# Tag Info

20

Consider all of the $6\times 5$ ways to pick two pieces of fruit.   That's $30$: $$\boxed{\begin{array}{|l|ccc:ccc|}\hline ~ & A_1 & A_2 & A_3 & O_1 & O_2 & O_3 \\ \hline A_1 & \times & \color{green}{A_1,A_2} & \color{green}{A_1,A_3} & \color{blue}{A_1,O_1} & \color{blue}{A_1,O_2} & \color{blue}{A_1,O_3} ... 18 The only reason you are multiplying by 2 in the second case is because you are using a shortcut due to the fact that the two scenarios that you are adding have a probability found with the same formula. You just need to add up the probabilities you are seeking. Case 1) \frac{3}{6}*\frac{2}{5} Case 2) \frac{3}{6}*\frac{3}{5}+\frac{3}{6}*\frac{3}{5} ... 17 the answer is indeed \frac 12 . As an alternative way to see that: let's pause just before Bob tosses his final (extra) toss. At this point, there are three possible states: either Bob is ahead, Alice is ahead, or they are tied. Let p be the probability that Bob is ahead. By symmetry, p is also the probability that Alice is ahead (so the ... 14 When all you have is the raw set structure, the only limit concept that really makes sense is: S is the limit of the sequence S_1, S_2, S_3,\ldots iff$$ \forall x\; \exists N\in\mathbb N\; \forall n>N : x\in S\Leftrightarrow x \in S_n $$In other words every possible element is either in all but finitely many S_n (in which case it is in the ... 14 In terms of the sample space of events Ω, an event E happens almost surely if P(E) = 1, whereas an event happens surely if E=Ω . An example: suppose we are (independently) flipping a (fair) coin infinitely many times. The event$$\{ \text{I will get heads infinitely often}\}$$is an almost sure event (because it is possible get only a finite ... 14 Let p_n be the nth prime. The events p_n\mathbb N are independent*, because$$P(p_n\mathbb N \cap p_m\mathbb N)=P(p_np_m\mathbb N)=\frac 1{p_np_m}=P(p_n\mathbb N)P(p_m\mathbb N)$$The sum of the reciprocals of the primes$$\sum_n \frac 1{p_n}$$famously diverges. So, by the second Borel-Cantelli lemma, the event that infinitely many of the events ... 12 You want to prove the statement:$$\lim_{n\to\infty}\sum_{i=1}^{n}c=c \implies c=0$$Instead, you can prove the equivalent statement:$$c\neq0 \implies \lim_{n\to\infty}\sum_{i=1}^{n}c \neq c$$And this is rather simple, as you can use the exact trick that you were trying to avoid:$$c\neq0 \implies ...

12

If the covariance matrix is not positive definite, we have some $a \in \mathbf R^n \setminus \{0\}$ with $\def\C{\mathop{\rm Cov}}\C(X)a = 0$. Hence \begin{align*} 0 &= a^t \C(X)a\\ &= \sum_{ij} a_j \C(X_i, X_j) a_i\\ &= \mathop{\rm Var}\left(\sum_i a_i X_i\right) \end{align*} So there is some linear combination of the $X_i$ which has ...

11

A direct proof: For any measurable set $A$, applying independence of the events $\{X\in A\}$ and $\{X\in A\}$ gives $$\mathbb P(X\in A)=\mathbb P(X\in A)\mathbb P(X\in A)\implies \mathbb P(x\in A)\in \{0,1\}.$$ Hence $\mathbb P$ is a dirac measure. Therefore $X$ is a.s. constant.

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 $P(X \leq Y)P(Y \leq Z) = \frac12 \times \frac12 = \frac14$.

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 ...

10

If we divide every roll by $n$, rolling the die and dividing by $n$ approximates the uniform distribution on $[0,1]$ for arbitrarily large $n$. We then are looking for the expected number of samples from a uniform distribution required to get a sum above $1$. For any integer $k \in \mathbb{N}$, Let $X_1, \ldots , X_k, \ldots$ be the random variables in ...

9

Define $X_n$ to be such that $X_n$ is $0$ with probability $1-\frac{1}{n}$ and $n^2$ with probability $\frac{1}{n}$. It is the case that $E[X_n]=n \to \infty$. But for any positive $k$ we have $\mathbb{P}(X_n > k) = \frac{1}{n} \to 0$. Thus showing a counter example.

9

Here is a proof of the first question using Zorn's lemma, and a counterexample to the second question using an ultrafilter. (So both cases used some form of the axiom of choice!) Theorem (Sierpinski): For a non-atomic probability space $(\Omega, \mathcal{F}, \mu)$, $\mu$ is surjective onto $[0,1]$. Proof: Let $x \in [0,1]$, and let $$\mathcal{G} = ... 8 In the context of probability theory, you can write$$ \Pr(\varnothing) = \Pr(\varnothing \cup \varnothing) = \Pr(\varnothing)+\Pr(\varnothing). $$Then if you subtract \Pr(\varnothing) from both sides, you get$$ 0 = \Pr(\varnothing). $$Thus you don't need infinite series. That \infty\cdot0=0 does not make sense in certain broad contexts, since for ... 8 You have some event, which you typically don't know when occurs, but that can/will occur some time in the future. The time that this event occurs is random, and it is a stopping time if, at any point in time, you know whether the event has occurred or not. A few quick examples. 1) Your own (a stopping time): Let \tau denote the time that I'm ruined (i.e. ... 8 Consider two normal distributions with the same variance and different means. 8 There is a difference between "almost surely" and "surely." Consider choosing a real number uniformly at random from the interval [0,1]. The event "1/2 will not be chosen" has probability 1, but is not impossible. I recommend reading the relevant Wikipedia article, which I found very clarifying when I was learning probability. 8 Definitely incorrect. Let F = X+Y. Suppose X and Y are IID normal. Then E[X|F] and E[Y|F] are both linear in F, and hence perfectly correlated. 8 Use the strong law of large numbers. Choose B\in M so that \lambda(B)\neq \mu(B), and consider the disjoint sets$$\left\{x\in X^\infty: {1\over n}\sum_{j=1}^n 1_{[x_j\in B]}\to \lambda(B)\right\}\mbox{ and }\left\{x\in X^\infty: {1\over n}\sum_{j=1}^n 1_{[x_j\in B]}\to \mu(B)\right\}.$$8 By definition: E[X] is only defined (EDIT: as a real number) for X such that E[|X|] < \infty. 8 Since X_n and X are both measurable (they are random variables after all) and the function f(x,y) = \lvert x-y \rvert preserves measurability (due to its continuity), \lvert X_n - X \rvert is measurable as well. Maybe it is better for your understanding if I argue in the following way. If X_n and X are measurable, then so is X_n - X. This is a ... 8 Hint: It is E(X(X-1))=E(X^2)-E(X). And Var(X)=E(X^2)- [E(X)]^2 \Rightarrow E(X^2)=Var(X)+[E(X)]^2 7 Hint:$$\exp(-I_{E_n}) = e^{-1} I_{E_n}+ I_{E_n^c}$$implies$$\mathbb{E}(\exp(-I_{E_n})) =(e^{-1}-1) \mathbb{P}(E_n)+1.$$Now use that$$1+x \leq e^x.$$7 If I toss a coin an infinite amount of times, can I be sure to get an infinite amount of heads? According to the Borel-Cantelli lemma, since each coin toss is an event of probability \frac12 and a sum of \frac12 diverges, the probability of \limsup_{n\to\infty}\{\text{heads at n-th flip}\} is 1. But the \limsup is precisely the event of ... 7 The vector \vec X = [X_1,\dots,X_N]^T has a rotationally invariant distribution. That is, if A is any orthogonal matrix, then the distribution of \vec X and A \vec X are the same. Hence by letting A be an orthogonal matrix that takes [1,\dots,1]^T to [\sqrt N,0,\dots,0]^T, your problem is the same as computing$$ \Pr(\sqrt N X_1 \mid \sum ...

7

Notice that if $A,B,C$ are pairwise disjoint, you have \begin{align} \Pr(A\cup B\cup C) & = \Pr(A\cup B)+\Pr(C) & & (\text{since }A\cup B\text{ is disjoint from }C) \\[6pt] & = \Pr(A) + \Pr(B) + \Pr(C) & & (\text{since }A\text{ is disjoint from }B) \end{align} and the same can be done with any finite sequence of pairwise disjoint ...

7

The $\sigma$-algebra generated by the events $\{\omega \in \Omega: \omega_n = W \}$ is the so-called Borel $\sigma$-algebra on $\Omega = \{H,T\}^\mathbb{N}$. One can show, by transfinite induction (so you need some set-theory background) that there are at most $|\mathbb{R}| = 2^{\aleph_0}$ many Borel sets, while the power set of $\Omega$ has $2^{|\Omega|} = ... 7 Simply follow the hint... First note that, since$E(X\mid Y)=Y$almost surely, for every$c$, $$E(X-Y;Y\leqslant c)=E(E(X\mid Y)-Y;Y\leqslant c)=0,$$ and that, decomposing the event$[Y\leqslant c]$into the disjoint union of the events$[X>c,Y\leqslant c]$and$[X\leqslant c,Y\leqslant c]$, one has $$E(X-Y;Y\leqslant c)=U_c+E(X-Y;X\leqslant c,Y\leqslant ... 7 I give 2 proofs, that show that there is indeed something more than the pure analytical formula, beyond its intrinsic beauty. First proof: Let us consider the formula under the form:$$\frac{\sin \pi x}{\pi x} = \prod_{k=1}^{\infty}\cos{\frac{\pi x}{2^k}}= \lim_{n \rightarrow \infty}{\prod_{k=1}^{n}\cos{\frac{\pi x}{2^k}}}$\$ Its Fourier Transform (using ...

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