# Tag Info

27

I think this is a slightly more intuitive way of looking at the question: Suppose $f (x)=k$. Once we have chosen $k$, there are $200$ possible values for $g (x)$, one of which is $k$, hence we get an answer of $\frac{1}{200}$.

20

Consider all of the $6\times 5$ ways to pick two pieces of fruit.   That's $30$: $$\boxed{\begin{array}{|l|ccc:ccc|}\hline ~ & A_1 & A_2 & A_3 & O_1 & O_2 & O_3 \\ \hline A_1 & \times & \color{green}{A_1,A_2} & \color{green}{A_1,A_3} & \color{blue}{A_1,O_1} & \color{blue}{A_1,O_2} & \color{blue}{A_1,O_3} ... 18 The only reason you are multiplying by 2 in the second case is because you are using a shortcut due to the fact that the two scenarios that you are adding have a probability found with the same formula. You just need to add up the probabilities you are seeking. Case 1) \frac{3}{6}*\frac{2}{5} Case 2) \frac{3}{6}*\frac{3}{5}+\frac{3}{6}*\frac{3}{5} ... 14 When all you have is the raw set structure, the only limit concept that really makes sense is: S is the limit of the sequence S_1, S_2, S_3,\ldots iff$$ \forall x\; \exists N\in\mathbb N\; \forall n>N : x\in S\Leftrightarrow x \in S_n $$In other words every possible element is either in all but finitely many S_n (in which case it is in the ... 13 In terms of the sample space of events Ω, an event E happens almost surely if P(E) = 1, whereas an event happens surely if E=Ω . An example: suppose we are (independently) flipping a (fair) coin infinitely many times. The event$$\{ \text{I will get heads infinitely often}\}$$is an almost sure event (because it is possible get only a finite ... 12 Start with x_1=0 and then just let x_{n+1}=1 whenever f(n)/n\leq p and x_{n+1}=0 whenever f(n)/n> p. It's easy to prove by induction that we will always have \left|\frac{f(n)}n-p\right|\leq\frac 1n and so f(n)/n\rightarrow p. 12 Assuming the selections are independent and uniform, there are 100 \times 200 possible equiprobable pairs of which 100 can be a match, so the probability is {100 \over 200 \cdot 100 }. 12 Let me get you started: if (X,Y) is i.i.d., (Y,X) is distributed like (X,Y) hence Y-X is distributed like X-Y hence E(u(Y-X))=E(u(X-Y)) for every u such that the expectations exist. For u(t)=t^3 (or any odd function), this yields E((X-Y)^3)=E((Y-X)^3)=-E((X-Y)^3) hence... Thus, the answer is "Yes, provided E(|X|^3) is finite" (otherwise ... 11 It can be shown that nonnegative random variables X and Y have the same distribution so long as \mathbb{E}[X^\alpha]=\mathbb{E}[Y^\alpha] is finite for all \alpha\in(a,b], any 0\le a <b. Setting U=\log X and V=\log Y, define the functions$$ f(\alpha)=\mathbb{E}[1_{\{X > 0\}}e^{\alpha U}],\ g(\alpha)=\mathbb{E}[1_{\{Y > 0\}}e^{\alpha ...

11

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 turns out the numerical findings about $\mathbb{E}[N_3]$ by David E is not a coincidence, it is exact! $$\mathbb{E}[N_3] = \frac{1}{1-\sin(1)}$$ Let $X_1, X_2, \ldots$ be a sequence of random variables. For each $n$, we will assume $X_n$ take value from the set $\langle n \rangle \stackrel{def}{=}\{ 1, 2, \ldots, n \}$ with uniform probability. After ...

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

Convergence in probability does not imply convergence almost surely: Consider the sequence of random variables $(X_n)_{n \in \mathbb{N}}$ on the probability space $((0,1],\mathcal{B}((0,1]))$ (endowed with Lebesgue measure $\lambda$) defined by \begin{align*} X_1(\omega) &:= 1_{\big(\frac{1}{2},1 \big]}(\omega) \\ X_2(\omega) &:= 1_{\big(0, ... 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 This is false, when \lim_{n \to \infty} p_n doesn't exist. For instance, consider p_{2n-1} = \dfrac1{2^n} and p_{2n} = 1-\dfrac1{2^n}. This gives us \sum_{n=1}^{2N} p_n = \sum_{n=1}^{2N} (1-p_n) = N, whereas\sum_{n=1}^{2N} p_n(1-p_n) < \sum_{n=1}^N \dfrac1{2^{n-1}} < 2$$However, the claim is true when \lim_{n \to \infty} p_n exists. ... 9 I'll try to give you an example that shows why it is very useful. Suppose that the random variable X follows the binomial distribution with parameters n and p. Then the expected value of X is given by$$ \operatorname EX=\sum_{k=0}^nk\binom{n}{k}p^k(1-p)^{n-k}. $$But we know that X=\sum_{j=1}^nY_j, where Y_1,\ldots,Y_n are independent and ... 9 Assuming that by "number" you mean "integer," and by "random" you mean "uniformly at random," AND ASSUMING THAT THE CHOICES ARE INDEPENDENT, then the desired probability is$$\sum_{k=1}^{200} \Pr[X = k]\Pr[Y = k] = \sum_{k=1}^{100} \frac{1}{100} \cdot \frac{1}{200} + \sum_{k=101}^{200} 0 \cdot \frac{1}{200} = \frac{1}{200}.$$9 Here is a proof of the first question using Zorn's lemma, and a counterexample to the second question using an ultrafilter. (So both cases used some form of the axiom of choice!) Theorem (Sierpinski): For a non-atomic probability space (\Omega, \mathcal{F}, \mu), \mu is surjective onto [0,1]. Proof: Let x \in [0,1], and let$$ \mathcal{G} = ...

9

Write $A \setminus B$ as $A \cap B^{c}$.

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

Since $\mathbb E[X^2] = \mathbb E[X\mathbb E[Y|X]] = \mathbb E[\mathbb E[XY|X]] = \mathbb E[XY]= \mathbb E[\mathbb E[XY|Y]] = \mathbb E[Y\mathbb E[X|Y]] = \mathbb E[Y^2],$ observe that $$\mathbb E[(X-Y)^2]=\mathbb E[X^2+Y^2-2XY]=0,$$ which implies that $X$ and $Y$ differ on a set of measure zero. For the weaker condition where $X$ and $Y$ are ...

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

The problem is the claim that all $x^2$ values between $0$ and $4$ are equally likely. That's just not true, except in the sense that each one of them has probability zero. But in any more reasonable sense, it fails. In particular, if all were equally likely, then the probability that the square was between 0 and 2 would be the same as the probability that ...

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

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 $Var(X)=E(X^2)- [E(X)]^2 \Rightarrow E(X^2)=Var(X)+[E(X)]^2$

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