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## Hot answers tagged probability-theory

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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