# Tag Info

20

Consider all of the $6\times 5$ ways to pick two pieces of fruit.   That's $30$: $$\boxed{\begin{array}{|l|ccc:ccc|}\hline ~ & A_1 & A_2 & A_3 & O_1 & O_2 & O_3 \\ \hline A_1 & \times & \color{green}{A_1,A_2} & \color{green}{A_1,A_3} & \color{blue}{A_1,O_1} & \color{blue}{A_1,O_2} & \color{blue}{A_1,O_3} ... 18 The only reason you are multiplying by 2 in the second case is because you are using a shortcut due to the fact that the two scenarios that you are adding have a probability found with the same formula. You just need to add up the probabilities you are seeking. Case 1) \frac{3}{6}*\frac{2}{5} Case 2) \frac{3}{6}*\frac{3}{5}+\frac{3}{6}*\frac{3}{5} ... 10 In order to give more strength to the induction hypothese let us prove more generally: \exists\alpha\in\left\{ 0,1\right\} ^{I}:\left\{ B_{i}^{\left(\alpha_{i}\right)}\right\} _{i\in I}\text{ is independent}\implies\forall\beta\in\left\{ 0,1\right\} ^{I}:\left\{ B_{i}^{\left(\beta_{i}\right)}\right\} _{i\in I}\text{ is independent} Assume that the ... 6 When you state that the chance of getting an apple first is 3/6 you are assuming that all the apples are interchangeable. If you want to label the apples you can. You can then say the chance of getting apple 1 followed by apple 2 is (1/6)(1/5)=1/30. Now you can find that there are six different permutations of two apples, so the total chance of ... 5 Let X = R \cos \Theta, Y = R \sin \Theta for random variables (R, \Theta) supported on \mathbb{R}_+ \times [0,2\pi). It's easy to see that this covers the entire support of (X,Y). The determinant of the Jacobian for (x,y) \mapsto (r\cos \theta, r\sin \theta)is r.$$f_{R\Theta}(r,\theta) = r \frac{1}{2 \pi} \exp(- r^2 \cos^2 \theta - r^2 ...

5

Questions and Answers in MSE and two other references: Ad 1) : How to derive these Lie Series formulas & chain references Ad 3) : Ambiguous matrix representation of imaginary unit? idem Extra : Where the exponent in the Laplace Transform comes from Ad 4) : Gaussian Blur as a sample application of the Extra item Updates.Ad 2). The Fourier transform ...

5

The series $\sum\limits_{n=1}^\infty\frac1{S_n}$ diverges almost surely, where $S_n=X_1+\cdots+X_n$ for some i.i.d. random variables $X$ and $(X_n)$ such that $P(X\geqslant k)=\frac1k$ for every positive integer $k$. The same result applies to every positive i.i.d. sequence $X$ and $(X_n)$ such that $x\cdot P(X\geqslant x)\leqslant c$ for some finite ...

4

Hint: Countable additivity shows that $$\lim_{N\to\infty}P(|X|\le N)=1.$$That takes care of the case of one random variable; the case of finitely many random variables follows. EDIIT: The discrepancy between this and the other answer is, it seems to me, explained by the fact that there's some confusion over what the word "tight" means. With the standard ...

4

The convergence in probability to a constant $\mu$ is equivalent to weak convergence to $\mu$. And the latter, by Lévy's criterion, is equivalent to the pointwise convergence of characteristic functions to $e^{i\mu t}$. Thus, $S_n/n \overset{P}{\longrightarrow} \mu$ $\iff$ $\varphi_{S_n/n}(t)\to e^{i\mu t}, t\in \mathbb{R}$ $\iff$ ...

4

Take $(X_n)_{n\geqslant 1}$ an i.i.d. sequence where $X_1$ is a non-degenerated zero-mean random variable and $\mathcal F_n :=\sigma(X_1,\dots,X_n)$ for $n\geqslant 1$. Then $\mathbb E[X_n\mid\mathcal F_{n-1}]=0$ for each $n$.

4

Since $B_t$ is Gaussian with mean $0$ and variance $t$, the moments of $B_t$ can be calculated explicitly. For any $k \in \mathbb{N}$ we have $$\mathbb{E}(B_t^{2k+1}) = 0 \tag{1}$$ and $$\mathbb{E}(B_t^{2k}) = t^k \frac{2^k \Gamma(k+1/2)}{\sqrt{\pi}},$$ in particular, $$\mathbb{E}(B_t^2) = t \qquad \mathbb{E}(B_t^4) = 3t^2 \qquad \mathbb{E}(B_t^6) = 15 ... 4 Because of the weak convergence, there is a probability space (\Omega,\mathcal F,\Bbb P) and random variables X_1,X_2,\ldots,X defined thereon, with values in S, such that (i) P_n is the distribution of X_n (i.e., \Bbb P(X_n\in B)=P_n(B) for all Borel B\subset S), (ii) P is the distribution of X, and (iii) \lim_{n\to\infty} ... 3 Fix \epsilon>0. By the uniform continuity of g, there exists \delta>0 such that$$|x-y| \leq \delta \implies |g(x)-g(y)| \leq \epsilon.$$This implies$$\{\omega; |g(X_t(\omega))-g(X_s(\omega))| > \epsilon\} \subseteq \{\omega; |X_t(\omega)-X_s(\omega)| >\delta\}.$$Hence,$$\mathbb{P}(|g(X_t)-g(X_s)| > \epsilon) \leq ...

3

Note that $$\int_n^{\infty}\frac{dx}{x^2\log x}<\frac{1}{\log n}\int_n^{\infty}\frac{dx}{x^2}=\frac{1}{n\log n}$$

3

Fisher information is related to the asymptotic variability of a maximum likelihood estimator. The idea being that higher Fisher Information is associated with lower estimation error. Shannon Information is totally different, and refers to the content of the message or distribution, not its variability. Higher entropy distributions are assumed to convey ...

3

The original renewal process has independent and identically disributed (i.i.d.) inter-renewal times $\{T_1, T_2, T_3, ...\}$. Thus, starting from time 0, renewals occur at times $\{T_1, T_1+T_2, T_1+T_2+T_3, ...\}$. Now fix a probability $p >0$. Independently place each renewal time to $P_1$ with prob $p$. So we get new inter-renewal times $\{Z_1, ... 3 Let$A$be the event that$X>E(X)+t$, and let$B$be the event that$e^{uX}>e^{u(E(X)+t)}$. But since$x \mapsto e^{ux}$is an increasing function,$X>E(X)+t \implies e^{uX}>e^{u(E(X)+t)}$and therefore$A \subseteq B$. Also, since$x \mapsto \frac1u\ln x$is an increasing function,$e^{uX}>e^{u(E(X)+t)} \implies X>E(X)+t$and therefore$B ...

3

I have some not very checked ideas which may help someone to obtain the bounty. I have two approaches to the sufficient condition. The first is to show that the family $\mathcal A=\{A_y: A_y$ is a measurable subset of $X\}$ is sufficiently big. For instance, Theorem 17.10 from “Classical Descriptive Set Theory” by Alexander S. Kechris, implies that for ...

3

Assume that $(X_i)_{i \in I}$ are independent random variables. Let $Y_i := X_i \cdot 1_{|X_i|\leq C_i}$, where $C_i>0$ is chosen so large that $Y_i \not\equiv c_i$ for some constant $c_i$. This is possible since the $X_i$ are non-constant. Then the $(Y_i - \Bbb{E}(Y_i))_i$ form a family of independent and hence orthogonal random variables. Note ...

3

Note that the Baire category theorem implies that if $U_n$, $n\in \Bbb {N}$ are open dense sets, then $\bigcap U_n$ is dense. In particular, the intersection is uncountable, since if it was of the form $\{x_n \mid n\}$, we could set $V_n = [0,1]\setminus \{x_n\}$ (which is open and dense), so that Baire again implies that $\bigcap U_n \cap V_n ... 3 For$n\in\mathbb{N},n\geq 0$, we have $$\Pr(Y\leq n)=\Pr(X< n+1)=1-e^{-n-1}.$$ For$z\in[0,1), we have \begin{align*} \Pr(Z\leq z)&=\sum_{m=0}^\infty\Pr(X\in[m,m+z])\\ &=\sum_{m=0}^\infty(\exp(-m)-\exp(-m-z))\\ &=(1-e^{-z})\sum_{m=0}^\infty(1/e)^m\\ &=\frac{e-e^{1-z}}{e-1} \end{align*} Finally, \begin{align*} \Pr(Y\leq n,Z\leq ... 3X_T$is the random variable with value$X_{T(\omega)}(\omega)$at each point$\omega\in\Omega$where$T(\omega)<\infty$. For$\omega\in\{T=\infty\}$one can take$X_T(\omega)$to equal$\lim\limits_{n\to\infty}X_n(\omega)$for those$\omega$for which the limit exists.$|X_{T\wedge n}|=\sum\limits_{k=0}^{n-1}1_{\{T=k\}}|X_k|+1_{T\ge n}|X_n|\le ...

3

This is a neat proof: suppose I have a number (between 0 and 1, say) whose decimal expansion eventually repeats: $$x=a_1. . . a_nb_1. . . b_mb_1. . . b_mb_1. . . b_m . . .$$ where $a_i, b_i$ are decimal digits. (This includes the case when the decimal expansion terminates: then we just have $m=1, b_1=0$.) Then we have$^*$ $$10^nx-a_1a_2. . . a_n=0.b_1. . . ... 3 Picking two sweets are independent events, so the probability is equal to the product of the probability of the two events separately, then you have:$$1/3=\frac{6}{n} \cdot \frac{5}{n-1}$$Which is equal to:$$n^2-n = 90$$3 By Markov's inequality, for each \varepsilon>0,$$ \Pr\{|X_n|\ge\varepsilon\}\le\frac{\operatorname EX_n^2}{\varepsilon^2}=\frac1{\varepsilon^2n}\to0\quad\text{as}\quad n\to\infty. $$Hence, X_n\to0 in probability as n\to\infty 3 Is this the sort of thing you had in mind? Take Lebesgue measure on [0,1]. For each positive integer n, let X_n = 2^n on an interval of measure 2^{-1-n}, -2^{n} on an interval of measure 2^{-1-n}, 0 everywhere else, where all these intervals are pairwise disjoint (note that the sum of the measures of the intervals is 1). The pointwise sum ... 3 T=\min(t,n) is a bounded stopping time. By Optional Stopping, the expectation is$$E(B^2_T -T) = E(B^2_0 -0) = 0$$Conclude by linearity of expectation. 3 In the Kolmogorov formalism, you could call the entire sample space a "hidden variable", since it is usually not observed directly but only through actual random variables. The reason it doesn't conflict with Bell's inequality is that is is not local -- on the contrary it is the most global hidden variable conceivable, governing both which measurement the ... 3 Let Z_n=a_n(X_n-X). We are told that \frac{1}{a_n}\to 0 so, in particular, \frac{1}{a_n}\overset{P}{\to}0. So by Slutsky's Theorem, X_n-X=\frac{1}{a_n}\times Z_n\overset{d}{\to}0\times Z=0. But convergence in distribution to constant is equivalent to convergence in probability to that same constant, so we conclude that X_n-X\overset{P}{\to}0. 3 I assume you are confused as to where to start, so here are some guidelines. Fill in the details and do the calculations, or ask if any particular points are unclear. To find the mean value E(Y\ |\ X=m), you can use that$$E(Y\ |\ X=m) = \sum_{n=0}^m P(Y=n\ |\ X=m)n, so you need to calculate the conditional probability $P(Y=n\ |\ X=m)$. You can find ...

Only top voted, non community-wiki answers of a minimum length are eligible