Tag Info

New answers tagged

1

I suggest you calculate the cumulative distribution function of $\frac{k}{\hat{k}}$.


1

These questions originated from an assignment (due tomorrow) of UNSW - course code, MATH2901. Please put this on hold. Thanks.


1

By definition: $Pr(A|B) := \frac{Pr(A\cap B)}{Pr(B)}$. By multiplying each side by $Pr(B)$ we get the multiplication principle and our base case for induction: $$Pr(A\cap B) = Pr(A|B)\cdot Pr(B)$$ Or, equivalently, by the commutativity of $\cap$ and $\cdot$, a more convenient form for our purposes: $$Pr(A\cap B) = Pr(A)\cdot Pr(B|A)$$ Suppose for our ...


2

Yes, I think your thought is correct. They want to see an inductive proof using the definition of conditional probability.


3

The first part is right, but the second part is wrong (although not an absolutely terrible estimate.) Basically, when you do the trial $n$ times, the value you computed is the expected number of occurrences of HAMLET. This is a case where using the negative probability is better. The probability that you do not get HAMLET from one monkey is: ...


1

Since $g$ is nonnegative, $g(X)\geqslant0$ almost surely. So by Markov's inequality, $$ \mathbb P(g(X) > r) \leqslant \mathbb P(g(X) \geqslant r) \leqslant \frac{\mathbb E[g(X)]}r.$$ If you aren't familiar with Markov's inequality, I will include a proof. Suppose $Y$ is a random variable with $\mathbb P(Y\geqslant 0)=1$ with $\mathbb E[|Y|]<\infty$, ...


1

The first statement is written incorrectly. Here is the correct result, which follows from the inclusion-exclusion principle; $B = A_1\cup A_2\cup A_3$, so $$\begin{align*} 1_B &= 1_{A_1\cup A_2\cup A_3} \\ &= 1_{A_1} + 1_{A_2} + 1_{A_3} - (1_{A_1\cap A_2} + 1_{A_1\cap A_3} + 1_{A_2\cap A_3}) + 1_{A_1\cap A_2\cap A_3}\\ &= \sum_{i=1}^3 1_{A_i} - ...


1

Let pdf of $Y=g(X)$ be $h(y)$. $\begin{array}{rcl}\mathbb{E}Y&=&\int_0^\infty y\ h(y)\ dy\\ &=&\int_0^r y\ h(y)\ dy+\int_r^\infty y\ h(y)\ dy\\ &\le&\int_0^r y\ h(y)\ dy+\int_r^\infty r\ h(y)\ dy\\ &=&\int_0^r y\ h(y)\ dy+r P(Y\ge r)\\ &\le&r P(Y\ge r)\\ \end{array}$ So $P(Y\ge r)\le \frac{\mathbb{E}Y}{r}$


1

To be a lot simpler than the other answers, just use the fact that the sum of two Gaussian random variables is gaussian and the fact that the scalar multiple of a Gaussian random variable is also Gaussian. The first of these facts can be worked out by computing the convolution of two arbitrary Gaussian distributions. The second requires some thinking about ...


1

Yes. Write this in matrix form, so you have $\mathbf {AX} = \mathbf Z$ where $\mathbf Z$ is bivariate normal. Then $\mathbf X = \mathbf {A^{-1}Z}$. The RHS is a linear combination of Gaussians, therefore the LHS is Gaussian.


2

Let $U$ be standard normal. Let $V=RU$, where $R$ is a Rademacher random variable that takes values $-1$ and $1$, each with probability $1/2$. Suppose $U$ and $R$ are independent. Let $X=\frac{U+V}{2}$ and $Y=\frac{U-V}{2}$. Then $X+Y$ and $X-Y$ are each normal. But $X$ is not normal, for it takes on value $0$ with probability $\frac{1}{2}$. Remark: ...


0

Im not sure whether I fully understood what you are asking, but note that the limit you have must necessarily be from the right (i.e $\lim s \downarrow t$), so the paths are almost surely right-continuous. So how can the process have jumps on almost every path despite the condition? Well, look at for example the NIG process which has jumps at every instant ...


2

Let $(\Omega,\mathcal A,P)$ be a probability space and let $\mathcal B$ denote the Borel $\sigma$-algebra on $\mathbb R$. Any random variable $X:\Omega\rightarrow\mathbb R$ induces a probability $P_X$ on measurable space $(\mathbb R,\mathcal B)$. This by: $$P_X(B)=P(\{X\in B\})$$ for $B\in\mathcal B$. This $P_X$ is the distribution of $X$ and is ...


0

$\newcommand{\E}{\operatorname{E}}$If $\E(|X|)<\infty$ then, since $\lceil |X|/c\rceil\le |X|/c+1$, you have $\E\lceil |X|/c\rceil <\infty$. Let $W=\lceil |X|/c\rceil$. Then use the fact that for a positive integer-valued random variable $W$, $$ \E W = \sum_{w=1}^\infty \Pr(W\ge w). $$ A somewhat similar argument handles the "only if" case.


2

Let $\varepsilon>0$, then since $\int_{\Omega}|X|dP<\infty$, there exists an $n>0$ such that: $$nP(|X| > n) = \int_{\{\omega\in\Omega : |X(\omega)|>n\}}ndP \leq \int_{\{\omega\in\Omega : |X(\omega)|>n\}}|X(\omega)|dP < \varepsilon.$$ Therefore we have $$nP(|X| > n) < \varepsilon.$$


2

$$\int_\Omega |X|\, dP < \infty \implies \int_{\{|X| > n\}} |X|\, dP \to 0$$ by the dominated convergence theorem. Since $nP(|X|>n)$ is bounded above by the last integral, we have it.


0

Use the fact that $\mathbb P[(A \cup B)^c] + \mathbb P[A] + \mathbb P[B] - \mathbb P[A \cap B] = 1$. In particular, $$\mathbb P[A \cap B] = 0.95+0.03+0.04-1= 0.02$$ so you can plug this into your method and you are done.


0

$P(A\cap B)=P(A\cup B)-P(A)-P(B)$ then you can the fact that 95% of the goods are not deflected.


1

The transition matrix is an $N \times N$ matrix given by $\mathbf{P} = \begin{pmatrix} 1/2 & 0 & 0 & \cdots & 0 &1/2 \\ 1/2 & 1/2 & 0 &\cdots &0 &0 \\ 0 & 1/2 & 1/2 & \cdots &0&0 \\ \vdots & & \ddots & \ddots && \vdots \\ \vdots &&&\ddots&\ddots &\vdots\\ 0 ...


1

If there are currently $0$, then you either go to $N$ with probability $1/2$ or stay at $0$ with probability $1/2$. If there are currently $0<n\leq N$, then you either go to $n-1$ with probability $1/2$ or stay at $n$ with probability $1/2$. (I honestly don't know how to give just a hint for this part, since I basically just read off the content of the ...


-3

If two RVs $X,Y$ are independent we have the following relation: $$E[XY]=E[X]E[Y]$$ So from what we've got here we can already see that $e^{it_1 X_1}$ and $e^{it_2X_2}$ are independent. These are the moment generating functions and for independent RVs we know that the moment generating functions have the following property $M(X+Y) = \langle e^{t(X+Y)} ...


1

Some steps: The intersections of two algebras $\mathcal F_1$ and $\mathcal F_2$ still is an algebra. Indeed, the whole set belongs to the intersection of the algebras, as well as the complement of an element of $\mathcal F_1\cap \mathcal F_2$. Stability by finite intersections also holds. Therefore, the question is actually equivalent to the following ...


1

No, continuity on $\bigcup_{n \in \mathbb{N}} \text{Img}(X_n)$ is not enough. (Counter)Example: Consider $([0,1],\mathcal{B}[0,1])$ endowed with the Lebesgue measure, $$g(x) := 1_{\{0\}}(x)$$ and $$X_n := \frac{1}{n}.$$ Then $X_n \to X := 0$ almost surely, $g$ is continuous on $(0,\infty) \supseteq \bigcup_n \text{Img}(X_n)$, but $g(X_n)=0$ does not ...


1

$$ \hat k >x \text{ if and only if }[X_1>x\ \&\ \cdots\ \&\ X_n>x] $$ and the probability of that is $\Big(\Pr(X_1>x)\Big)^n$. $$ \Pr(X_1>x) = \int_x^\infty \frac{\alpha k^\alpha}{u^{\alpha +1}} \, du = \left(\frac k x\right)^\alpha, $$ so $$ \Pr(\hat k>x) = \left(\frac k x\right)^{n\alpha}. $$ Thus $$ \Pr(x_1<\hat k < x_2) = ...


4

Suppose $i$ is an integer between $1$ and $24$, and let $A_i$ be the event that you roll an $i$ at least $4$ times in ten rolls. Notice that two of the $A_i$ could occur, but not three, since that would be twelve different rolls. By the inclusion/exclusion principle, you get that $$\mathbb{P}(\cup A_i) = \sum \mathbb{P}(A_i) - \sum_{i\ \neq j}\mathbb{P}(A_i ...


0

Outline: We use Inclusion/Exclusion. Find in one of the usual ways the probability of say $1$ coming up $4$ or more times. Multiply by $24$, and call the result $a$. However, this double-counts the ways in which $2$ faces come up $4$ or more times. (We cannot have more than $2$ faces coming up $4$ or more times.) So we will have to subtract the ...


1

Continuity and the local martingale property should (hopefully) be clear. For the quadratic variation/covariation, look at the quantities $(dW_1)^2$, $(dW_2)^2$, $dW_1dW_2$. The first two should be $dt$, and the third $0$. (Use the fact that $dB_1dB_2=\rho_t$). Then integrate to get the required answer.


0

The even $p$ norms corresponds directly to the $p$th moment of the random variable. $$E[X^p] = ||X||_p^p$$ The association for odd $p$ is not as clear because the definition of the norm requires taking the absolute value of the outcomes. At best you can note: $$E[|X|^p] = ||X||_p^p$$ In this way, for a symmetric distribution, you can ...


1

One of the characterizations of Multidimensional Brownian Motion $B_i(t)$ is that the covariations satisfy: $$[B_i, B_j]_t = \delta_{ij}t$$


0

You have the right idea, except for a little issue with notation (which is likely to have been the cause of your confusion): you should be looking at a sequence of functions of $t$, i.e. $$M_{Y_n}(t)=\sum_{k=1}^n e^{tk/n} \cdot \frac{1}{n}$$ which does not go to $0$. This is a geometric series which would converge to the mgf of $U$, but to do this without ...


2

We can show and use the following Lemma. Let $(X_n)_{n\geqslant 1}$ be a sequence of random variables such that $X_n\to X$ in probability and the cumulative distribution function of $X$ is continuous. Then for each $t\in\mathbf R$, the following convergence holds: $$\lim_{n\to\infty}\mathbb P(X_n\leqslant t)=\mathbb P(X\leqslant t). $$


1

Hint: Satisfaction of condition $X_2=m$ means that a number in $\{1,2,3,6\}$ was thrown exactly $n-m$ times. What is the probability that $k$ of these times it was a number in $\{1,2,3\}$?


2

$P(X=5)=\dfrac{\binom{4}{4}}{\binom{100}{5}}$ $P(X=6)=\dfrac{\binom{5}{4}}{\binom{100}{5}}$ $P(X=7)=\dfrac{\binom{6}{4}}{\binom{100}{5}}$ $\dots$ And in general: $$\forall{n\in[5,100]}:P(X=n)=\dfrac{\binom{n-1}{4}}{\binom{100}{5}}$$ In words: Take ball #$n$, and choose another $4$ balls out of balls #$1,\dots,n-1$.


2

Count how many ways $n$ can be the largest number. If you replace the balls, there are $n^5$ ways they can be $\leq n$, minus $(n-1)^5$ ways they are all less than $n$. If you don't replace the balls, there are $n-1\choose4$ ways that the largest is $n$.


1

example Without assuming some "joint" measurability of $X_t(\omega)$ in $t$ and $\omega$ you are out of luck. We use the Continuum Hypothesis. There is a set $A \subseteq [0,1] \times [0,1]$ such that: $\qquad$For each $t \in [0,1]$,$\qquad \{\omega \in [0,1] : (t,\omega) \in A\}$ is countable, $\qquad$For each $\omega \in [0,1]$,$\qquad \{t \in ...


1

All of these statements can be false. What follows is more or less the standard counterexample of a local martingale that is not a martingale. I've stolen the details from an MO post of mine which constructs something slightly different. (To avoid search-and-replace I'm keeping my process called $Y$ instead of $\beta$.) Set $T=1$. Let $r(t)$ be any ...


1

Set $\DeclareMathOperator \gcd{gcd}$ $$N_x := \{n \in \mathbb{N}; p^n(x,x)>0\} \qquad \quad d_x := \gcd(N_x).$$ state 4: Since $p(4,4) = \frac{1}{2}>0$, it follows that $p^n(4,4)>0$ for all $n \in \mathbb{N}$; hence, $N_4 = \mathbb{N}$ and $d_4 = \gcd(\mathbb{N})=1$, i.e. state $4$ is aperiodic. state 3: We have $$p^2(3,3) = \mathbb{P}^3(X_2 = ...


0

"Is this even right?" Yes. {A2r∣r=k,k+1,…}⊆{Ar∣r=k,k+1,…} and consequently σ({A2r∣r=k,k+1,…})⊆σ({Ar∣r=k,k+1,…}). – drhab 2 days ago Thanks @drhab (again) !! Btw, any idea if the proof you gave implies, or at least, has some relation to, lim supA2n⊆lim supAn ? – BCLC 2 days ago 1 up voted {A2r∣r=k,k+1,…}⊆{Ar∣r=k,k+1,…} implies ⋃r≥kA2r⊆⋃r≥kAr. This for every ...


2

The problem is equivalent to Coupon collector's problem. The expected value is $$\mathbb{E}(X) = N H_N \approx N \, ln \, N$$ where $H_N$ is $N$-th harmonic number. Here $\mathbb{E}(X) \approx 2364.64$ The idea in solving this problem is calculating the expected number of people such that the number of different birthdays we have written increases ...


0

If $Z$ is independent of $X$ and independent of $Y$, then $(X,Z )\sim (Y,Z)$. Indeed, if $s,t\in\mathbf R$, then \begin{align} \mathbb P(X\leqslant s,Z\leqslant t)&=\mathbb P(X\leqslant s)\mathbb P(Z\leqslant t)\quad\mbox{ since $X$ and $Y$ are independent}\\ &= \mathbb P(Y\leqslant s)\mathbb P(Z\leqslant t)\quad \mbox{ because }X\sim Y \\ ...


1

You got indeed the idea, and your proof looks correct. Some comments: In the second line of your attempt, the expression of the p.d.f. of a uniform distribution in the unit interval is not correct if $t\notin [0,1]$. How do you define the random variable $f(x)$? Do you mean $f(U_1)$? Actually, you don't need to use the weak law of large numbers, but ...


4

As you have guessed, the answer is NO in general. And a single counter example is enough: take $\{X_n\}$ i.i.d. uniform $(0,1)$ then $|X_n|< n$ for all $n\geq 1$ so that $$ Y_n=X_n\implies\text{Var}{Y_n}=\text{Var}{X_n}=\frac{1}{12}\implies\sum_{n=1}^\infty\frac{\text{Var}{Y_n}}{n}=\frac{1}{12}\sum_{n=1}^\infty\frac{1}{n}=\infty. $$


1

Expanding on @Did's comment, we have $$ \psi_{X_\lambda}(t) = e^{\lambda\left(e^{itb(\lambda)}-1-itb(\lambda)\right)} $$ For sufficiently large $\lambda$, this is approximately equal to $$ e^{\lambda\left(-\frac12 t^2b^2(\lambda)\right)}.$$ Choosing $b(\lambda) = \lambda^{-\frac12}\sigma$ (where $\sigma>0$), we have ...


1

Since the function $m'_X$ is continuous, we can find $\delta\lt h$ such that if $|s|\lt 2\delta$, then $m'_X(s)-m'_X(0)\leqslant -\mathbb E(X)/2$ (definition of continuity with $\varepsilon:=-\mathbb E(X)/2\gt 0$). We thus have $$|s|\lt 2\delta\Rightarrow m'_X(s)\leqslant \frac{ \mathbb E(X)}2,$$ hence for $0\lt t\lt\delta$, we have $$\tag{*} ...


1

Define $$A_n := \left\{ \left| \frac{X_n}{n} \right| \geq 1 \right\}.$$ The set $$A := \limsup_{n \to \infty} A_n$$ satisfies, by the Borel-Cantelli lemma, $\mathbb{P}(A)>0$ and $\frac{X_n(\omega)}{n}$ does not converge to $0$ for each $\omega \in A$.


0

Write $S_n=X_1+...+X_n$, where the $X_i$'s are your i.i.d. jumps with $\mathbb E[X_i]=0, \mathrm{Var}(X_i)=\sigma^2$ for $i=1,...,n$. Then the Central Limit Theorem tells you that $$\frac{S_n}{\sqrt {\sigma^2 n}} \to Z \sim \mathcal N(0,1)$$ from which you know that $Z$ is unbounded (as the tails of the normal distribution extend to infinity).


1

The Fisher information is essentially the negative of the expectation of the Hessian matrix, i.e. the matrix of second derivatives, of the log-likelihood. In particular, you have $$l(\alpha,k)=\log\alpha + \alpha \log k - (\alpha + 1) \log x$$ from which you compute the second-order derivatives to create a $2 \times 2$ matrix, which you take the expectation ...


1

For 1), recall that $\mathbb P=\delta_0$ just means that $\mathbb P(\{0\})=1$. So $X_n(0) = (1-0)^n = 1$, and $\mathbb P(X_n=1)=1$. Hence trivially $X_n\stackrel{a.s.}{\longrightarrow}1$ and $X_n\stackrel{d}{\longrightarrow}1$ For 2), recall that $\mathbb P=\frac12\delta_0 + \frac12\delta_1$ means $\mathbb P(\{0\})=\mathbb P(\{1\})=\frac12$. So $X_n(0) = 1$ ...


1

Model: Set $A:=0$, $B:=50$. Let $\xi_j \sim \frac{1}{6} \delta_{-1} + \frac{1}{3} \delta_0 + \frac{1}{2} \delta_1$, $j \geq 1$, be independent identically distributed random variables. Then $$X_n := k + \sum_{j=1}^n \xi_j$$ is the position of the person after $n$ steps if the person starts at $X_0=k$. We would like to calculate the probability that the ...


2

The expected value of the number of purchases until you have obtained the complete collection is $E_n=\sum\limits_{i=1}^{n}\frac{n}{n-i+1}=n\sum\limits_{k=1}^{n}\frac1k$. For reasons of symmetry $E(X_i)=\frac{E_n}{n}=\sum\limits_{k=1}^{n}\frac1k$ for all $i$.



Top 50 recent answers are included