# Tag Info

51

We see that in probability, we represent the event $A$ as a set of elements in our sample space, and $\neg A$ as the complement of $A$ in our sample space. Thus, in probabilistic terms, $$P(A \wedge \neg A) = P\left(A \cap \overline{A}\right) = P(\emptyset)$$ by the definition of the complement. And by the Kolmogorov Axioms, we see that $$P(\emptyset) = 0.... 20 Consider all of the 6\times 5 ways to pick two pieces of fruit. That's 30:$$\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} \...

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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) $\frac{3}{6}*\frac{3}{5}+\frac{3}{6}*\frac{3}{5}$ Or,...

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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

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 ...

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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

Let $p_n$ be the $n$th 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 $... 12 If the covariance matrix is not positive definite, we have some$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 theX_iwhich has ... 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: $$c\neq0 \implies \lim_{n\to\infty}\sum_{i=1}^{n}c=c\lim_{n\to\infty}... 12 We have this sum:$$ \sum_{k=0}^n k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k} \tag 1 $$First notice that when k=0, the term k\dfrac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k} is 0, and next notice that when k\ne0 then$$ \frac k {k!} = \frac 1 {(k-1)!} $$so that$$ k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k} = \frac{n!}{(k-1)!(n-k)!} p^k (1-p)^{n-k}. $$The two expressions ... 11 He probably meant that either: 1. The space of continuous functions which are nowhere monotonic has probability one using Wiener measure. I.e. a continuous function is "almost surely" nowhere monotonic using the standard probability measure for that space. 2. That set of nowhere monotonic continuous functions is of the first category (Baire category ... 10 It would generally not be true even if they were independent. For example if X,Y,Z were identically and independently continuously distributed then they can come in any order with equal probability so P(X \leq Y \leq Z) = \frac16 but similarly P(X \leq Y)P(Y \leq Z) = \frac12 \times \frac12 = \frac14. 10 If we divide every roll by n, rolling the die and dividing by n approximates the uniform distribution on [0,1] for arbitrarily large n. We then are looking for the expected number of samples from a uniform distribution required to get a sum above 1. For any integer k \in \mathbb{N}, Let X_1, \ldots , X_k, \ldots be the random variables in ... 10 In order to give more strength to the induction hypothese let us prove more generally: \exists\alpha\in\left\{ 0,1\right\} ^{I}:\left\{ B_{i}^{\left(\alpha_{i}\right)}\right\} _{i\in I}\text{ is independent}\implies\forall\beta\in\left\{ 0,1\right\} ^{I}:\left\{ B_{i}^{\left(\beta_{i}\right)}\right\} _{i\in I}\text{ is independent} Assume that the ... 10 The trick is the using the identity k { n \choose k} = n {n-1 \choose k-1}.$$\begin{align*} &\sum_{k=1}^n k { n \choose k } p^k (1-p)^{n-k}\\ &= \sum_{k=1}^n n { n-1 \choose k-1} p^k (1-p)^{n-k}\\ &=np \sum_{k=1}^n {n-1 \choose k-1} p^{k-1} (1-p)^{(n-1)-(k-1)}\\ &=np (p+(1-p))^n\\ &=np \end{align*}$$10 The event \{H,T\} is the event that the coin turns up heads or tails. This event always happens and thus has probability 1. The empty event, \varnothing, should not be thought of as the event that the coin lands on its edge. It is assumed that the coin always lands heads or tails. Rather the event \varnothing is the event that there is no outcome, ... 10 No. We might have P(X=n)=2^{-n}. Then \Bbb E[X]=2 and \Bbb E[e^X]=\sum_n 2^{-n}e^n=\infty 9 Simply follow the hint... First note that, since E(X\mid Y)=Y almost surely, for every c,$$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 ... 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 $$\mathcal{G} = \{... 9 Define X_n to be such that X_n is 0 with probability 1-\frac{1}{n} and n^2 with probability \frac{1}{n}. It is the case that E[X_n]=n \to \infty. But for any positive k we have \mathbb{P}(X_n > k) = \frac{1}{n} \to 0. Thus showing a counter example. 8 Definitely incorrect. Let F = X+Y. Suppose X and Y are IID normal. Then E[X|F] and E[Y|F] are both linear in F, and hence perfectly correlated. 8 There is a difference between "almost surely" and "surely." Consider choosing a real number uniformly at random from the interval [0,1]. The event "1/2 will not be chosen" has probability 1, but is not impossible. I recommend reading the relevant Wikipedia article, which I found very clarifying when I was learning probability. 8 By definition: E[X] is only defined (EDIT: as a real number) for X such that E[|X|] < \infty. 8 In the context of probability theory, you can write$$ \Pr(\varnothing) = \Pr(\varnothing \cup \varnothing) = \Pr(\varnothing)+\Pr(\varnothing). $$Then if you subtract \Pr(\varnothing) from both sides, you get$$ 0 = \Pr(\varnothing). $$Thus you don't need infinite series. That \infty\cdot0=0 does not make sense in certain broad contexts, since for ... 8 I think that's a misunderstanding. The passage you quote is a bit informally phrased. By "the mean time to failure of some disk", they don't mean the mean time to failure of each individual disk but the mean time it takes until any one of the 100 disks fails. If the failures form a Poisson process, then having 100 disks instead of 1 disk will increase ... 8 It might be easier to think about if we abstract a little. Brownian motion is just a continuous, time dependent random variable determined by some probability space (X,\mathcal F, P). Let us view it as a function of two variables, f(x,t). If x is fixed, we have a continuous function, f_x(t)=f(x,t), and we can compute it's integral F(x,t)=F_x(t)=\... 7 If I toss a coin an infinite amount of times, can I be sure to get an infinite amount of heads? According to the Borel-Cantelli lemma, since each coin toss is an event of probability \frac12 and a sum of \frac12 diverges, the probability of \limsup_{n\to\infty}\{\text{heads at n-th flip}\} is 1. But the \limsup is precisely the event of infinite ... 7 Let (B_t)_{t \geq 0} be a Brownian motion on a probability space (\Omega_1,F_1,P_1) and let (\Omega_2,F_2,P_2) be an arbitrary probability space. Define a new probability space (\Omega,F,P) by$$\Omega := \Omega_1 \times \Omega_2 \qquad F := F_1 \otimes F_2 \qquad P := P_1 \otimes P_2.$$If we set$$\tilde{B}_t(\omega_1,\omega_2) := B_t(\omega_1) \... 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 X_i^... 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|} = ...

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