# Tag Info

10

It would generally not be true even if they were independent. For example if $X,Y,Z$ were identically and independently continuously distributed then they can come in any order with equal probability so $P(X \leq Y \leq Z) = \frac16$ but similarly $P(X \leq Y)P(Y \leq Z) = \frac12 \times \frac12 = \frac14$.

6

Let $X=Z$, and $Y$ such that $P(X \le Y) = \frac{1}{2}$ and $P(X=Y)=0$, then in order for the equality to hold we must have \begin{eqnarray} 0 &=& P(X=Y) \\ &=&P(X \le Y \le Z) \\ &=& P(X \le Y) P(Y \le X)\\ &=& P(X \le Y) \left( 1- P(X<Y) \right) \\ &=& \frac{1}{4} \end{eqnarray} So the equality does not hold in ...

5

Assuming at least that $\Bbb E|X_j|<\infty$: Clearly yes if $\sum|a_j|<\infty$; then Chebyschev says that $\sum P(|a_jX_j|>\epsilon)<\infty$ for every $\epsilon>0$. No in general. Assuming $X_j$ is not essentially bounded there exist $a_j\to0$ such that $\sum P(|a_jX_j|>1)=\infty$, so the less trivial half of Borel-Cantelli says that ...

5

Let $A_n$ be independent events with $\mathbb{P}(A_n)=1/n$, and define $X_n=1_{A_n}$. Then $X_n\to0$ in probability, but $X_n$ does not converge almost everywhere. Apply the second Borel-Cantelli lemma twice; once to the sequence $A_n$ and also to the sequence $A_n^c$, to conclude that $$P([X_n=1\mbox{ infinitely often}] \cap [X_n=0\mbox{ infinitely ... 4 Perhaps the most commonly used example is the natural filtration of Brownian motion, i.e. \mathcal{F}_t=\sigma(W_s:s \in [0,t]). This sort of thing is usually how filtrations are created. 4 Let x\leq x^\prime be two continuity points of G. If there were a subsequence with a_{n(k)}\leq 0 for all k, then a_{n(k)}x^\prime +b_{n(k)}\leq a_{n(k)}x +b_{n(k)} and hence F_{n(k)}(a_{n(k)}x^\prime +b_{n(k)}) \leq F_{n(k)}(a_{n(k)}x +b_{n(k)}). Letting k\to\infty gives G(x^\prime)\leq G(x) so that G is a non-increasing function, ... 3 Yes, it is. Here is an outline. There are quite a few details to fill in, however, so you may have to brush up your functional analysis. Lemma 1. If (\Omega,\mu) is any measure space, and H is any separable Hilbert space, let L^2(\Omega, \mu; H) be the space of Borel measurable functions f : \Omega \to H such that \|f\|^2_{L^2(\Omega,\mu;H)} ... 3 Let the event space be$$\Omega=\{(r,r),(r,w),(w,r),(w,w)\}$$the corresponding probabilities are$$\frac{12}{72},\frac{20}{72},\frac{20}{72},\frac{20}{72}.$$Then$$Pr(A\cap B)=Pr((w,r))=\frac{20}{72}$$and$$Pr(B)=Pr(\{(r,r),(w,r)\})=\frac{12}{72}+\frac{20}{72}=\frac{32}{72}$$and$$Pr(A)=Pr(\{(w,r),(w,w)\})=\frac{40}{72}$$so,$$Pr(A\mid ...

3

The assumption underlying Benford's law is that since the physical units are arbitrary, the distribution of physical constants should be invariant under multiplication by constant, which corresponds to a change of units (though no such distribution can really exist). Another way of stating this is that the distribution of the logarithm of a physical ...

3

Clearly $F'(x)=f(x)$ (wherever CDF is continuous). Substitute $F(x)=t$. Then $dt=f(x)dx$. $$\int_{-\infty}^a F(x)~f(x)~dx \rightarrow \int_{0}^{F(a)} t~dt= \frac{(F(a))^2}{2}$$

3

As I mentioned in my comment, there is an analogue of this for $\sigma$-fields but it is a lot more complicated. To slightly simplify the discussion, I'm going to pretend the only operations we care about are union and intersection, and ignore complements. In fact, because of the identities $(\bigcap A_i)^c=\bigcup A_i^c$ and $(\bigcup A_i)^c=\bigcap ... 2 Just make a small change: Instead of$\mathbb{P}(X<Y)$consider$\mathbb{P}(X+2/m \leq Y). Then, by a very similar argumentation, we find \begin{align*} \mathbb{P}\left(X+2/m \leq Y \right) &= \mathbb{P}\left(X+2/m \leq Y, |X_n-X| \leq \frac{1}{m} \right) + \mathbb{P}\left( X+2/m \leq Y, |X_n-X| > \frac{1}{m} \right) \\ &\leq ... 2 Hints: Fix x \in [a,b]. Suppose that f \in C_b^2 solves the differential equationx f'(x) + \frac{\sigma^2}{2} f''(x)=0. \tag{1}$$Show, using Itô's formula, that (f(X_t))_{t \geq 0} is a martingale. Define$$f(x) := \Phi \left( \frac{2}{\sigma} x \right)$$where$$\Phi(x) := \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x \exp \left(- \frac{z^2}{2} \right) ... 2 What is the chance there are exactlyk$boxes with a single ball? For$1\leq i\leq N$, define$A_i$to be the event that the$i$th box gets exactly one ball. These are exchangeable events whose intersections have multinomial probability $$P(A_1\cdots A_j)={M \choose 1,1,1,\dots,1,M-j}\left({1\over N}\right)^{1}\cdots \left({1\over ... 2 The interpretation here is that the balls are distinguishable as are the boxes. This gives the combinatorial species$$\mathfrak{S}_{=M}(\mathcal{E} + \mathcal{U}\mathcal{Z} + \mathfrak{P}_{\ge 2}(\mathcal{Z})).$$This translates into the bivariate generating function with z representing the number of balls and u the number of boxes with a ... 2 Define \tilde\tau_x=\inf\{t\ge 0:\tilde B(t)=x\} where \tilde B(0)=0. Using the translation invariance, symmetry, and the reflection principle for BM$$P\{\tau_0>T\}=P\{\tilde\tau_1>T\}=\frac{2}{\sqrt{\pi}}\int_{0}^{\frac{1}{\sqrt{2T}}} e^{-u^2}du$$... 2 This is a problem of "reality", not mathematics. If you have n different, indistinguishable objects and you pick one at random, than the probabilty for every object to be picked is p=\frac{1}{n}. However, you're problem arises, when you assigning the picking process to a person. A person might not pick fully at random. An observation from a magicians ... 2 I will follow the outline that I described in my first comment. Let's start with random walk. Fix \alpha \in \mathbf{R} and consider the process (M_j) defined as$$M_j = (\cosh{\alpha})^{-j}\cosh{(\alpha S_j)}$$We want to show that (M_j) is a martingale with respect to the natural filtration (\mathcal{F}_j) generated by the constituent random ... 2 See my comments under the question. What I write below will assume my guesses there are right and that a=1. Your symmetric matrix C will have one row corresponding to each unordered pair \{i,j\}. The entry in row \{i,j\} and column \{k,\ell\} will be 2 if \{i,j\}=\{k,\ell\}. Those are just the diagonal entries in the matrix. In row ... 2 First show the independence hypotheses imply that \Bbb E(e^{itY}\mid N=k)=\varphi_X(t)^k. This means that \Bbb E(e^{itY}\mid N)=\varphi_X(t)^N. Then \Bbb E(e^{itY})=\Bbb E(\Bbb E(e^{itY}\mid N))=\Bbb E(\varphi_X(t)^N), as noted by Didier in the comments. Finally$$\Bbb E(s^N)=\sum_{k\geqslant 0} s^k P(N=k)=\sum_{k\geqslant 0} s^k e^{-\lambda} ... 2 The characteristic function of$Y$is$\varphi_Y(t)=\mathbb{E}[\varphi_X(t)^N]=\mathbb{E}[\cos(t)^N]=e^{\lambda(\cos(t)-1)},$and so the characteristic function of$Y/\sqrt{\lambda}$converges, as$\lambda\to\infty, $$\varphi_{Y/\sqrt{\lambda}}(t)=e^{\lambda(\cos(t/\sqrt{\lambda})-1)}\to e^{-t^2/2}=\varphi_{N(0,1)}(t).$$ 2 I think barak manos is doubly counting some selections. Following Marconius's tip, we have \begin{align} \binom{6}{2}(7)(3) + (6)\binom{7}{2}(3) + (6)(7)\binom{3}{2} & = (15)(7)(3) + (6)(21)(3) + (6)(7)(3) \\ & = 315 + 378 + 126 = 819 \end{align} different ways to select chips to satisfy the condition. Note that there are $$\binom{6}{4} ... 2 You can look at$$\mathcal B_r=\Big\{A\subseteq [0,1]\times[0,1]\ \bigg|\ A\cap [0,r]\times[0,1]\text{ is Borel, }\land (A\cap(r,1]\times[0,1]\in\{(r,1]\times[0,1],\varnothing\})\Big\}$$For r\in[0,1]. Where Borel means the standard Borel sets of [0,1]\times[0,1]. Then for r\leq s we have \cal B_r\subseteq B_s, since whenever A\in\cal B_r, we ... 2 An example of a filtered probability space is ([0,1), \mathbf{B}_{[0,1)}, \{\mathcal{F}_t\}_{t\geq 0} , \mathbb{P}), where, for each t\in \mathbb{R}, t\geq 0, define$$ \mathcal{F}_t = \mathbf{B}_{\left[0\,,\,1-\frac{1}{t+1}\right]}=\textrm{ the Borel\sigma-algebra defined in } \left [0\,,\,1-\frac{1}{t+1} \right] $$Note that, by definition of ... 1 Counterexample: Let n=2 and \theta=1; \epsilon_1,\epsilon_2\sim_{i.i.d}\text{Bernoulli}(0.5). Take x_1=\epsilon_2-\epsilon_1, and x_2=\epsilon_1-\epsilon_2. Then$$y_1=\epsilon_2\text{; }y_2=\epsilon_1$$and$$\frac{1}{2}=P\{y_1=0,y_2=0|x_1=0,x_2=0\}\ne P\{y_1=0|x_1=0,x_2=0\}P\{y_2=0|x_1=0,x_2=0\}=\frac{1}{2}\times\frac{1}{2}$$So, y-s ... 1 Since Y^i = \theta^T X^i + E^i, by the substitution principle$$(Y^i \mid X^i =x^i) = \theta^T x^i + E^i,where x^i is a constant. Since the E^i are independent, the Y^i are conditionally independent. More explicitly, conditional pairwise independence holds because \begin{align} P(Y^i \in{A_i}, Y^j \in{A_j}\mid X^i=x^i,X^j=x^j) &= P( E^i ... 1 Is your question just about \sigma-algebras or about filtration?? If you are just looking for an example of an uncountable increasing family of \sigma-algebras, there are many examples. One of the simplest examples is: for each t\in \mathbb{R}, t\geq 0, define \mathcal{F}_t = \mathbf{B}_{[0,t+1]}\textrm{ the Borel \sigma-algebra defined in } ... 1 Extended Comment, not Answer. I do not believe all of the speculations in the Question and Comments are exactly correct and relevant. Here are some things I believe are correct, assuming you are dealing with an M/M/1 queue. The formula W = \lambda/(\mu - \lambda) = \rho/(1 - \rho), where \rho = \lambda/\mu is for the average number of customers in an ... 1 Let's model your Markov chain as follows. Let (U_n)_{n\geq 0} be i.i.d. random variables uniformly distributed on the unit disk in \mathbb{R}^2. Set Z_0 take values in \bar D, and for n\geq 0 letZ_{n+1}=Z_n+{1\over 2} \varphi(Z_n)\, U_n,$$where \varphi(z)=\inf\{\|x-z\|: x\in D^c\}. Since \begin{eqnarray*}\mathbb{E}(Z_{n+1}\mid {\cal F}_n) ... 1 Using Hoeffding's inequality$$\text{Pr}\left(\left|\bar X\right|>\epsilon\right)\le 2\exp\left(-\frac{1}{2}\epsilon^2n\right)\$

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