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8

I think that's a misunderstanding. The passage you quote is a bit informally phrased. By "the mean time to failure of some disk", they don't mean the mean time to failure of each individual disk but the mean time it takes until any one of the $100$ disks fails. If the failures form a Poisson process, then having $100$ disks instead of $1$ disk will increase ...


4

Compare it to lottery. If you own one ticket, the probability to win is very low. If you own 100 tickets it's much more likely that you win (something). Likewise, the probability that one disk fails is quite low but the probability that any of the 100 disks fails is higher, and therefore the mean time to failure is lower.


4

$S_{n}=n\bar{X}_{n}$ where $\bar{X}_{n}:=\frac{1}{n}\left(X_{1}+\cdots+X_{n}\right)$ and $P\left(\bar{X}_{n}\to\mu\right)=1$ according to the strong law of large numbers. If $\bar{X}_{n}\left(\omega\right)\to\mu>0$ then $S_n(\omega)=n\bar{X}_{n}\left(\omega\right)\rightarrow+\infty$. So: $$\left\{ \bar{X}_{n}\to\mu\right\} \subseteq\left\{ ...


4

Note that $Y=|X|$ is a random variable that only takes on non-negative values. So the lemma holds for the random variable $Y$. Because $\lim_{y\to\infty}y\Pr(Y\gt y)=0$, there is a $B\ge 1$ such that if $y\ge B$ then $\Pr(Y\gt y)\le \frac{1}{y}$. Let $p=1-\epsilon$. By the lemma, $$E(Y^{1-\epsilon})=\int_0^B (1-\epsilon)y^{-\epsilon}\Pr(Y\gt y)\,dy+ ...


4

This is a very valid question, and one that I didn't really touch until my second time learning undergraduate probability. I am by no means an expert; this is merely the intuition I've developed. $X$ is a function. What they don't tell you in undergraduate-level probability is that $$X: \Omega \to \mathbb{R}$$ is a "random variable," where $\Omega$ is the ...


4

Change variables $s=tu$: $$\int_0^t e^{B_s} \, ds = \int_0^1 e^{B_{t u} }t \, du=(*)$$ Use scaling: $(B_{tu}:u \ge 0) \overset{\mbox{dist}}{=}\sqrt{t}( B_{u}:i \ge 0)$. Therefore $$ (*) \overset{\text{dist}}{=} \sqrt{t} \int_0^t e^{\sqrt{t} B_u} \, du $$ Take $\sqrt{t}$-th root: $$t^{1/\sqrt{t}} \left(\int_0^1 e^{\sqrt{t} B_u} \, ...


3

$$ P\{|X_n-Y_n|>\epsilon\}\le P\{|X_n-X|+|Y_n-Y|+|X-Y|>\epsilon\} \\ \le P\{|X_n-X|>\epsilon/3\}+P\{|Y_n-Y|>\epsilon/3\}+P\{|X-Y|>\epsilon/3\} $$


3

Partial answer: Simple cases: If we generate only two points, this is like fixing one as north pole and generating the other randomly. Here, the expected angular distance is "clearly" $\frac\pi2$ because points at angular distances $\theta$ are just as likely as points at distance $\pi-\theta$. The same argument holds for the $m$th closest neighbour when ...


3

Note that $X$ is a finite set. So given $x\in X$, the set $B_x= \{ S \in \mathcal{F} : x\in S\}$ is finite. Since $\mathcal{F}$ is a $\sigma$-algebra, we have $X\in B_x$ ($B_x$ is not empty) and $$A_x=\bigcap_{S\in B_x} S \in \mathcal{F}$$ Clearly $x\in A_x$, and $A_x$ is the smallest element in $\mathcal{F}$ containing $\{x\}$. Note that if $y\in X$ and ...


3

The mean time to failure of some disk is not the mean time to failure of those disks: it is the mean time before at least one of those 100 disks fails. In other words, the time to failure of some disk $T$ is the minimum of the time to failure of all disks, i.e. $$T = \min_{i \leq i \leq N} \{X_i\}.$$ A standard of model is that aging has no memory ...


3

"Theory of Stochastic Processes", Gusak et al., Springer, 2010: Problem 1.10: Prove that it is impossible to construct on the probability space $\Omega = [0, 1]$, $\mathcal{F}=\mathcal{B}([0,1])$, $\mathsf{P}=\lambda$ a family of independent identically distributed random variables $\{\xi_t, t\in[0,1]\}$ with a nondegenerate distribution. ($\lambda$ ...


3

Basically no, because this is sensitive to the distribution of $X_1$, which must be given initially. However, typically you could do it if you instead had $X_N,X_{N+1},\dots,X_{N+M}$, where $N$ and $M$ are both large. Then under generic (but not universal) circumstances, there is a unique invariant distribution, the chain converges to this distribution, and ...


3

You meant $\{X_n\}$ are uniformly integrable if $$ \lim_{M\to\infty} \sup_n E[ |X_n|\chi_{\{|X_n|>M\}}]=0$$ (e.g. https://en.wikipedia.org/wiki/Uniform_integrability) Now for your question. Answer is NO. Why ? In general the random variables are not in $L^p$. Easy example to remember $X_n = X$, where $X$ is a random variable in $L^1$ but not in ...


2

Recall the following two statements: Lemma 1 (subsequence principle): A sequence $(a_n)_{n \in \mathbb{N}}$ converges to a limit $a$ if, and only if, any subsequence of $(a_n)_{n \in \mathbb{N}}$ contains a further subsequence which converges to $a$. Lemma 2: If $Y_n \to Y$ in probability, then there exists a subsequence $(Y_{n(k)})_{k \in ...


2

The answer depends on your choice of parametrization. The way you have characterized the survival function $$S_X(x) = \Pr[X > x] = \exp(-\alpha x^{\beta - 1}), \quad x \ge 0,$$ is a little different than other parametrizations. Without seeing the source document to determine how the authors themselves chose to parametrize the distribution, it is ...


2

The answer is yes. As a concrete example of (1), suppose $X_1, X_2, X_3$ are exchangeable. We show $X_1, X_3$ are exchangeable by filling in the omitted slots with the 'everything' event: $$\begin{align} P(X_1\in A, X_3\in B)&=P(X_1\in A, X_2\in{\mathbb R}, X_3\in B)\\ &\stackrel{(*)}=P(X_3\in A, X_2\in {\mathbb R}, X_1\in B)\\ &=P(X_3\in A, ...


2

Note that for each fixed $n$, $X_i^n$ are iid. $$E(X_1^n)=\sum_{k:2^k\leq n\log n}2^k\dfrac{1}{2^k}\sim\log_2 n+\log_2\log n$$ therefore $$E(\dfrac{S_n'}{n\log_2n})\sim\dfrac{n\log_2 n+n\log_2\log n}{n\log_2 n}\to1$$ Finally, $$Var(S_n')=nVar(X_1^n)$$ due to iid-ness and $$Var(X_1^n)\leq E(X_1^n)^2=\sum_{k:2^k\leq n\log ...


2

For any $r\ge 0$ $$ \mathbb{E}(M_{\infty}-M_n)^2=\mathbb{E}(M_{\infty}-M_{n+r})^2+\mathbb{E}(M_{n+r}-M_n)^2 \\\overset{(d)}=\mathbb{E}(M_{\infty}-M_{n+r})^2+\sum_{k=n+1}^{n+r}\mathbb{E}(M_k-M_{k-1})^2. $$ Taking $r\to\infty$, the first term converges to $0$ by (f).


2

Let $a = m -1$, $b = n - m - 1$, $d = a + b = n - 2$. $\theta_{ij}$, $1 \le i \ne j \le n$ be the angular separation between point $x_i$ and $x_j$. $\ell_m$ be the expected angular separation among a pair of $m^{th}$ nearest neighbors. $\mathcal{E}_m$ be the event that $x_2$ is the $m^{th}$ nearest neighbor of $x_1$. $\mathbf{1}_m$ be the indicator ...


2

Write $\cos(a_j t) = (\exp(i a_j t) + \exp(-i a_j t))/2$, and expand the product. We get $$ 2^{-n} \sum_{x \in \{-1,1\}^n} \exp \left( i \sum_{j=1}^n x_j a_j t\right) $$ Now note that $$\dfrac{1}{2\pi} \int_0^{2\pi} \exp(ikt)\; dt = \cases{0 & if $k$ is a nonzero integer\cr 1 & if $k = 0$\cr}$$ Thus, assuming the $a_j$ are all integers, your right ...


2

No. By your definition, $U_1,U_2$ will be both uniform on the unit interval. Then, by Holder's inequality $E[U_1U_2]\leq (E[U_1^p])^{1/p}(E[U_2^{(1-1/p)^{-1}}])^{1-1/p} = \left( \frac{1}{p+1} \right)^{1/p} \left( \frac{1}{\left(1-1/p \right)^{-1} +1} \right)^{1-1/p}$ for all $p\geq 1 $. Now, plot the upper bound, and see that there exists a $p$ such that ...


2

Let $C := \bigwedge_{B \in S_1}\bar{B}$ and $D := \bigwedge_{B \in S_2}\bar{B}$. The statement now reads like this: $$\Pr(A\mid C \wedge D) =\frac{\Pr(A \wedge C \mid D)}{\Pr(C \mid D)}.$$ If $\text{Pr}_D(X)$ denotes $\text{Pr}(X \mid D)$, then the statement is equivalent to $$\text{Pr}_D(A\mid C) =\frac{\text{Pr}_D(A \wedge C)}{\text{Pr}_D(C)},$$ which ...


1

Let $D:=\{(t,\omega):f(t,B_t(\omega)\not=0\}$ and $C=\{(t,x):f(x,t)\not=0\}$. Observe that $D=\varphi^{-1}(C)$, where $\varphi(t,\omega) =(t,B_t(\omega))$ (mapping $[0,T]\times\Omega$ into $[0,T]\times\Bbb R$). Consequently, writing $\lambda$ for Lebesgue measure on $[0,T]$, $$ 0=\lambda\otimes\Bbb P(D) =\int\int 1_{C}(t,x){1\over\sqrt{2\pi ...


1

For any measurable set $A \subseteq [0,T] \times \Omega$ it holds that $$\begin{align*} \{(t,\omega); (t,B_t(\omega)) \in A\} &= \int_0^T \! \int 1_A(t,B_t) \, d\mathbb{P} \, dt \\ &= \int_0^T \! \int 1_A(t,x) \frac{1}{\sqrt{2\pi t}} e^{-x^2/2t} \, dx \, dt. \end{align*}$$ Since $e^{-x^2/2t}$ is strictly positive, this implies $$(\lambda|_{[0,T]} ...


1

Any rotationally invariant $\alpha$-stable Lévy process can be written as a subordinate process. More precisely, if $(B_t)_{t \geq 0}$ is a ($d$-dimensional) Brownian motion and $(S_t)_{t \geq 0}$ an $\frac{\alpha}{2}$-stable subordinator such that $(B_t)_{t \geq 0}$ and $(S_t)_{t \geq 0}$ are independent, then $$X_t := B_{S_t}, \qquad t \geq 0,$$ is a ...


1

By definition $$ \mathbb{E}[Y^p] = \int_0^{M} y^p f_Y(y)dy\ , $$ where $f_Y(y)$ is the pdf of $Y$, and thus equal to $f_Y(y)=\frac{d}{dy}\mathbb{P}[Y<y]=-\frac{d}{dy}\mathbb{P}[Y>y]$. Hence $$ \mathbb{E}[Y^p] = -\int_0^{M} y^p \frac{d}{dy}\mathbb{P}[Y>y]dy\ , $$ and using integration by parts $$ \mathbb{E}[Y^p] = -\left[y^p ...


1

The density $g$ of the ratio $Z=X/Y$ is given by the following integral: $$\int_{\mathbb R} |y|f(zy,y)\ \mathsf dy.\tag1 $$ Here $f(x,y)=2(x+y)\mathsf 1_{(0,y)}(x)\mathsf 1_{(x,1)}(y)$, so because $0<\frac xy < y$ we have \begin{align} f(zy,y)&=2y(1+z)\mathsf1_{(0,y)}(zy)\mathsf1_{(zy,1)}(y)\\ &=2y(1+z)\mathsf1_{(0,1)}(y)\mathsf1_{(0,1)}(z). ...


1

It is not that complicated. The conditional expectation, of random variable given an event, is defined in terms of conditional probability. $$\begin{align}\mathsf E(Z\mid Z\geq 0) ~=~& \int_\Bbb R z~f_Z(z\mid Z\geq 0)\operatorname d z\\[1ex]~=~& \int_0^1 z f_Z(z)\operatorname d z~\Big/~\int_0^1 f_Z(z)\operatorname d z \end{align}$$ Also you have ...


1

Similar to what @user1952009 was saying, we can fix a positive integer M and then define the finite linear transformation: $$(Y_t^M,\dots,Y_{t+h}^M) = (\sum_{|j|\le M}\psi_jX_{t-j},\dots, \sum_{|j|\le M}\psi_jX_{t+h-j})$$ Here we can without problem make the argument: $$(\sum_{|j|\le M}\psi_jX_{t-j},\dots, \sum_{|j|\le M}\psi_jX_{t+h-j}) \stackrel{d}{=} ...


1

Indicating the dependence of $T$ on $c$ explicitly, you have $$\{T_c=\infty\}=\{\sup_n|M_n|\le c\}$$ Therefore, $$\bigcup_{c>0,c\in\Bbb Q}\{T_c=\infty\}=\{\sup_n|M_n|<\infty\}$$



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