# Tag Info

0

Since $F$ is right continuous increasing function, $F$ is discontinuous for at most countable points. Moreover $F=f+h$, where $f$ is a continuous increasing function and $h$ is a step function. Thus $h$ is discontinuous for at most countable points. Note that $h$ is Lebesgue integrable and $h'=0$ almost everywhere for $h$ is step function. Since $F'=0$ ...

0

Hint (Which is almost the answer): Imagine the balls being all in a row, with boundaries between them (like in the balls and boxes problems in your handout). Then, there are $n!$ different ways of putting the discrete balls in a row changing their order. But within each box, you don't care about order. Hence you have to divide $n!$ by the permutations of ...

1

Thanks Andre, I found it. For any event A, let IA be the indicator random variable of A, i.e. IA equals 1 if A occurs and 0 otherwise. Then

1

The Chebyshev inequality is $$\mathbb{P}(|x - \mu| \geq a) \leq \frac{\sigma^2}{a^2}$$ .Substituting $$a = k\sigma$$gives the answer.

0

In the Kolmogorov formalism, you could call the entire sample space a "hidden variable", since it is usually not observed directly but only through actual random variables. The reason it doesn't conflict with Bell's inequality is that is is not local -- on the contrary it is the most global hidden variable conceivable, governing both which measurement the ...

0

I think I solved it. First of all we shift and rescale the random variables, so we denote by $Y_i = \frac{X_i + 1}{2}.$ It is not difficult to check that $\mathbb{E}[Y_{i_1} \cdots Y_{i_p}] = \mathbb{E}[Y_{i_1}] \cdots \mathbb{E}[Y_{i_p}]$ and $\mathbb{P}[Y_i = 0] = \mathbb{P}[Y_i = 1] = \frac{1}{2}.$ But on the other hand, $\mathbb{P} [Y_{i_1} = 1, ... 1 Try to use the fact that$B_t/\sqrt{t}$has the same distribution as$B_1$and the fact that$\sup_{s\le t }B_s/\sqrt{s}$is monotonic w.r.t.$t$1 The density function is just the Radon derivative satisfying $$\int 1_A(x)dP_X(x)=P_X(A)=\int_A f(x)dx$$ Then when$g(x)$is simple function we have $$\int g(x)dP_X(x)=\int g(x)f(x)dx$$ By monotone convergence and monotone convergence respectively, the above equation is also true when$g(x)$is positive measurable function and bounded measurable function. ... 1 Let$\{X_n\}$be a Markov chain on$\{0,1,2,3,4,5,6\}$with transition matrix $$P = \begin{pmatrix} 1-p & p & 0 & 0 & 0 & 0 & 0\\ 1-p & 0 & p & 0 & 0 & 0 & 0\\ 0 & 0 & p & 1-p & 0 & 0 & 0\\ p & 0 & 0 & 0 & 1-p & 0 & 0\\ 1-p & 0 & 0 & 0 ... 0 Based on Lost1's comment: In the first place, to have conditional expectation, we need integrability. A product of integrable random variables is not necessarily integrable: Let X, Y \in \mathscr L^{1}(\Omega, \mathscr F, \mathbb P). Consider X and X - Y w/ X having an infinite second moment but finite first moment. Then$$E[X(X-Y)] = E[X^2] - ... -1 Are the flips independent? Consider a probability space$(\Omega, \mathscr F, \mathbb P)$where$\Omega = \{H,T\}^{\mathbb N}$So we have$\omega = (\omega_1, \omega_2, ...)$where$\omega_n \in \{H, T\} \ \forall n \in \mathbb N\mathscr{F} = \sigma(\omega_n = W | \ W \in \{H, T\})$like here (because I guess$\mathscr{F} = 2^{\Omega}$doesn't ... 0 Firstly, If X and Y are random variables with the same distribution, prove that f(X) and f(Y) are random variables that have the same distribution. Secondly, is$h$Borel-measurable on$(\mathbb R, \mathscr B (\mathbb R))$? If the random variables are in$(\Omega, \mathscr F, \mathbb P)$and take values in$(\mathbb R, \mathscr B (\mathbb R))$, I think we ... 0 If we have $$\lim X_n = c \ \text{a.s.}$$ let us take the expectation of both sides to get: $$E[\lim X_n] = E[c] = c$$ Now do we have $$E[\lim X_n] = \lim E[X_n]?$$ Note that boundedness of$|X_n|$'s means that$\exists M > 0$s.t. $$|X_n| \le M \ \forall n \ge 1$$ Note that$M$is integrable$\because \ E[|M|] = E[M] = M < \infty$By the ... 2 Take a look only on one of these restaurants, hence the number of students that enters it distributed Binomial with$n=200$and$p=1/2$. Denote it by$Y$. Then use the continuous correction and the Normal approximation to compute the probability of interest: $$P(Y> 120) = P(Y\ge 120.5) \approx 1 - \phi \left(\frac{120.5 - np}{\sqrt{npq}} \right),$$ ... 2 For a fixed$t$, $$\int_{-\infty}^\infty g(x)\mathrm{d}x=\int_{-\infty}^\infty g(x-t)\mathrm{d}x,$$ as can be verified with a variable substitution for$x-t$. This integral is a constant, so you are indeed justified in this step. 1 I think I figured it out. Since Did's comments below the stated question illustrates how to get $$0<y_1^2<y_2$$ and $$0<y_2^2<y_1.$$ By rewriting the second equation we get $$0<y_2<\sqrt(y_1)$$ which we can add to the first equation above. This gives us: $$0<y_1^2<y_2<\sqrt(y_1)$$ Now, the relationship that${ ...

1

Assume that $|X_n| < C$ for some constant $0 < C < \infty$. Also assume that $Y_n$ converge to $0$ in probability. Let $\epsilon, \delta > 0$ be arbitrary constants. Then for $n$ large enough we have $$\mathbb p \{|X_n Y_n| > \epsilon\} \le \mathbb p \{|Y_n| > \epsilon/C\} \le \delta.$$ Thus $X_nY_n$ converges to $0$ in probability.

1

To address the question of showing $P(B_{k+1}\cap B_k)=1/(k(k+1))$, it may be easier to use a combinatorial argument than to integrate. Suppose our first $k+1$ values for the $Y_i$ are: $y_1,\ldots,y_{k+1}$. Since $Y_i$ have a continuous distribution, these $k+1$ values are distinct almost surely, so we assume that here on. The $Y_i$ are iid so all ...

0

You are right, this probability is not given without any reason. Since $n_i$ is the number of transactions involving $i$ items, the rate at which $n_i$ goes to $n_i+1$ is the rate at which orders involving $i$ products arrive. The total rate at which orders arrive is $\lambda$, and the probability that any order contains $i$ items is $1/M$, so that ...

-1

Interesting. I always thought this was the definition of $\liminf$. The proof so far is right. I would prefer to add to the last sentence 'Since this holds for some $m \ge 1$' and then use m in the indices instead of n. The reverse inclusion can be proven as follows: Suppose $\omega \in \bigcup_{m\ge1}\bigcap_{n\ge m} A_n$. Then $\omega \in ... -1 I'm not sure the problem makes sensea. It seems like you are asked to prove: Given a probability space$(\Omega, \mathscr F, \mathbb P)$, let$A_1, A_2, ...$be events. If$\limsup A_n = \liminf A_n := A$and$\limsup 1_{A_n} = \liminf 1_{A_n}$, then$\lim 1_{A_n} = 1_A$If$\lim 1_{A_n} = 1_A$, then$\limsup A_n = \liminf A_n := A$If we ... 1 What you write doesn't make sense and I think you have a lot of misunderstandings. Your double integral makes no sense, I don't know where you got that. You don't define$T$or$X$. I have no idea where you get the idea of "temporal" or "spatial" random variable. Let me try to clear things up for you.$X_t$is a collection of random variables. Each$t$... -1 We must show two things:$P(A_n) = 1/n$.$(A_n)_{n \ge 1}$is an independent collection of events. From these two, we can use BCL2. The short version is symmetry.$X_n$has the same chance of being the max of$X_1, X_2, ..., X_n$as any of the others Continuity, independence and equal distribution are essential. See the proof here.$A_n$'s are ... 1 I was thinking say$m = 2$and$Y_1 = \infty$, then$\sup_{n \ge m} Y_n < \infty$, but$\sup_{n \ge 1} Y_n = \infty ?$You almost answered you own question. You need further assumption for$Y_n$with$n\geq 2$though. Say,$Y_n\equiv 1$for each$n$. Then the$Y_n$'s are independent. (Why?) Then you can check that the quoted statement is not true. 1 Assuming that$(\sup\limits_n Y_n)(x)=\sup\limits_{n}\{Y_n(x)\}$then indeed, there is nothing guaranteeing that the first random variable$Y_1$should be finite a.s and therefore there is no reason that your claim should be true in general. The example you posited disproves it, but you need to fix values for all the later$Y_n$as well (i.e. with$n\geq2$). ... 1 Let$X, Y$be random variables in$(\Omega, \mathcal B, \mathbb P)$. If$A \in \mathcal B$, then$1_A$and$1_{A^C}$are random variables. Note that $$Z = X1_A + Y1_{A^C}$$ Since sums or products of random variables in$(\Omega, \mathcal B, \mathbb P)$are random variables in$(\Omega, \mathcal B, \mathbb P)$,$Z$is a random variable in$(\Omega, ...

1

'if' Let $A_n^c := \{X_n > M\}$. By BCL1, we have $$P(\limsup A_n^C) = 0$$ $$\to P(\liminf A_n) = 1$$ $$\to \lim P(A_n) = 1$$ $$\to P(\bigcap_{n=1}^{\infty} A_n) = 1$$ $$\to \prod_{n=1}^{\infty} P(A_n) = 1 \ \text{Why?}$$ $$\to \forall n \in \mathbb N, P(A_n) = 1$$ $$\to \forall n \in \mathbb N, P\{X_n \le M\} = 1$$ $$\to P( \sup_{n \ge 1} (X_n) ... 1 Let's use standard notation \partial for boundary. Since B is close set, \bar{B}=B and B^o=(-\infty, x). So \partial B=\bar{B}-B^o=\{x\}. Since P(X\leqslant x) is continuous at x,$$ \lim_{\Delta x\to0}P(X\leqslant x+\Delta x)=P(X\leqslant x) And thus \begin{align} P(X \in \partial B)&=\lim_{\Delta x\to0}P(x<X\leqslant x+\Delta x) \\ ... 1 The first assertion is wrong. For the simplest case, take d = 2, and X_n \equiv (X, X) for n = 1, 2, 3, \ldots, where X \sim U[0, 1]. It can be easily seen that the support of X_n is \{(x, y) \in [0, 1] \times [0, 1]: x = y\}, which has Lebesgue measure 0. Thus X_n cannot converge weakly to a U([0, 1] \times [0, 1]) random vector, whose ... 0 Let p \equiv P(S). The probability of all n draws satisfying the condition is, in fact, p^n. The probability of exactly k out of the n draws satisfying the condition is given by\binom n k p^k(1-p)^{n-k}$$In particular, the probability of exactly 1 success is given by$$ \binom n 1 p(1-p)^{n-1}$$Finally the probability of at least one ... 1 Correct. The presence of a density function implies that the distribution is absolutely continuous with respect to Lebesgue measure, and in that case the density (i.e., the Radon-Nikodym derivative of the distribution) is only defined up to sets of Lebesgue measure zero. 1 Since p>\frac{1}{2}, we can write p= \frac{1}{2}+q for some q>0. Now$$\frac{(n-1)^p}{\sqrt{n}} = \sqrt{\frac{n-1}{n}} (n-1)^q \geq \frac{1}{2} (n-1)^q, \qquad n \in \mathbb{N},$$implies$$\mathbb{P} \left(B_1 > C \frac{(n-1)^p}{\sqrt{n}} \right) \leq \mathbb{P} \left(B_1 > \frac{C}{2} (n-1)^q\right).$$Choose k \in \mathbb{N} ... 1 You want to know something about \sum_iX_i. However, X_i has a different distribution for different values of i, so this is difficult. Estimating \sum_i X_i+Y_i is easier, since X_i+Y_i has a Bin(n,p) distribution for all i. So Y_i is just an auxiliary variable to show that X_i is dominated by a Bin(n,p) variable. I would say that ... 0 It seems to me the answer should be about \frac{1}{8}. You're correct that the valid potential lengths are 14:01 - 15:59 and 15:01 - 16:59 for day 1 and day 2, respectively. Just to make the numbers nicer let's recenter to 0, so that the times are 0:00 - 1:58 for day 1 and 1:00 - 2:58 for day 2. Note the possible time in seconds for day 1's length is ... 0 Let S_n = X_1 + \cdots + X_n. Since N is a.s.-finite, it follows that$$ S_N = S_N \sum_{n=0}^{\infty} \mathbf{1}_{\{N = n\}} = \sum_{n=0}^{\infty} S_N \mathbf{1}_{\{N = n\}} = \sum_{n=0}^{\infty} S_n \mathbf{1}_{\{N = n\}} \qquad \Bbb{P}\text{-a.s.} (This really shows that \Bbb{E}[S_N | N] = N \Bbb{E}[X_1], but we need to check integrability ... 1 \begin{align} \mathsf E(\sum_{n=1}^N X_n) & = \mathsf E(\mathsf E(\sum_{n=1}^N X_n\mid N)) & \textsf{Iterated Expectation} \\[1ex] & = ~ & \textsf{Linearity of Expectation} \\[1ex] & = ~ & \textsf{Independence of \{X_n\} from N} \\[1ex] & = ~ & \textsf{Identical Distribution of \{X_n\}} \\[1ex] & = ~ & ... 2 Hint:E \left [ \sum_{i=1}^N X_i \right ] = \sum_{n=1}^\infty E \left [ \left. \sum_{i=1}^N X_i \right | N=n \right ] P(N=n) \\ = \sum_{n=1}^\infty \sum_{i=1}^n E[X_i|N=n] P(N=n).$$Can you compute the inner expectation? 1 The first Borel-Cantelli lemma yields$$\mathbb P\left(\limsup_{n\to\infty}|X_n|>n\right)=0. $$As for each n$$\{|X_n|>n\}\subset \bigcup_{k=n}^\infty \{|X_k|>k\}, it follows that \begin{align} 0&=\mathbb P\left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty \{|X_k|>k\}\right) \\&=\mathbb P\left(\lim_{n\to\infty}\bigcup_{k=n}^\infty ... 0 I used BCL1 to deduce thatP(\liminf [X_n = 2]) = 1$$Let \omega \in \Omega. If \omega \in \liminf [X_n = 2], then \exists m \ge 1 s.t.$$2 = X_m(\omega) = X_{m+1}(\omega) = ...\to S_{m+k}(\omega) = S_m(\omega) + 2k \ \forall k \ge 0\to \lim _{k \to \infty} S_{m+k}(\omega) = \infty\to \lim _{k \to \infty} S_{k}(\omega) = \infty... 0 An example is a random variable X having a student-t distribution with \nu = 2 degrees of freedom Its E[X] = 0 for \nu > 1. Its E[X^2] - Var[X] = \infty for 1 < \nu \le 2 1 We have four times involved so define an r.v. for each. The range for each of the times is 1 minute and it is simpler to use minutes as our units. We define the random variables as: \begin{align} X &= \text{Day 1 Start Time - 5:31pm (in minutes)} \\ Y &= \text{Day 2 Start Time - 5:31pm (in minutes)} \\ Z &= \text{Day 1 End Time - 5:46pm (in ... 3 Because t is fixed and will play no role, I'll drop it from the story. Thus we have an integrable random variable X such that E[\varphi(X)|\mathcal F]=\varphi(E[X|\mathcal F]) for all bounded and continuous \varphi. I will allow \varphi to be complex-valued. (Look at the real and imaginary parts of \varphi separately and then add.) Let's define ... 0 Hint: Consider n^{-1}\log Z_n as n\to\infty. 1 To use the Radon-Nikodym theorem we need that: (1) P_{(X,Y)} \mbox{ is absolutely continuous w.r.t. the Lebesgue measure} \lambda^2 on \mathbb{R}^2 (2) P_{(X,Y)} is \sigma-finite which is satisfied since it is even finite: P_{(X,Y)}(\mathbb{R}^2) = 1 < \infty. (3) The Lebesgue measure \lambda^2 is \sigma-finite, for this we define for ... 0 If an arbitrary centered Gaussian measure is given on a product of a Banach space X with itself, then it is not guaranteed to be the tensor product of two centered Gaussian measures on X. As an example, consider the one-dimensional real Banach space R. On the product space R x R (the Euclidean vector plane) consider the measure with density (up to some ... 1 Let g:\mathbb R^2\to\mathbb R be the map (a,b)\mapsto a+b. Clearly for any t\in\mathbb R,g^{-1}((-\infty,t])=\{(a,b)\in\mathbb R^{2}:a+b\leqslant t\} is a measurable set. Therefore $g$ is a measurable map, so $X+Y=g(X,Y)$ and $\sigma(g(X,Y))\subset \sigma(X,Y)$. It follows that if $E\in\sigma(g(X,Y))$, $F\in\sigma(Z)$, then $E\in\sigma(X,Y)$ so ...

3

Let $U$ be a random variable which is uniform on $(0,1]$. Let $X=Y=U^{-1/2}$. Then $E[X]=E[Y]=2$ but $E[XY]=+\infty$. You can make lots of examples of functions like this, which are not integrable but their square root is integrable, because it diverges "more slowly". On an infinite measure space you can also have the opposite phenomenon with tailing: ...

1

No: let $X=\pm 1$ with probability $1/2$, $X_n:=(-1)^nX$ and $Y_n:=X$ for each $n$. Then $X_n\to X$ in distribution, $Y_n \to X$ in probability but $X_nY_n=(-1)^n$ which does not converge (in any sense).

0

Along the lines of my last comment in your previous question, you could do a reverse convolution type thingy: Suppose we have a renewal process $N(t)$ with rate $\lambda>0$. Let $X$ be a random variable with the same distribution as the inter-arrival times of $N(t)$. Then $E[X]=1/\lambda$. Fix $\epsilon>0$. We want to add another process $A(t)$ ...

0

The problem (as was pointed out by @PhoemueX) is that there is no warranty that $D_n$ are in the semi-algebra. Consider the following example: \begin{array}{lll} B_1 & = & \left[ \frac{1}{3},\frac{3}{4} \right], \\ B_2 & = & \left[ 0,1 \right]. \end{array} Where clearly, \$B_1,B_2 \in \mathcal{J} = \left \{ \textrm{ all intervals in } [0,1] ...

Top 50 recent answers are included