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5

A first observation is that $|x+y|\geqslant|x-y|$ if and only if $xy\geqslant 0$, so defining $$f(x,y):=|x+y|-|x-y|,$$ the positivity of $f$ is linked to that of $xy$. We would like to find a more tractable expression for $f$. Assume that $x\gt 0$ and $ y\gt 0$. Then $f(x,y)=x+y-|x-y|=2\min\{x,y\} =\min\{|x|,|y|\}$. Since $f(-x,-y)=f(x,y)$ we get ...


4

As you have guessed, the answer is NO in general. And a single counter example is enough: take $\{X_n\}$ i.i.d. uniform $(0,1)$ then $|X_n|< n$ for all $n\geq 1$ so that $$ Y_n=X_n\implies\text{Var}{Y_n}=\text{Var}{X_n}=\frac{1}{12}\implies\sum_{n=1}^\infty\frac{\text{Var}{Y_n}}{n}=\frac{1}{12}\sum_{n=1}^\infty\frac{1}{n}=\infty. $$


4

Suppose $i$ is an integer between $1$ and $24$, and let $A_i$ be the event that you roll an $i$ at least $4$ times in ten rolls. Notice that two of the $A_i$ could occur, but not three, since that would be twelve different rolls. By the inclusion/exclusion principle, you get that $$\mathbb{P}(\cup A_i) = \sum \mathbb{P}(A_i) - \sum_{i\ \neq j}\mathbb{P}(A_i ...


4

The first part is right, but the second part is wrong (although not an absolutely terrible estimate.) Basically, when you do the trial $n$ times, the value you computed is the expected number of occurrences of HAMLET. This is a case where using the negative probability is better. The probability that you do not get HAMLET from one monkey is: ...


3

Let's compute the probability that we eventually get $x=0$. This event is independent of the vertical motion, so we can restrict to 1-d motion along the $x$-axis. Assume we start at a positive integer $x>0$. For $i \in \{0, 1, 2, \ldots\}$ define: $$ p_i = \mbox{Probability we eventually reach 0, given we start at $x$-location $i$} $$ Then $p_0=1$, ...


3

Let $\varepsilon>0$, then since $\int_{\Omega}|X|dP<\infty$, there exists an $n>0$ such that: $$nP(|X| > n) = \int_{\{\omega\in\Omega : |X(\omega)|>n\}}ndP \leq \int_{\{\omega\in\Omega : |X(\omega)|>n\}}|X(\omega)|dP < \varepsilon.$$ Therefore we have $$nP(|X| > n) < \varepsilon.$$


3

If you are guessing randomly, then yes, the probability of getting the sequence correct is just $6^{-4}$. You can think of guessing each peg one at a time. The probability of getting any peg correct is $6^{-1}$, and as an individual peg does not give you information on any other peg, the probability of getting all $4$ correct is just $6^{-1}\times ...


3

Let $U$ be standard normal. Let $V=RU$, where $R$ is a Rademacher random variable that takes values $-1$ and $1$, each with probability $1/2$. Suppose $U$ and $R$ are independent. Let $X=\frac{U+V}{2}$ and $Y=\frac{U-V}{2}$. Then $X+Y$ and $X-Y$ are each normal. But $X$ is not normal, for it takes on value $0$ with probability $\frac{1}{2}$. Remark: ...


2

$P(X=5)=\dfrac{\binom{4}{4}}{\binom{100}{5}}$ $P(X=6)=\dfrac{\binom{5}{4}}{\binom{100}{5}}$ $P(X=7)=\dfrac{\binom{6}{4}}{\binom{100}{5}}$ $\dots$ And in general: $$\forall{n\in[5,100]}:P(X=n)=\dfrac{\binom{n-1}{4}}{\binom{100}{5}}$$ In words: Take ball #$n$, and choose another $4$ balls out of balls #$1,\dots,n-1$.


2

Count how many ways $n$ can be the largest number. If you replace the balls, there are $n^5$ ways they can be $\leq n$, minus $(n-1)^5$ ways they are all less than $n$. If you don't replace the balls, there are $n-1\choose4$ ways that the largest is $n$.


2

Yes, I think your thought is correct. They want to see an inductive proof using the definition of conditional probability.


2

First, the graph must be undirected (if it's directed and there exists an edge from $a$ to $b$ but not from $b$ to $a$, then if $X_1 = a$ and $X_2 = b$, it is impossible to have $X_{n-1} = b$ and $X_n = a$. Let $d(x)$ be the out degree of node $x$: $$P(X_i = x_1, \ldots, X_n = x_n | X_0 = x_0) = \prod_{i \in [0, n)} d(x_i)^{-1}$$ $$P(X_i = x_{n-1}, ...


2

To show convergence in probability, $$ \mathbb E[X_n] = \frac{n+1}{2(n+1)\log(n+1)} - \frac{n+1}{2(n+1)\log(n+1)} = 0$$ and $$ \mathrm{Var}(X_n)=\mathbb E[X_n^2] = \frac{2(n+1)^2}{2(n+1)\log(n+1)} = \frac{n+1}{\log(n+1)}. $$ Hence $\mathbb E[S_n]=0$, and $$\begin{align*} \frac1{\varepsilon^2}\mathbb E\left[\left(\frac{S_n}n\right)^2\right] &= ...


2

We can show and use the following Lemma. Let $(X_n)_{n\geqslant 1}$ be a sequence of random variables such that $X_n\to X$ in probability and the cumulative distribution function of $X$ is continuous. Then for each $t\in\mathbf R$, the following convergence holds: $$\lim_{n\to\infty}\mathbb P(X_n\leqslant t)=\mathbb P(X\leqslant t). $$


2

Yes, the stopped compensated Poisson process is uniformly integrable: By Doob's maximal inequality, we have $$\mathbb{E} \left( \sup_{t \in [0,K]} |\bar{N}(t \wedge T)|^2 \right) \leq 4 \mathbb{E}(|\bar{N}(K \wedge T)|^2)$$ for any $K>0$. Since $(\bar{N}_t^2-\lambda t)_{t \geq 0}$ is a martingale, we obtain $$\mathbb{E} \left( \sup_{t \in [0,K]} ...


2

In order to prove measurability of $\Lambda(m)$, it suffices to show that the processes $$(t,\omega) \mapsto \max_{(t-1/m)^+ \leq s \leq t} W_s(\omega) \qquad \quad (t,\omega) \mapsto \min_{t \leq s \leq t+1/m} W_s(\omega)$$ are progressively measurable. Since any continuous (adapted) process is progressively measurable, we are done if we can show that ...


2

Let $(\Omega,\mathcal A,P)$ be a probability space and let $\mathcal B$ denote the Borel $\sigma$-algebra on $\mathbb R$. Any random variable $X:\Omega\rightarrow\mathbb R$ induces a probability $P_X$ on measurable space $(\mathbb R,\mathcal B)$. This by: $$P_X(B)=P(\{X\in B\})$$ for $B\in\mathcal B$. This $P_X$ is the distribution of $X$ and is ...


2

In general $\mathbb{E}[X \cdot Y] = \mathbb{E}[X] \cdot \mathbb{E}[Y]$, which you are trying to do, applies only if $X,Y$ are independent.


2

Not without further assumptions, I guess (your last equation holds in general only if $X$ and $\mathbf{1}_A$ are independent). Take $X$ the random variable equal to $0$ with probability $9/10$, and $10$ with probability $1/10$; and $A=\{X=0\}$. Then $$\mathbb{E}[X] = 1$$ but $$\mathbb{E}[X\mathbf{1}_A] = 0$$ while $\mathbb{P}(A) = 9/10$.


2

The problem is equivalent to Coupon collector's problem. The expected value is $$\mathbb{E}(X) = N H_N \approx N \, ln \, N$$ where $H_N$ is $N$-th harmonic number. Here $\mathbb{E}(X) \approx 2364.64$ The idea in solving this problem is calculating the expected number of people such that the number of different birthdays we have written increases ...


2

Work on a compact interval. call the function $f$. We then have a constant $L$ for that interval. If $t_0 < t_1 < ... < t_n,$ then $$ \sum |f(t_j)-f(t_{j+1})|^2 \leq L \sum |t_j-t_{j+1}|^2 $$ As the partition size goes to zero, the RHS goes to zero. In particular if the max distance between two of the points is $\delta$ then it is $$ \leq \delta ...


2

Consider $(\mu_n)_{n\geqslant 1}$ a Cauchy sequence for the metric $\rho$. Then for each $f$ (measurable) and bounded by $1$, the sequence $\left( \int_X f(x)\mathrm d\mu_n(x)\right)_{n\geqslant 1} $ is Cauchy. In particular, for each measurable subset $A$ of $X$, the sequence $(\mu_n(A))_{n\geqslant 1}$ is convergent. By the Vitali–Hahn–Saks theorem, we ...


2

$$\int_\Omega |X|\, dP < \infty \implies \int_{\{|X| > n\}} |X|\, dP \to 0$$ by the dominated convergence theorem. Since $nP(|X|>n)$ is bounded above by the last integral, we have it.


2

The expected value of the number of purchases until you have obtained the complete collection is $E_n=\sum\limits_{i=1}^{n}\frac{n}{n-i+1}=n\sum\limits_{k=1}^{n}\frac1k$. For reasons of symmetry $E(X_i)=\frac{E_n}{n}=\sum\limits_{k=1}^{n}\frac1k$ for all $i$.


2

No, in general the covariance does not converge to $0$. Just consider $([-1,1],\mathcal{B}([-1,1]))$ endowed with the Lebesgue measure and $$X_n(\omega) := Y_n(\omega) := -n 1_{[-1/n,0)}(\omega) + n 1_{(0,1/n]}(\omega), \qquad \omega \in [-1,1].$$ Since $X_n \to X:=0$ almost surely and $Y_n \to Y := 0$ almost surely, we have in particular $X_n \to 0$ and ...


2

Let us use your idea. If the sequence of consecutive heads begins at Position 1, then the next term must be T, and the last term can be chosen freely, $2$ choices. If the sequence begins at Position 2, everything is forced, we only have THHHT, $1$ choice. And if the consecutive heads start at Position 3, our sequence must be of shape XTHHH, $2$ ...


1

Set $\DeclareMathOperator \gcd{gcd}$ $$N_x := \{n \in \mathbb{N}; p^n(x,x)>0\} \qquad \quad d_x := \gcd(N_x).$$ state 4: Since $p(4,4) = \frac{1}{2}>0$, it follows that $p^n(4,4)>0$ for all $n \in \mathbb{N}$; hence, $N_4 = \mathbb{N}$ and $d_4 = \gcd(\mathbb{N})=1$, i.e. state $4$ is aperiodic. state 3: We have $$p^2(3,3) = \mathbb{P}^3(X_2 = ...


1

Since the function $m'_X$ is continuous, we can find $\delta\lt h$ such that if $|s|\lt 2\delta$, then $m'_X(s)-m'_X(0)\leqslant -\mathbb E(X)/2$ (definition of continuity with $\varepsilon:=-\mathbb E(X)/2\gt 0$). We thus have $$|s|\lt 2\delta\Rightarrow m'_X(s)\leqslant \frac{ \mathbb E(X)}2,$$ hence for $0\lt t\lt\delta$, we have $$\tag{*} ...


1

Model: Set $A:=0$, $B:=50$. Let $\xi_j \sim \frac{1}{6} \delta_{-1} + \frac{1}{3} \delta_0 + \frac{1}{2} \delta_1$, $j \geq 1$, be independent identically distributed random variables. Then $$X_n := k + \sum_{j=1}^n \xi_j$$ is the position of the person after $n$ steps if the person starts at $X_0=k$. We would like to calculate the probability that the ...


1

If you want to know whether $\lim_{n\rightarrow\infty}X_n<\infty$ then you can just pick out some $k\in\mathbb N$ and have a look at sequence $X_{k+1},X_{k+2},\dots$. The values taken by $X_1,\dots,X_k$ are simply irrelevant when it comes to this question, and this is the case for any $k\in\mathbb N$. This observation allows the conclusion that event ...



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