# Tag Info

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There are 3 ways to chose the 1st and 5th digits, 4 ways to choose the 2nd and 4th digits and 3 ways to choose the 3rd, 6th and 7th digits. Thus there are $3^2\cdot 4^2 \cdot 3^3 = 3^5\cdot 4^2 = 3888$ different cards with the special property you described. In total there are $10^7$ possible card numbers, and assuming each number is equally likely the ...

3

By the axioms of probability theory, you have $$P(A \cap B) \le P(A)$$ (since $(A \cap B) \subset A)$ and similarly $$P(A \cap B) \le P(B)$$ Adding these equations and dividing by 2 should give the desired result..

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There's a major misconception here. First, what do we really mean by $P(X=Y)=1$? Well, we mean exactly what it says: We mean that $P(E)=1$, where $E$ is the event $$E=\{s\in S\,:\,X(s)=Y(s)\}.$$ I imagine you can concoct an example where that set $E$ is not measurable. (EDIT: For example, let $S=T=\{0,1\}$, and give both $S$ and $T$ the trivial ...

1

In the title and the post, the question asked whether it is true that if $A$, $B$, and $D$ are pairwise independent, then $A\cap B$ and $B\cap D$ are also independent. (There is also a puzzling $C$ around that plays no role in the question. But we have added a remark in case $C\cap D$ is intended, and not $B\cap D$.) Not necessarily. Toss a fair coin ...

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Have a look here: S. J. Taylor, "Exact asymptotic estimates of Brownian path variation", Duke Math. J. vol. 39 (1972), pp. 219–241.

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For the first, $X-Y$ we need $x\in [0;\infty)$ and $(x-z)\in[0;1]$ where $z\in[-1;\infty)$ That's $-1\leq \max(0,z) \leq x\leq 1+z <\infty$, so the integration is: $$f_{X-Y}(z) = \mathbf 1_{z\in[-1;0)}\int\limits_0^{1+z}f_X(x)f_Y(x-z)\operatorname d x +\mathbf 1_{z\in[0;\infty)}\int\limits_z^{1+z}f_X(x)f_Y(x-z)\operatorname d x$$ For the second, $Y-X$, ...

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If you want to impose just uncorrelatedness between the variables in the sequence rather than independence, then you need to impose a stronger condition than $\sum_{i=1}^{\infty} \operatorname{Var}[X_i]/i^2 < \infty$. To be more precise, a sufficient condition in this case that will guarantee SLLN is the following: $$\sum_{i=1}^{\infty} ... 0 One convolution formula is:$$f_Y(y) = \iint\!\!\ddots\!\!\iint_{\Bbb R^{n-1}} f\left(x_1, x_2 , x_3, \ldots,x_{n-1}, y-\sum_{k=1}^{n-1} x_k\right)\operatorname d x_{n-1} \cdots \operatorname d x_1$$0 Not really answering the question, but there is a section of this paper on non-Archimedean probability http://arxiv.org/pdf/1106.1524.pdf on fair lotteries over the rationals. Some insight on the nature of your question can probably be gained by this, even though your question is about ordinary probability distributions. 2 The first equality that you don't understand is indeed a MacLaurin series expansion of the characteristic function, using the link between the moments of a random variable and the derivatives of the characteristic function. See https://en.wikipedia.org/wiki/Characteristic_function_%28probability_theory%29#Moments . So Shalop is right, the \sigma is a ... 0 Hint: From the earlier line, we see that \frac{a_n}{n} \uparrow \implies a_{kn}>ka_n Lets start at n=1:$$a_k>ka_1$$Therefore,$$a_2>a_1 \implies ka_1<a_k<\sum_{i=1}^{k}a_i$$Then,$$Pr(|X_1|\ge ka_1)\geq Pr(|X_1|\ge a_k) \geq Pr\left(|X_1|\ge \sum_{i=1}^{k}a_i\right)$$But,$$Pr\left(|X_1|\ge \sum_{i=1}^{k}a_i\right) \leq ...

2

Your intuition is correct. That does not sound, and it isn't, right. Let $x$ be the conditional probability that a hair is left at the scene given that somebody else did it. By the law of total probability, the probability that a hair is left at the scene is $$0.99 = 0.99 \times 0.1 + 0.01 \times x$$ So $x=89.1>1$, which cannot be a probability. The ...

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You have used indicators to do: \begin{align} \mathsf E[X] & = \sum_{j=1}^k \mathsf P[X_j{=}1] \\ & = \sum_{j=1}^k (1-\mathsf P[X_j{=}0]) \\ & = k (1-(1-\frac{1}{k})^n)\end{align} So continue: \begin{align}\mathsf E[X^2] & = \sum_{j=1}^k \mathsf P[X_j{=}1] + \mathop{\sum\sum}_{\substack{j\in\{1;k\}\\i\in\{1;n\}\setminus\{j\}}}\mathsf ... -1 Let's say M_n is the number of ways to pair 2n people together (n couples). Now let's count these pairings by singling out one person. There are 2n-1 ways to pair this person off. Once paired there are M_{n-1} ways to pair the remaining 2n-2 people. This leads to the recurrence relation M_n = (2n-1)M_{n-1}, \;\; M_1=1. $$Now assume that ... 0 If X and Y are independent, \Bbb{E}(Y\mid X)=\Bbb{E}(Y)=c (not Y) which is surely measurable with respect to \{\emptyset, \Omega\}. This should be obvious from elementary probability. If X and Y are independent, the expectation of Y is just the expectation of Y, it doesn't care what X does. To show that this is the right random variable ... 2 You are right, It is \frac{1}{2n-1}. It is convenient to assume that in every couple, one of the members of the couple is wearing blue, and the other is wearing red. Let the people wearing blue be labelled 1 to n, and let X_i=1 if blue-wearer i is partnered with the person he/she couples with, and 0 otherwise. We have ... -1 Fix 0<\varepsilon<1 and write A_n:=\{X_n>\varepsilon\}. In order to show that X_n doesn't converge a.s. to 0 it suffice to prove that X_n remains "far" from 0 infinitely many times a.s. i.e. P(\limsup_nA_n)=1. By Borel Cantelli lemma, being A_n indipendent, it suffices to show that \sum_nP(A_n)=+\infty, which is true (see above). 1 Here is a proof by contradiction: Suppose that p_t(x,x) does not converge to 1 as t\downarrow 0 . Then there exist a sequence of times t_n \downarrow 0 and an \epsilon >0 such that p_{t_n}(x,x)\le 1- \epsilon for all n. Moreover, the latter implies, by taking complements, that P_x(X_{t_n}\ne x) \ge \epsilon for all n. As the space is ... 0 I don't think your Claim is correct in the generality you have stated it in. Consider a Markov chain on the state space E=\{1, 1/2, 1/3,\ldots\}\cup\{0\} with the topology it inherits as a subspace of the real line. The process X_t is to be a sort of "pure birth" process: If started in state 1/n, X_t holds there for an exponential random time with ... 1 Let \bar x = (x_1+\cdots+x_n)/n and recall from algebra that$$ \sum_{i=1}^n (x_i-\theta)^2 = n(\bar x-\theta)^2 + \sum_{i=1}^n (x_i-\bar x)^2. Then deal with the density: \begin{align} f_{X_1,\ldots,X_n}(x_1,\ldots,x_n) \propto {} & \prod_{i=1}^n\frac 1 \theta \exp\left( \frac{-1} 2 \left( \frac{x_i-\theta}{\theta} \right)^2 \right) = \frac 1 ... 0 I think it is not obvious that the third and fourth term (in your calculation) converge to their prospective limits; however, the idea of your proof is basically correct. By definition,[B_{\ell}]_t = \mathbb{P}-\lim_{|\Pi| \to 0} \sum_{j=1}^n (B_{\ell_{t_j}}-B_{\ell_{t_{j-1}}})^2;$$here \Pi = \{0=t_0<\ldots<t_n = t\} denotes a partition of ... 1 By the definition of the conditional probability,$$ \sum_{k=0}^n P(A \subset B | N_B = k) P(N_B = k) = \sum_{k=0}^n P(\overbrace{\{A \subset B\} \cap \{N_B = k\}}^{\text{disjoint sets}}) = P(\{A \subset B\} \cap \bigcup_{k=1}^n \{N_B = k\}) = P(A \subset B). $$Furthermore,$$ P(N_B = k) = \frac{1}{2^n} \binom{n}{k} = \frac{1}{2^n} \frac{n (n-1) \cdots ...

0

Your LHS is a real number: $P(M\leq t)$, but your RHS is a random variable: $t^Y$. This is the way to do it: $$P\left(M\leq t\right)=\sum_{n=1}^{\infty}P\left(M\leq t\mid Y=n\right)P\left(Y=n\right)=\sum_{n=1}^{\infty}t^{n}P\left(Y=n\right)=g_{Y}\left(t\right)$$

2

Data $\rho\left(X,Y\right)=\frac{1}{2}$ enables you to find $\mathbb{E}XY=P\left(XY=1\right)=P\left(X=1\wedge Y=1\right)$ on base of: $$\rho\left(X,Y\right)=\frac{\mathbb{E}XY-\mu_{X}\mu_{Y}}{\sigma_{X}\sigma_{Y}}$$ This on its turn enables you to find the probabilities $P\left(X=0\wedge Y=1\right)$, $P\left(X=1\wedge Y=0\right)$ and $P\left(X=0\wedge ... 0 HINT: It is the convolution of two distributions. Check this PDF (page no. 294). Check this answer to get an idea how to do it for Normal distribution- Ratio Distribution. 1 It is not clear what you wanted to express by $$E[(X)_{2}]=E[x!/(x-2)!].$$ By the definition of the$r^{th}$factorial moment of a random variable$X$is $$E[X(X-1)(X-2)\cdots (X-r+1)].$$ We assume that the expectation in question exists. In the case of the binomial distribution with parameters$n,p$the$2^{d}$factorial moment, by definition is ... 2 I don't know if there is any technical reason behind this identity. There is a simple intuition behind the identity. We want to know if the variables X and Y are linearly related or not.i.e, can we say something like Y= m X where m is a constant. If m is positive, then Y increases as X increases. If m is negative, the Y decreases as X increases. First we ... 1 If you want to impose just uncorrelatedness between the variables in the sequence rather than independence, then you need to impose a stronger condition than$\sum_{i=1}^{\infty} Var[X_i]/i^2 < \infty. To be more precise, a sufficient condition in this case that will guarantee SLLN is the following: $$\sum_{i=1}^{\infty} Var[X_i] \left(\frac{log \, ... 0 By definition of i.o., P(X_n>a \text{ i.o. })=1 means$$ P\left(\bigcap_{k\geqslant 1}\bigcup_{n\geqslant k}\{x:X_n>a\}\right)=1 By definition of \limsup, we have \begin{align} \{x:\limsup_{n\rightarrow\infty} X_n \geqslant a\}&=\{x:\inf_{k\geqslant1}\sup_{\:n\geqslant k} X_n \geqslant a\} \\ &\supset\bigcap_{k\geqslant ... 1 If infinitely many terms of a sequence y_1,y_2,y_3,\ldots are >a then \limsup\limits_{n\to\infty} y_n \ge a. That statement can be made without knowing anything about probability. Here we have to say \text{“ }\ge\text{ ''} in the conclusion even though we say \text{“ }>\text{''} in the hypothesis, since, for example, we could have a ... 0 Let me assume that you mean that the probability that X_n is more than a for infinitely many natural n is one. Then the limit supremum is more than a, otherwise for any positive ε there is a point after which the sequence is bounded above by a+ε, which implies that X is more than a+ε with probability 0, and hence X is in the interval ... 0 What is meant here is that for each given \epsilon>0 there exists \delta>0 such that \forall i, \forall x,\forall y,\qquad |x-y|<\delta\Longrightarrow|X_i(x)-X_i(y)|<\epsilon. Suggestions: (i) Show that Y(x):=\sup_{i\in I}X_i(x) is a (uniformly) continuous function of x. In particular, Y is \mathcal B(\Bbb R)) measurable; i.e a ... 1 To elaborate on @John Dawkin's answer, we have \begin{align} \int_A\int_\mathbb R h(y)g(y,X)\ \mathsf dx \ \mathsf d\mathbb P &= \int_\Omega \mathsf 1_B(X)\int_{\mathbb R} h(y)g(y,X)\ \mathsf dy\ \mathsf d\mathbb P\\ &= \int_{\mathbb R}\int_{\mathbb R} h(y)f_{X,Y}(x,y)\mathsf 1_B(x)\ \mathsf dy\ \mathsf dx\\ &= \int_{\mathbb R} \int_{\mathbb R} ... 0 The common distribution of the pair of random variables (X,Y) is uniform over the triangle given in the OP. As a resultf_{X,Y}(x,y)=\begin{cases} \frac13,& \text{ if }& (x,y) \in A\\ 0,& \text{ otherwise } \end{cases}$$where A is as shown below The cdf of X at x equals the red surface area times \frac13 because X<x if ... 0 Let X denote the number of exams that must be graded before the teacher finds one with all answers correct and E is the event that the first exam graded has all answers correct then:$$\mathbb EX=\mathbb E(X\mid E)P(E)+\mathbb E(X\mid E^c)P(E^c)$$Note that:$$\mathbb E(X\mid E)=1\text{ and }\mathbb E(X\mid E^c)=1+\mathbb EX$$This provides you a way ... 1 You can integrate it, it's just that the result is not expressible in elementary functions. You can express it using the error function, or using the standard normal CDF. 2 Not only on an intuitive level: Define the map \Phi\colon L^2(\Omega)\otimes H\longrightarrow L^2(\Omega;H),\,f\otimes \xi\mapsto f(\cdot)\xi. This is an isometric isomorphism: \Phi is isometric:$$\langle f(\cdot)\xi,g(\cdot)\eta\rangle_{L^2}=\int_\Omega f(\omega)g(\omega)\langle \xi,\eta\rangle_H\,dP=\langle \xi,\eta\rangle_H\int_\Omega ... 3 Remember that the typical elementA\in\sigma(X)$has the form$X^{-1}(B)$, where$B$is a Borel subset of$\Bbb R$. Thus$1_A=1_B(X)$, and your left-hand integral can be written $$\int_\Omega 1_B(X)\int_{-\infty}^\infty h(y)g(y,X)\,dy\,d\Bbb P=\int_{-\infty}^\infty \int_{-\infty}^\infty 1_B(x)h(y)g(y,x)f_X(x)\,dy\,dx.$$ Now use the connection between$g$... 0 Yes this is correct, and you're right it's true for any distribution. 0 Note that $$\frac{1}{2} \log 3 - \log 2 < 0$$ and therefore $$\log \left( \prod_{j=1}^n X_j \right) = \sum_{j=1}^n \log (X_j) = n \left( \frac{1}{n} \sum_{j=1}^n \log(X_j) \right) \to - \infty$$ almost surely. Now the continuity of$\exp$entails $$\prod_{j=1}^n X_j = \exp \left[ \log \left( \prod_{j=1}^n X_j \right) \right] \to 0 \qquad \text{a.s.}$$ 1 Let $$A:=\left\{X_1=x_1,\ldots,X_n=x_n\right\}\;.$$ As Calvin Khor pointed out, we can assume, that$x_i\ne x$. Thus, $$\left\{\tau_x^1<\infty\right\}\cap A=\underbrace{\left\{\exists k>n:X_k=x\right\}}_{=:B}\cap A\;.$$ Let $$\tilde X:=\left(X_{k+n}\right)_{k\in\mathbb N_0}$$ and$f$be the indicator function of$\bigcup_{n\in\mathbb ...

0

I believe this is what they had in mind. For brevity $A_n:=[X_1=x_1,…,X_n = x_n]$. \begin{align}P_x(τ_x^1 = ∞ ∩ A) &= P_x(τ_x^1 = ∞ |A) P(A) \\&\overset{\star}{=} P_x(τ_x^1 = ∞ |X_n = y)P(A) \\&= P_y(τ_x^1 = ∞)P(A) \end{align} Where $\star$ is where the markov property is needed. I'm not sure how to fully justify this. Regarding the choice of ...

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On this website http://economictheoryblog.com/2015/02/19/ols_estimator/ you can find the proof that the beta estimator of OLS is indeed an unbiased estimator of the true beta in the population. HTH

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It means that $X$ and $Y$ have finite second moments, and so $\mathbb E[|XY|]<\infty$ by Hölder's inequality. Therefore the covariance (and hence the correlation) of $X$ and $Y$ exists. Recall that $$\operatorname{Var}(X+Y) = \operatorname{Var}(X) + \operatorname{Var}(Y) +2\operatorname{Cov}(X,Y),$$ and so $$\operatorname{Cov}(X,Y) = ... 0 i think it's because of the fact (\cup_{i=1, i\neq j}^{k} A_i)\cap A_j=\phi  2 Say that |f(x)| \le M is not satisfied a.e. and pick \epsilon > 0 and \delta > 0 such that if G = \{|f(x)| \ge M + \epsilon\} then \mu(G) \ge \delta. Now notice that$$ 0 = \lim_{n \to \infty}\mu(\{|f(x) - f_n(x)| \ge \epsilon\}) \ge \lim_{n \to \infty}\mu(\{|f(x)| - |f_n(x)| \ge \epsilon\}) \ge \mu(G) \ge \delta.  You can apply the ...

1

In case @joriki's "mere arithmetic rearrangement" is still unclear, \begin{align*} \mathbf{P}[E_{i+1}]\cdot p-\mathbf{P}[E_i]&=-(1-p)\cdot\mathbf{P}[E_{i=1}]\\ \mathbf{P}[E_{i+1}]\cdot p+(p-1-p)\cdot\mathbf{P}[E_i]&=(p-1)\cdot\mathbf{P}[E_{i=1}] \end{align*} Basically we've just added and subtracted a $p$ from the coefficient of $\mathbf{P}[E_i]$ on ...

1

No, it is not true that rectangular support means that each $x$ variable have the same number of $y$ variables." It means that the pdf is defined and non-zero on some rectangular region. For example, if your pdf lives on the region where $0<x<1$ and $0<y<5$, it lives on a rectangular region. If your pdf lives on $0<x<1$ and $y>0$, ...

1

In the second line, there's no conceptual leap, just two typos – if you correct the $P$ at the beginning to a $p$ and add a closing parenthesis at the end, this line becomes a mere arithmetic rearrangement of the previous one. In the first line, the idea is that if player $A$ has $i$ units and heads is flipped, then player $A$ has $i+1$ units, so the ...

1

For this very special case, note that starting at 0 it must go 0,1,2,3,4 and then from 4 it might go back to 0, making the least $n>0$ with $P_{00}^{(n)}>0$ to be $n=5.$ But again resuming from 4 it might go on to 5, then it must go to 6,7, then 0. That means that also $P_{00}^{(8)}>0.$ The gcd of 5 and 8 is 1, so if I'm following the definition ...

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