq (\lambda + c)^2$$The second relation ... 3 By the Borel-Cantelli lemma, if the series$$ \sum_{n=1}^\infty P\{|Y_n|>\varepsilon\} $$converges for each \varepsilon>0, Y_n\to0 almost surely as n\to\infty. Using Chebyshev's inequality,$$ \sum_{n=1}^\infty P\{|Y_n|>\varepsilon\} \le\frac1{\varepsilon^2}\sum_{n=1}^\infty\operatorname E|Y_n|^2$since$\operatorname EY_n=0$. By ... 3 If the distribution is discrete, there may be a problem. For example, imagine tossing a fair coin that has a$0$on one side and a$1$on the other. Then all possibilities$(0,0)$,$(0,1)$,$(1,0)$, and$(1,1)$are equally likely. When we order, the result$(0,1)$is twice as likely as either$(0,0)$or$(1,1)$. Remark: If the underlying distribution is ... 3 I think that almost everything you might want to know about this space ($\mathrm{MALG}_\lambda$, called the Lebesgue measure algebra) is in exercises (17.42)-(17.46) in Kechris' book. This metric$d$turns$\mathrm{MALG}_\lambda$into a Polish space, and the Boolean operations ($\cup$,$\cap$, and relative complement in$[0,1]$) are well defined and ... 3 If you've already proved the base case$P(A_1\cup A_2) \le P(A_1)+P(A_2), then you can do this: \begin{align} & P(A_1 \cup \cdots \cup A_n \cup A_{n+1}) \\[10pt] = {} & P( \Big( A_1 \cup \cdots \cup A_n \Big) \cup A_{n+1}) \\[10pt] \le {} & P(A_1 \cup \cdots \cup A_n) + P(A_{n+1}) & & \text{by the base case applied to these two events,} ... 3 Through the Spitzer identity, it is possible to find some kind of transform of the distribution ofM_n$. Well, not exactly. The Spitzer identity involves the expressions$M^+_n = \max_{0\le k\le n} S_k$, where$S_0 = 0$,$S_k = X_1 + \dots + X_k$,$k\ge 1$. So this translates to the positive part of expression you are interested in. But it is possible to ... 3 Call three heads and three tails when tossing six coins a success. Then the probability of success is$\frac{\binom{6}{3}}{2^6}$, which simplifies to$\frac{5}{16}$. Let$X$be the number of trials until the first success. Then$X$has geometric distribution with parameter$p=\frac{5}{16}$. It is a standard result that if$X$has geometric distribution ... 3 Note that your limit is$\lim_{n\to\infty} n \, c$. If$c \ne 0$, the sequence diverges, so the limit cannot be$c\$.