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8

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.

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, $$E_n = ... 5 Note that E[X+Y]=E[X]+E[Y] holds in full generality, even if X and Y are not mutually independent. Proofs of linearity of expectation do not assume independence of X and Y. Here's one for example. In other words you do not need to impose that restriction. 5 If you can argue that \Pr(A \cap B) \le \Pr(A), then you can also argue that \Pr(A \cap B) \le \Pr(B) by symmetry, giving the desired result. If you need a formal argument to show that \Pr(A \cup B) \ge \Pr(A), consider writing A \cup B as a union of the disjoint events A and B \setminus A. 4 Let X_L be L times a Bernoulli random variable (that is, it takes values 0 and L with equal probability). Its expectation is L/2 which tends to infinity with L. However, P(X_L>0)=1/2 for all L. 4 The variable X is the sample mean and the variable Y is the sample variance times (n-1). So Basu's theorem implies that they are independent. The distribution of Y is \chi^2_{n-1} as the sum of the squares of the n iid normal random variables X_i-X, (where X is used and so there are n-1 degrees of freedom instead of n). 4 Since$$\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*}$$for all n \geq m, we find by the Cauchy Schwarz inequality$$\begin{align*} \left\| ...

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 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 of $M_n$. 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

First, let us observe that the distribution is symmetric since \begin{align*} P(-X_1\le x) &=P(X_1\ge-x)\\ &=c\int_{-x}^\infty(1+t^2)^{-1}(\log(1+t^2))^{-1}dt\\ &=-c\int_{-x}^\infty(1+(-s)^2)^{-1}(\log(1+(-s)^2))^{-1}ds\\ &=c\int_{-\infty}^x(1+s^2)^{-1}(\log(1+s^2))^{-1}ds\\ &=P(X_1\le x). \end{align*} Since the distribution is symmetric, ...

3

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 ...

3

Draw a picture. We sketch how to find the cdf $F(s,t)$ in other cases. Let $T$ be the triangle where our joint density "lives." It has corners $(0,0)$, $(1,1)$, and $(0,1)$. We will be referring to it several times. Suppose first that $t\gt 1$. Draw the point $P=(s,t)$. We want the probability of falling in the region below and to the left of $P$. If $s\le ... 2 It is nondecreasing as a function of sets, i.e., if one of the events$A_i$is replaced with a superset then the probability cannot decrease. But the probability$P(A_1\cup\ldots\cup A_n)$is not even a function of the individual probabilities$P(A_i)$as numbers, let alone a nondecreasing function of them. Here is a counterexample in$[0,1]$with the ... 2 Note that the$X_n$are bounded by$1$, so the absolute convergence follows from the convergence of the sum$\sum \limits_{n = 1}^\infty \frac{2}{3^n}$. 2 In thise case, the sequence of random variables is given by$X_n = \frac{X}{n}$. You just need to apply the definition of convergence in probability resp. almost sure convergence to solve this exercise. 2 Let the sample space be$[0,1]$and let each element of the filtration be the borel$\sigma-algebra$. Let$I$be some non-measurable subset and$\tau_x=\mathbb{1}(x), x\in [0,1]$. Then$\{\tau_x<t\}$is the complement of a singleton for$x\le 1$and$[0,1]$otherwise so it belongs to each$F_t$, but$\{sup_{x\in I}\tau_x<t\}=\cap_{x\in I}\{\tau_x ...

2

If $(X,Y)$ are jointly Gaussian with mean zero, then $P(X>0,Y>0)={\arccos(-\rho)\over 2\pi}$ where $\rho$ is the correlation of $X$ and $Y$. Therefore, for Brownian motion and $0<s<t$ $$P(B_s>0, B_t>0)={\arccos(-\sqrt{s/t})\over 2\pi}.$$

2

To answer your question about infinity specifically, one fixes an infinite hypernatural $H$, and works with the collection $$\{1,2,\ldots,n-1,n,n+1,\ldots,H-1, H\}.$$ As you suggested, one can assign probability $\frac{1}{H}$ to the occurrence of each individual number in this collection. This is the basic idea behind using infinitesimals in probability. ...

2

Not in general, if I understood correctly your question. Take $E=F=\{0,1\}$, and $X,Y$ be independent Bernoulli$(\frac{1}{2})$ random variables. Take $f(x,y) = x+y\! \mod 2$. Then $f(X,Y)$ and $Y$ are independent, and no function $f^\prime(X)$ of $X$ only can be a.s. equal to $f(X,Y)$ (which is a uniform r.v. on $\{0,1\}$ independent of $X$).

2

Consider an $1>\epsilon>0$. For all $n$, $|X_n-0|=0$ with probability $1-1/n$, and $|X_n-0|=1$ with probability $1/n$. So $$Pr(|X_n-0|>\epsilon)=\frac{1}{n}$$ since $|X_n-0|>\epsilon$ only when $X_n=1$. $1/n$ converges to 0 as $n \to \infty$. If $\epsilon \geq 1$, then $Pr(|X_n-0|>\epsilon)=0$ for all $n$.

2

Start with $$E|X_n-c|\le M\cdot P[|X_n-c|>\epsilon]+\epsilon\cdot P[|X_n-c|\le\epsilon]\le M\cdot P[|X_n-c|>\epsilon]+\epsilon.$$ Because $X_n$ converges in probability to $c$, $\lim_n P[|X_n-c|>\epsilon]=0$ for each $\epsilon>0$. It follows from the inequality displayed above that $$\limsup_nE|X_n-c|\le\epsilon,$$ for each $\epsilon>0$. ...

2

Using that $P(A\cup B) = P(A)+P(B)-P(A\cap B)$, you get that $P(A\cap B) \leq P(A)$ and $P(A \cap B) \leq P(B)$ (simply because $P(A \cup B)\geq P(A)$). So you can say that $P(A\cap B) \leq \operatorname{min}\{P(A),P(B)\}$. Now, if $A=B$ then you get the equality, so that is the largest possible value.

2

$A \cap B\subseteq A$; therefore $\Pr(A\cap B)\le\Pr(A)$. In the same way, prove $\Pr(A\cap B)\le\Pr(B)$.

2

Hint: Use the summation by parts formula (also known as Abel's summation formula): $$\sum_{k=m}^n f_k (g_{k+1}-g_k) = (f_{n+1} g_{n+1}-f_m g_m) - \sum_{k=m}^n g_{k+1} (f_{k+1}-f_k).$$

2

Hint: use the Borel—Cantelli lemma. In more detail: fix any $\varepsilon > 0$, and let $A_n=A_n(\varepsilon)$ be the event $\{ \lvert Y_n\rvert > \varepsilon \}$. By Borel—Cantelli, to show that $Y_n \xrightarrow[n\to\infty]{\rm a.s.} 0$, it is sufficient to show $$\sum_{n=1}^\infty \mathbb{P} A_n < \infty.$$ In even more detail: (place your ...

2

This is not a correct MGF, because $M_X(0)$ always exists and is equal to $1$ but here $$M_X(0)=-0(1-e^0)=0\neq1$$

2

${\bf X}=(X_1,\dots, X_n)^\prime$ has a multivariate normal distribution with $\mu_{\bf X}=\mu {\bf 1}$ and $\Sigma_{\bf X}=\sigma^2 I$. Here ${\bf 1}$ is the column vector of all $1$s, while $I$ is the $n\times n$ identity matrix. Let ${\bf e}_1=(1,0,0,\dots,0)^\prime$, and let $A$ be the matrix of an orthogonal transformation that takes the vector $\bf ... 2 Partial solution here.$Y$is a quadratic form. In particular, let$\mathbf{1} \in \mathbb{R}^n$be a vector of ones, and define $$P_\mathbf{1} = \mathbf{1}\left(\mathbf{1}^{T}\mathbf{1}\right)^{-1}\mathbf{1}^{T} = \mathbf{1}\left(\dfrac{1}{n}\right)\mathbf{1}^{T}\text{.}$$ Let $$\mathbf{X}=\begin{bmatrix} X_1 \\ X_2 \\ \vdots\\ X_n \end{bmatrix}\text{.}$$ ... 2 You are given that$P(A^c\cap B)=P(A^c)P(B)=0.1$and that$P(C\cap B)=P(C)P(B)=0.2\$ which you can combine (as you almost have done by yourself) to obtain that $$P(C)=2P(A^c)$$ From this you can also deduce that $$0.2=P(B\cap C)\le P(C)=2P(A^c)\tag{1}$$ So, \begin{align}P(A^c\cup C)&=P(A^c)+P(C)-P(A^c\cap ...

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