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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 ...


3

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 ...


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 ...


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

If $n$ random variables $X_1,\ldots,X_n$ are independent, then their joint density $f_{X_1,\ldots,X_n}(x_1,\ldots,x_n)$ is equal to $f_{X_1}(x_1)\cdots f_{X_n}(x_n)$. In this case, for every index $i\in\{1,\ldots,n\}$, and every value of $x>0$, we have $$ f_{X_i}(x) = \begin{cases} 1/\theta & \text{if }x<\theta, \\ 0 & \text{if }x>\theta. ...


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: $$P_X(A)=P(\{X\in A\})$$ for $A\in\mathcal B$. This equality can also be written as: $$\int ...


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

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

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

$(255(f(n)-1)+g(n)-1)\,\mathrm{mod}\,256+1$ also is close but not perfect, as Ross Millikan says, you need to reroll if the distribution is supposed to be perfect, afaik.


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

$P(X=5)=\dfrac{\binom{4}{4}}{\binom{100}{5}}$ $P(X=6)=\dfrac{\binom{5}{4}}{\binom{100}{5}}$ $P(X=7)=\dfrac{\binom{6}{4}}{\binom{100}{5}}$ $\dots$ 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

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

Let $\Omega = \{a,b\}$, $\mathcal F=2^{\Omega}$, $\mathbb P(\{a\})=0$, $\mathbb P(\{b\})=1$. Define $X(a)=0$, $X(b)=1$. Then $\mathbb P(X=1)=1$ so $X$ is almost surely constant, but $$X^{-1}(\{1\})=\{b\}\notin\{\varnothing,\Omega\},$$ so $X$ is not trivial $\sigma$-algebra measurable. The converse is true. Suppose $\sigma(X)=\{\varnothing,\Omega\}$ and ...


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

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, $2$ ...


2

Consider we have an arrangement of 20 balls and 5 dividers - representing six variables to contain 20 integer solutions. How many ways are there to place the dividers between the balls, when more than one divider can be placed between any two balls ? This is the permutation of 25 items composed of a group of 20 and a group of 5 indistinguishable items. ...


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

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

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


2

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 $$nP(|X| > n) < \varepsilon.$$


1

$$ \hat k >x \text{ if and only if }[X_1>x\ \&\ \cdots\ \&\ X_n>x] $$ and the probability of that is $\Big(\Pr(X_1>x)\Big)^n$. $$ \Pr(X_1>x) = \int_x^\infty \frac{\alpha k^\alpha}{u^{\alpha +1}} \, du = \left(\frac k x\right)^\alpha, $$ so $$ \Pr(\hat k>x) = \left(\frac k x\right)^{n\alpha}. $$ Thus $$ \Pr(x_1<\hat k < x_2) = ...


1

Some steps: The intersections of two algebras $\mathcal F_1$ and $\mathcal F_2$ still is an algebra. Indeed, the whole set belongs to the intersection of the algebras, as well as the complement of an element of $\mathcal F_1\cap \mathcal F_2$. Stability by finite intersections also holds. Therefore, the question is actually equivalent to the following ...


1

I am not sure this qualifies for an alternative proof anyway, define $$A_j:=\left \{ \liminf_p X_p^{ 1/p}\geqslant \liminf_p\left(\mathbb E[X_p]\right)^{1/p} +\frac 1j\right\}.$$ We have to prove that for each $j$, $\mathbb P(A_j)=0$. If not, then by Markov's inequality and Fatou's lemma, $$\liminf_p\left(\mathbb E[X_p]\right)^{1/p} +\frac 1j\leqslant ...


1

Show that everything, except perhaps $p_n$, is a multiple of $p_{n-1}$


1

Let $p_{n-1} \gt p_n$, then $k_n \gt k_{n-1}$. Write sum as $$p_n = 1 - \sum_{i=1}^{n-1}{p_i}$$ Multiply both parts by $2^{k_n}$. Then left part is $1$(odd), right ... You can finish now.


1

Yes, this is correct. Let $(\Omega, \beta, \mathbb P)$ be a probability space and $\beta_1\subset\beta$ a $\sigma$-algebra. If $X:\Omega\to\mathbb R$ is $\beta_1$-measurable, then for each Borel set $B$, $X^{-1}(B)\in\beta_1\subset\beta$, so $X$ is $\beta$ measurable. Clearly the converse is not true (just take any non-degenerate random variable and ...


1

If $I$ is uncountable and $A_i \in \mathcal{F}_t$, then it does in general not follow that $\bigcap_{i \in I} A_i \in \mathcal{F}_t$. We only know that $\mathcal{F}_t$ is stable under countable intersections. Recall the following lemma: Let $g: [0,t] \to \mathbb{R}$ be a continuous function. Then $$\max_{s \in [0,t]} g(s) = \max_{s \in [0,t] \cap ...



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