# Tag Info

4

The field of probability can be made mathematically rigorous. Introductory textbooks on probability tend to use terminology like 'sample space', 'outcome', 'event', and outcomes are given names like 'Heads' or 'HTT' or 'King of Spades', all in an attempt to keep things informal and intuitive. These texts often refrain from defining such concepts precisely, ...

4

There is a mistake in the definition, it should surely be $\rho(C) = \lim \limits_{n \to \infty} \frac{1}{n} \# \{k \mid 1 \le k \le n, k \in C\}$. You use the index $n$ twice, when you should actually use two different indices. $$\rho(C_n) = \lim \limits_{m \to \infty} \frac{1}{m}\#\{k \mid 1 \le k \le m, k \in C_n\}$$ Now note that $$\frac{1}{m}\#\{k ... 4 The CDF F of a random variable X is unique, since it is defined as$$F(x) = P(X \le x).$$3 The first thing you must do is find a,b from the given mean (7) and variance (4). Only then can you try to find \mathsf P(a\leq X\leq 6\mid 4\leq X\leq b) The mean and variance of a uniform discrete distribution, X\sim\mathcal U\{a..b\} are:$$\mathsf E(X) = \frac{a+b}{2} = 7\\\mathsf{Var}(X)= \frac{(a-b+1)^2-1}{12} = 4$$3 Take an element i. Let p_i denote the probability that i is not in the intersection of your two randomly chosen sets. It is in exactly 2^{n-1} subsets, hence the probability that it is in a randomly chosen subset is \frac 12. It follows that the probability that it is in both of two randomly chosen subsets is \frac 14. Therefore p_i =\frac ... 3 The integral is defined as the supremum of functions of bounded functions with finite support. I would work directly from this definition to show the result. If your function had been bounded and of finite support, then your above analysis is almost complete. Each f_n would be composed of a finite set of functions that as n goes to infinity, ... 2 The L.H.S. is increasing with n Use the equality$$E_{m,n}=E_{2m,n+1} \biguplus E_{2m+1,n+1}$$to get$$\sum_{m=1}^{\infty }\frac{m}{2^n} \mu (E_{n,m}) \le \sum_{m=1}^{\infty }\frac{m}{2^{n+1}} \mu (E_{n+1,m}).$$The limit is indeed the R.H.S. Prove that if f is any simple function, you will have \sum_{m=1}^{\infty }\frac{m}{2^n} \mu (E_{n,m}) \ge ... 2 Yes; though not in general, it is the case for identical and independent distributions (iid). (Sorry, missed that in the first reading.)$$\mathsf E[V_3\mid V_1<\max\{V_2,V_3\}, V_3=\max\{V_2, V_3\}] \;=\; \mathsf E[V_3\mid V_3=\max\{V_1,V_2, V_3\}]$$This is not necessarily the same thing as \mathsf E[\max\{V_1,V_2, V_3\}] Indeed:$$\begin{align} ...

2

Although it cannot be seen clearly, I think the steps you have problem with is this, $$\int{(f+g)}d\mu=\inf_{\psi\leq f+g}\int\psi d\mu$$ $$\leq\inf_{\psi_1\leq f, \psi_2\leq g}\left(\int\psi_1 d\mu+\int\psi_2 d\mu\right).$$ Just note that $\psi_1\leq f$ and $\psi_2\leq g \implies\psi=\psi_1+\psi_2\leq f+g$. Set-theoretically, we have the following ...

2

Note that $\mathbb{E}[f]$ is a constant (assuming it exists). All constant functions are measurable with respect to ANY $\sigma$-algebra. It has nothing to do with $f$ being independent of the $\sigma$-algebra.

2

Not sure if this is what you are asking, but: Given any two members of the equivalence class $E(X|\mathcal N)$, each is measurable with respect to $\mathcal N$, so the set where the members differ (which has probability zero) belongs to the sigma algebra $\mathcal N$.

2

You could require $f$ to only be continuous, but the idea is to have as few test functions as possible to satisfy the definition. Another way to say this is, the above possible values for $\{...\}$ are equivalent to the requirement that the equality holds for continuous functions. Here is a proof of that for the continuous bounded characterization. On one ...

2

I think you misspoke somewhat in your first statement: first hitting times are stopping times, but there are many other kinds of stopping times. What you've written is a last exit time, which is not a stopping time, because as you've said we cannot know whether $\tau \leq t$ by observing the process up to time $t$. This creates some difficulties in ...

2

Consider a "spinner": an object like an unmagnetized compass needle that can pivots freely around an axis, and is stable pointing in any direction. You give it a spin and see where it comes to rest, measuring the resulting angle (divided by $2\pi$) as a number from $0$ to $1$.

2

Take a look further to the front of the book, maybe you can find the definition of the generated sigma-algebra that the author uses. Typically, the generated sigma-algebra is defined just in the way you described. In this case you only need to prove the second assertion of this exercise. However, there are also other definitions. One possible definition is ...

2

Without brute force: \begin{aligned} E[X_1X_2|X_2X_3] &= E[E[X_1X_2|X_2X_3,X_2]|X_2X_3] \\ &= E[X_2E[X_1|X_2X_3,X_2]|X_2X_3] \\ &= E[X_2E[X_1]|X_2X_3]\\ &= E[X_2\cdot 0 |X_2X_3]\\ &=0 \end{aligned} First equation is the tower property for conditional expectations, ...

2

I don't understand why you must split up the integral, why not say $$\int_{-\infty}^x f(u) du = \int_{-1}^x f(u) du = \left.\frac{u + \alpha u^2/2}{2}\right|_{-1}^x ...$$

2

You just need to look at more stopping times. For any $t\geq 0$, you have $${E}[\mathbf{1}_{\{\tau_0<T\}}X(T \wedge (\tau_0+t))] ={E}[\mathbf{1}_{\{\tau_0<T\}}E[X(T\wedge (\tau_0+t))|F_{\tau_0}]] \le {E}[\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)]=0.$$ This gives $$P\biggl(X(T\wedge(\tau_0+t))=0\mbox{ for all }t\in\mathbb{Q}\cap[0,T]; \tau_0<T\biggr) ... 2 You can generalize your derivation from:$$E[\mathbf{1}_{\{\tau_0<T\}}X(T)]=E[\mathbf{1}_{\{\tau_0<T\}}E[X(T)|F_{\tau_0}]]\le E[\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)]=0$$to:$$E[\mathbf{1}_{\{\tau_0=s\}}X(t)]=E[\mathbf{1}_{\{\tau_0=s\}}E[X(t)|F_{\tau_0}]]\le E[\mathbf{1}_{\{\tau_0=s\}}X(\tau_0)]=0$$where t > s. As you mentioned, using the fact ... 2 An atom in a measurable space (M,\mathcal{M}) is a nonempty set A\in\mathcal{M} such that B\in\mathcal{M} and B\subseteq A implies that either B=\emptyset or B=A. If \mathcal{M} is generated by a countable family of sets, so it is countably generated, every point in M is in a unique atom and the atoms form a partition of M. A countably ... 2 As suggested by @Dominik, we can adopt the usual proof to get the inequality even for non-centered random variables. The point is that the expectation "magically" cancels out. We assume (as before) that$$ \mathbb{E}\left|X_{\ell}\right|^{n}\leq\frac{n!}{2}\cdot R^{n-2}\cdot\sigma_{\ell}^{2} $$for all 1\leq\ell\leq m and n\geq2. In one direction, we ... 2 The vector \vec X = [X_1,\dots,X_N]^T has a rotationally invariant distribution. That is, if A is any orthogonal matrix, then the distribution of \vec X and A \vec X are the same. Hence by letting A be an orthogonal matrix that takes [1,\dots,1]^T to [\sqrt N,0,\dots,0]^T, your problem is the same as computing$$ \Pr(\sqrt N X_1 \mid \sum ...

2

For continuous random variables, the probability of a tie is immeasurably small.   So we can say: $$V_1<\max\{V_2, V_3\} \iff V_2=\max\{V_1,V_2, V_3\} \cup V_3=\max\{V_1, V_2, V_3\}\quad\text{a.s.}$$ Since the events in this union are disjoint (almost surely), then they partition the conditioned space: \begin{align}\mathsf E[\max\{V_2,V_3\}\mid ... 2 The random variable is not limited to values less than x. For instance, I can show you that the function X(t) = \frac{1}{t} $$is a measurable function on (0, 1). Here's how. Let's look at$$ \{ t \mid X(t) \le 11 \} $$That's the set of all points in the domain for which X(t) = 1/t is less than 11, which is exactly$$ A = \{t \mid \frac{1}{11} ...

2

A random variable $X$ on $\Omega$ is no more and no less than a function $X:\>\Omega\to{\mathbb R}$ satisfying the technical condition that it is measurable: For any $x\in{\mathbb R}$ the set $\{\omega\in\Omega\>|\>X(\omega)\leq x\}$ belongs to ${\cal F}$. This guarantees that for any two given values $a$, $b$ the probability $$P[a\leq X(\omega)\leq ... 1 The actual definition is that a function X:\Omega\to\mathbb R is called a random variable when X is (\mathcal F, \mathcal B(\mathbb R))-measurable. In other words, X^{-1}(B)\in\mathcal F for any B\in\mathcal B(\mathbb R), where \mathcal B(\mathbb R) is the Borel \sigma-algebra on \mathbb R. Now, it is sufficient that ... 1 I think for any realization of random variable :  X_{1},...,X_{n}, the empirical distribution F_{n}(x)=\frac{1}{n}\sum_{i=1}^{n}I_{\{X_{i}\le x\}} is just a discrete distribution function concentrated on these n value and attached weight \frac{1}{n} to each of them. Thus, the integral \int g(x)dF_{n}(x) is same as expected value of g(x) with ... 1 The player meets monsters at the rate \lambda, and with probability p they are winning monsters, so the player meets winning monsters at the rate \lambda p. We can extend the game up to t=1, with monsters appearing after the player's death meeting a dead player. Then, since the Poisson monsters appear uniformly in time, the probability distribution ... 1 Hint: the probability that X+Y\equiv n\pmod{10} is$$ \sum_{k=0}^9\Pr[X=k]\Pr[Y\equiv n-k\pmod{10}] $$Now consider the sum$$ \sum_{k=0}^9\Pr[Y\equiv n-k\pmod{10}] $$1 You have been given that$$f_{X,Y}(x,y) = \begin{cases}4xy & : 0 \leq x \leq 1,\; 0 \leq y \leq 1\\ 0 & : \text{ elsewhere}\end{cases}$$You wish to know where f_{X,Y}(x, z-x) is supported (ie: not zero) with respect to x, for values of z where 0\leq z\leq 2. (Since z=x+y, then 0+0\leq z\leq 1+1.)$$f_{X,Y}(x,z-x) = ...

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