# 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 $30$: $$\boxed{\begin{array}{|l|ccc:ccc|}\hline ~ & A_1 & A_2 & A_3 & O_1 & O_2 & O_3 \\ \hline A_1 & \times & \color{green}{A_1,A_2} & \color{green}{A_1,A_3} & \color{blue}{A_1,O_1} & \color{blue}{A_1,O_2} & \color{blue}{A_1,O_3} ... 18 The only reason you are multiplying by 2 in the second case is because you are using a shortcut due to the fact that the two scenarios that you are adding have a probability found with the same formula. You just need to add up the probabilities you are seeking. Case 1) \frac{3}{6}*\frac{2}{5} Case 2) \frac{3}{6}*\frac{3}{5}+\frac{3}{6}*\frac{3}{5} ... 14 In terms of the sample space of events Ω, an event E happens almost surely if P(E) = 1, whereas an event happens surely if E=Ω . An example: suppose we are (independently) flipping a (fair) coin infinitely many times. The event$$\{ \text{I will get heads infinitely often}\}$$is an almost sure event (because it is possible get only a finite ... 14 When all you have is the raw set structure, the only limit concept that really makes sense is: S is the limit of the sequence S_1, S_2, S_3,\ldots iff$$ \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 ... 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 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 ... 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

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 $f(n)/n\rightarrow p$.

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

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

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 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 ...

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

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

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 $$\mathcal{G} = ... 8 In the context of probability theory, you can write$$ \Pr(\varnothing) = \Pr(\varnothing \cup \varnothing) = \Pr(\varnothing)+\Pr(\varnothing). $$Then if you subtract \Pr(\varnothing) from both sides, you get$$ 0 = \Pr(\varnothing). $$Thus you don't need infinite series. That \infty\cdot0=0 does not make sense in certain broad contexts, since for ... 8 Definitely incorrect. Let F = X+Y. Suppose X and Y are IID normal. Then E[X|F] and E[Y|F] are both linear in F, and hence perfectly correlated. 8 There is a difference between "almost surely" and "surely." Consider choosing a real number uniformly at random from the interval [0,1]. The event "1/2 will not be chosen" has probability 1, but is not impossible. I recommend reading the relevant Wikipedia article, which I found very clarifying when I was learning probability. 8 You have some event, which you typically don't know when occurs, but that can/will occur some time in the future. The time that this event occurs is random, and it is a stopping time if, at any point in time, you know whether the event has occurred or not. A few quick examples. 1) Your own (a stopping time): Let \tau denote the time that I'm ruined (i.e. ... 8 Since \mathbb E[X^2] = \mathbb E[X\mathbb E[Y|X]] = \mathbb E[\mathbb E[XY|X]] = \mathbb E[XY]= \mathbb E[\mathbb E[XY|Y]] = \mathbb E[Y\mathbb E[X|Y]] = \mathbb E[Y^2],  observe that$$ \mathbb E[(X-Y)^2]=\mathbb E[X^2+Y^2-2XY]=0, $$which implies that X and Y differ on a set of measure zero. For the weaker condition where X and Y are ... 8 Consider two normal distributions with the same variance and different means. 8 Use the strong law of large numbers. Choose B\in M so that \lambda(B)\neq \mu(B), and consider the disjoint sets$$\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\}.$$8 Basically, your idea is correct, but you really should try to write this up more formally; otherwise it is hard to tell what you did. By definition, Y = \mathbb{E}(Z \mid \mathcal{A}) if and only if Y is \mathcal{A}-measurable and$$\int_A Y \, d\mathbb{P} = \int_A Z \, d\mathbb{P} \qquad \text{for all} \, \, A \in \mathcal{A}. \tag{1} Now, in the ...

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 ...

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