# 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 $\frac{1}{200}$.

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

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

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

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

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

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

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 $$f(\alpha)=\mathbb{E}[1_{\{X > 0\}}e^{\alpha U}],\ g(\alpha)=\mathbb{E}[1_{\{Y > 0\}}e^{\alpha ... 9 This is false, when \lim_{n \to \infty} p_n doesn't exist. For instance, consider p_{2n-1} = \dfrac1{2^n} and p_{2n} = 1-\dfrac1{2^n}. This gives us \sum_{n=1}^{2N} p_n = \sum_{n=1}^{2N} (1-p_n) = N, whereas$$\sum_{n=1}^{2N} p_n(1-p_n) < \sum_{n=1}^N \dfrac1{2^{n-1}} < 2$$However, the claim is true when \lim_{n \to \infty} p_n exists. ... 9 Assuming that by "number" you mean "integer," and by "random" you mean "uniformly at random," AND ASSUMING THAT THE CHOICES ARE INDEPENDENT, then the desired probability is$$\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

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

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

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

8

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 ...

8

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, ... 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\}.$$7 The vector \vec X = [X_1,\dots,X_N]^T has a rotationally invariant distribution. That is, if A is any orthogonal matrix, then the distribution of \vec X and A \vec X are the same. Hence by letting A be an orthogonal matrix that takes [1,\dots,1]^T to [\sqrt N,0,\dots,0]^T, your problem is the same as computing$$ \Pr(\sqrt N X_1 \mid \sum ...

7

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

The probability is $0$, assuming that each point is chosen with uniform distribution on the circle. For a right-angled triangle you need two of the points to be the two endpoints of a diameter. The probability for that is obviously $0$, since there isn't any range of possibilities, and the distribution is continuous. For anyone who is not satisfied so ...

7

Draw a Venn diagram to help you understand $P(A|B)=P(A\cap B)/P(B)$. Then use this to relate the quantities $P(A|B)$ and $P(B|A)$ algebraically. Let's discuss the first point. Suppose we have a finite sample space so we can count the number outcomes in each possible "event." To determine $P(A|B)$, we're essentially asking what the probability of getting an ...

7

It can be proved by induction that $x_n = \frac{1}{(n-1)!}$ $x_1$ and $x_2$ satisfy this, and by doing induction, $$x_{m+1} = x_{m}-\frac{1}{m}x_{m-1} = \frac{1}{(m-1)!} - \frac{1}{m}\frac{1}{(m-2)!} = \frac{1}{m!}$$ Thus you series reduces to the ordinary series $\sum_{n=0}^{\infty}\frac{1}{n!}$ for $e$

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