New answers tagged

0

Define $X_t = -\int_0^t\beta_s\,dW_s$ and $Y_t = -\frac{1}{2}\int_0^t\beta_s^2\,ds$. Then $Z_t = e^{X_t+Y_t}$. Even though you did not mention it I am guessing that there is a condition on $(\beta_s)_{s\geq 0}$ that makes $X$ a local martingale. Since Brownian motion has continuous paths, $X$ is continuous as well, which implies $Z$ is continuous as well. ...


1

The problem is actually already solved by Jyrki's comment, since the only way to satisfy the given conditions is that the $3$ voters have the cyclical preferences given in his example, or the opposite preferences. Thus you just need to count the number of times each of the candidates wins under the proposed "solution", given such a preference set. As you ...


1

Suppose that $p_i$ and $p_j$ are prime. Then $$ A_i\cap A_j=\{X=p_i^2p_j^2k\mid k\in\mathbb{N}\}. $$ So, $$ \begin{align*} P(A_i\cap A_j)&=\sum_{k=1}^{\infty}P(X=p_i^2p_j^2k)\\ &=\frac{1}{\zeta(s)p_i^{2s}p_j^{2s}}\sum_{k=1}^{\infty}\frac{1}{k^s}\\ &=\frac{1}{p_i^{2s}p_j^{2s}}. \end{align*} $$ On the other hand, $$ \begin{align*} P(A_i)&=P\{X=...


0

Make a Venn or Dodgson diagram for $A, B, C,$ and $D$. The probability of the universe (the entire diagram including the part that is outside of $A \cup B \cup C \cup D$ is one. Write the probability in each area. By area, I mean a space you cannot leave without crossing a boundary of one of the sets. The total of all the values will be one. The equation ...


-1

Here is what I am thinking so far, although I need help in the final step of seeing whether or not the Yoneda lemma is applicable in this setup. Consider that any monoid can be represented as a category with a single object, representing the monoid itself, and the morphisms of the category are the elements of the monoid, the associativity of their ...


0

$\newcommand{\P}{\mathbb{P}}$I am not sure what is wrong with the following, that seems to imply that the events are not independent. If the $A_i$ are independent, in particular they are pairwise independent so it must hold that $\P(A_i \cap A_j) = \P(A_i) \P(A_j)$ So we know that $$\P(A_i) = \frac 1{\zeta(s)}\sum_{n=1}^\infty \frac 1{({np_i^2})^s} $$ ...


1

By the stationarity of the increments, this is equivalent to $$\min_{s \in [0,t]} \mathbb{P}(|X_s| \leq \epsilon)>0$$ for all $t>0$. Without any additional assumptions this does, in general, not hold true. Consider for example $X_t := t$ (i.e. a deterministic drift process), then $$\mathbb{P}(|X_t| \leq \epsilon) = 0$$ for all $t>\epsilon$. ...


2

We give a very weak result: sometimes the only functions $g$ are the trivial ones. Let $X$ take on values $0$ and $1$, and let $\Pr(X=1)=p$, where $p$ is transcendental. Let $g(0,0)=a$, $g(0,1)=b$, $g(1,0)=c$, $g(1,1)=d$. Here $a,b,c,d$ only take on values $0$ and $1$. The corresponding probabilities are $(1-p)^2, p(1-p), p(1-p),p^2$. There are two ways the ...


2

No, this fails especially when $\mu$ is atomic. Take $\mu_n(\{0\})=1-1/n$ and $\mu_n(\{n\})=1/n$. Then $\mu_n$ converges weakly to the point mass at 0. Now take $f(x)=x$. Then $\sup_n \int_X fd\mu_n=1$ but $\int_X fd\mu_n=1\neq 0=\int_Xfd\mu$. Usually we speak of measures losing mass in the limit, however in this case it's a loss of expectation for ...


0

1-) The meaning of ∫P(Q)*dQ is the probability of finding the value of Q, according to the limits of the integration. 2-) The use of dR in the proof is fundamental to simplify the description of the shield of the sphere. Since we are talking about a sphere, nothing more appropriate than use spherical coordinates. In cartesian coordinates, it would be very ...


2

Consider the following random variables: $$ P(X = -1) = P(X = 1) = 1/2\\ P(Y = 0 \mid X = -1) = 1 \\ P(Y = 1 \mid X = 1) = P(Y = -1 \mid X = 1) = 1/2 $$ Then $E(XY) = E(X)E(Y) = 0$.


1

Let X be a normally-distributed random variable with $\mu=0$. Let $Y = X^2$. Clearly X and Y are not independent. But when you calculate their correlation, you will get zero.Because: $$Cov(X,Y) = E(XY) - E(X)E(Y) = E(X^3) - 0*E(Y) = E(X^3)= 0$$ So X and Y are uncorrelated but not independent.


1

Hint: $ +\infty = \mathbb{E}[|X_n|] = \int_0^{+\infty} \mathbb{P}[|X_n > y|] \ dy = \sum_{n=0}^{+\infty} \int_n^{n+1} \mathbb{P}[|X_n| > y] \ dy \leq \sum_{n=0}^{+\infty} \int_n^{n+1} \mathbb{P}[|X_n| > n] dy = \sum_{n=0}^{+\infty} \mathbb{P}[|X_n| > n] \int_n^{n+1} 1 dy = \sum_{n=0}^{+\infty} \mathbb{P}[|X_n| > n] $


3

No, there is no such random variable $\xi$ and constant $c \neq 0$ so that $\xi$ and $\xi + c$ are distributed the same. Why? If these two random variables were distributed the same, then their CDF's (say $F_1$ and $F_2$) would also be the same. Now $$ F_1(z) = P(\xi \leq z) = P(\xi + c \leq z + c) = F_2(z+c) = F_1(z+c). $$ Therefore, the CDF is periodic ...


0

For any events $A$ and $E$ we can write $$ \mathbb{P}(E)=\mathbb{P}(A\cap E)+\mathbb{P}(A^c\cap E) $$ If $\mathbb{P}(A)=1$ then $\mathbb{P}(A^c)=1-1=0$ and $$ 0\leq \mathbb{P}(A^c\cap E)\leq \mathbb{P}(A^c)=0 $$ so $\mathbb{P}(E)=\mathbb{P}(A\cap E)$. Similarly, if $\mathbb{P}(A)=0$ then $0\leq \mathbb{P}(A\cap E)\leq \mathbb{P}(A)=0$, hence $\mathbb{P}(A\...


0

You're right, this is no hard, although is quite tricky. The only thing you need is self-similarity. First note that $x^qa^{-q} = q\cdot\sup_{\lambda>0}(\lambda x - p^{-1}\lambda^p a^p)$. Then $$ \sup_{t\ge 0} \left(\frac{X_t}{1+t^{p/2}}\right)^q = q\cdot\sup_{t\ge 0}\sup_{\lambda>0}\left(\lambda X_t - p^{-1}\lambda^p (1+t^{p/2})^p\right)\\ = \sup_{\...


2

As it was hinted in the comments, the result is not true unless you assume uniform convergence somewhere. Here is a simple example. Let $X_n = X$ have non-degenerate Bernoulli distribution and $Y_{un} = X \mathbf{1}_{u\ge n}$. Then $P(Y_{un}\neq X_n)\to 0$, $u\to\infty$, and $e^{itY_{un}} \to 1$, $n\to\infty$, in all senses. However, $e^{itX_n} = e^{itX}\...


1

We can endow $\mathcal A$ with a pseudo-metric defining $\rho\left(A,B\right):=\mu\left(A\Delta B\right)$, where $\Delta$ denotes the symmetric difference operator. In this way, $\left(\mathcal A,\rho\right)$ is a complete pseudo-metric space. For the details, see this thread.


2

Define random variables $X_1, X_2,\ldots,X_n$ where $X_i$ equals $1$ if voter $i$ votes for Candidate A, and $-1$ otherwise; define $S_k=X_1 + \cdots + X_k$. Then $S_n>0$ means that Candidate A is the winner, and $S_{\alpha n}>0$ means Candidate A is leading after $\alpha n$ votes. Calculate for any $\alpha\in(0,1)$ that $S_{\alpha n}$ has mean $0$ and ...


0

$X = 165+ 6 Z_1$ Where $Z_n$ are standard normal random variables. $Y = 175 + A Z_1 + B Z_2$ a component the depends on X and a component that is independent from X $A^2 + B^2 = 64$ the variance of Y $6A = cov(X,Y)\\ \rho = \frac {cov(X,Y)}{\sigma_x\sigma_y} = 0.5\\ \frac {6A}{48} = 0.5\\ A = 4\\ B = 4 \sqrt 3\\ Y = 175 + 4 Z_1 + 4 \sqrt 3 Z_2\\ \frac 43 ...


1

You simply threw away the fact that $\sigma_j^2:=Var(X_j)/j\to0$ and used only the fact that it's bounded by $C$. Do everything just the way you did til the very end. Then say $$\frac1n\sum_{j=1}^n\sigma_j^2\le\frac1n\left(\sum_{j=1}^{n_0}C+\sum_{j=n_0+1}^n\epsilon\right)\le\frac{C n_0}{n}+\epsilon.$$Now letting $n\to\infty$ shows that $$\limsup_{n\to\infty}\...


0

Hint: note that two invariant integrable functions are equal almost everywhere if and only if their integrals are equal on every invariant set; why? :)


3

Well, there's not much leeway, so it's probably most efficient to count by hand: $8+8+4=20$, $8+7+5=20$, $8+6+6=20$, $7+7+6=20$ – that's it, $4$ ways.


0

I think a more rigorous proof than previously given as answers uses more axioms of a probability measure. Of the three axioms, we need: 1)P($\Omega$)=1 2)Given $A_i$ are mutually disjoint events in a sample space $\Omega$, P($\cup_{i=1}^{\infty}A_i$)=$\sum_{i=1}^{\infty}P(A_i)$ So we say $A_i = \emptyset$, because these are mutually disjoint, we can use ...


0

The order of the inner summation is reversed. In other words, it is a substitution $$n_1\ \rightarrow\ n_2-n_1.$$


0

Yes, ergodicity implies stationarity. Consider an ensemble of realizations generated by a random process. Ergodicity states that the time-average is equal to the ensemble average. The time-average is obtained by taking the average of a single realization, giving you a particular number. To obtain the ensemble average, you take the average across the ...


1

(1) $\iff$ (2) Since $f \geq 0$ is measurable, there exists a sequence $(f_n)_{n \in \mathbb{N}}$ of simple functions (aka elemtary functions/step functions) such that $0 \leq f_n \uparrow f$. Each $f_n$ can be written in the form $$f_n(x) = \sum_{j=1}^{N_n} c_{j,n} 1_{B_{j,n}}(x).$$ for suitable constants $c_{j,n} \geq 0$ and $B_{j,n} \in \mathcal{B}(X)$....


2

The probability that some two men choose the same number is the complement of the probability that no two men choose the same number. The number of ways to choose the three men's numbers such that they are all different is $$\binom{20}{3}\binom{17}{3}\binom{14}{3}\binom{11}{3} = 1140\cdot 680\cdot 364\cdot 165$$ The number of ways to choose the same ...


0

I don't think I have enough of your problem to answer that question, but if you look at Huber's Robust Statistics or Keener's Theoretical Statistics, the sections on M-estimators might help. For example, in Keener, the maximum likelihood estimator is characterized as when the derivative of the likelihood equals 0. They are able to do proofs using a Taylor ...


2

If I'm understanding correctly, I think you can prove the result by first proving the result when $g$ is an indicator function of sets, then for simple functions $g$ (by linearity), then for positive functions $g$ (by monotonicity), and then finally for integrable functions $g$ (by splitting the positive and negative portions and applying the result ...


0

How about this: if $Y=X$ (that is, they represent the same random variable). Then $X/Y=1$ and $Y$ and $1$ are uncorrelated.


0

Let's write this out: The CLT says that if $X_j$ are iid with finite expected value $\mu$ and finite variance $\sigma^2$, then the sequence $$Y_n = \sqrt{n}\cdot\left({\frac{X_1+\ldots+X_n}{n}-\mu}\right)$$ converges in distribution to $N(0, \sigma^2)$. Now $n$ here is not actually a variable, it is merely an index. So saying that $Y_n/n$ converges in ...


0

The problem in the presented proof is that it only show that for any fixed $k$, $\lim_{n\to +\infty}\mathbb P\left(\left|S_n-S_k\right|/n\gt\varepsilon \right)= 0$, which is not hard to see from the convergence in probability of $\left(S_n/n\right)_{n\geqslant 1}$ to $0$. We actually have to prove that $\lim_{n\to +\infty}\color{red}{\max_{1\leqslant k\...


1

It's almost correct; except you're taking the factor $\frac14$ into account twice. Either $A$ is meant to be an area; then the result is correct, $\frac{4-2A}4$, but then the integrand should be $1$, not $\frac14$; or $A$ is meant to be a probability; then the integrand $\frac14$, the constant density, is correct, but then the result should be $1-2A$, not $\...


2

Here is a proof that $\lim_{n\rightarrow\infty} q_n = \sqrt{2} -1$ with probability 1: Let $\liminf q_n$ and $\limsup q_n$ represent the random variables $\liminf_{n\rightarrow\infty} q_n$ and $\limsup_{n\rightarrow\infty} q_n$, respectively. Let $\{a_k\}_{k=0}^{\infty}$ be the deterministic sequence that satisfies $a_0=1/2$ and: $$a_{k+1} = \frac{1}{2}-\...


6

Suppose $Y$ is a non-constant postive-valued random variable. Suppose $\operatorname{E}(Y^2)<\infty$ so that correlations can make sense. Suppose $U$ has a standard normal distribution and it independent of $Y$. Let $X=UY$. Then $U=\dfrac X Y$ is independent of $Y$; hence uncorrelated with $Y$.


6

One simple case where your identity is true: Let $Y$ be some nonconstant and positive RV, and let $X:=cY$ for some nonzero $c$. Then $E(X)=cE(Y)=E(X/Y)E(Y)$.


0

No, that is not true. If a random variable can take any value on the real line, but it is exceedingly likely that said variable will be between $0$ and $0.1$, then the pdf will have values that are, if not over, then at least close to $10$ in that small interval. As another example, if $X\sim\mathcal{N}(0,\sigma^2)$ with $\sigma$ very small, then $f_X(0)$ ...


0

The underlying idea is extremely important in analysis; a simple example is that you can embed any inner product space into the space of linear functionals on its conjugate by the linear transformation $v \mapsto \langle -, v\rangle$. It's not an application of Yoneda lemma, but it's clearly the same sort of idea. In fact, there's a good analogy between ...


2

It seems that when we talk about Poisson random measure, we often assume that the characteristic measure $\nu$ is $\sigma$-finite. Then the random measure $\tilde{N}(t,dx)$ is also $\sigma$-finite. Of course, $\tilde{N}(t,A)$ is not a martingale if $\nu(A)=\infty$. To define the stochastic integral $$\int^T_0 f(t,x)\tilde{N}(dt,A),$$we first consider the ...


2

Since $$\left\{ \left| \frac{S_n-S_k}{n} \right|< \epsilon \right\} \supseteq \left\{ \left| \frac{S_n}{n} \right|< \frac{\epsilon}{2} \right\} \cap \left\{ \left| \frac{S_k}{n} \right| < \frac{\epsilon}{2} \right\}$$ we have $$\mathbb{P} \left( \left| \frac{S_n-S_k}{n} \right| < \epsilon \right) \geq \mathbb{P} \left( \left| \frac{S_n}{n} \...


1

Let $p_n$ be the probability the $n$-th toss is a $6$. Note that $p_1=1$. We obtain a recurrence for $p_n$. The probability the $n$-th toss is not a $6$ is $1-p_n$. Given that, the probability the $n$-th toss is a $6$ is $\frac{1}{5}$. Thus $$p_{n+1}=\frac{1}{5}(1-p_n)=\frac{1}{5}-\frac{1}{5}p_n.$$ The homogeneous recurrence $p_{n+1}=-\frac{1}{5}p_n$ ...


1

The Markov Chain has 6 states labelled 1 through 6. For each state, the transition probability to each of the other states is $1/5$. So the probability transition matrix is $M=\dfrac{1}{5}(J-I)$ where $J$ is the all ones matrix, and $I$ is the identity matrix. The distribution vector for the $n$th score is given by $(1,0,0,0,0,0)M^n$, i.e. by the first row ...


2

Assuming that by $D(X)$, you mean the variance, $Var(X)$, we have: $Var(X+Y)=Var(X)+Var(Y)+2Cov(X,Y)$. So you need to add $-3/2$ to the value that you already have. Note that $Var(X+Y)$ is equal to $Var(X)+Var(Y)$ if and only if $X$ and $Y$ are independent, i.e. when $Cov(X,Y)=0$.


0

Here is another example. Notice first that $$ (\Bbb{E}(XY))^2 - \Bbb{E}X \Bbb{E}(XY^2) = \operatorname{Cov}(X, Y)^2 - (\Bbb{E}X)^2 \operatorname{Var}(Y) - \Bbb{E}[(X - \Bbb{E}X)(Y - \Bbb{E}Y)^2].$$ Now let $(X, Y)$ be a multivariate Gaussian variable. Then the last term vanishes and we have $$ (\Bbb{E}(XY))^2 - \Bbb{E}X \Bbb{E}(XY^2) = \operatorname{Cov}(...


0

Note that $$\mathbb E(X)\mathbb E(XY^2)-\mathbb E(XY)^2=V(XY)-coV(X,XY^2) $$ If $X$ and $Y$ are independent $$\mathbb E(X)\mathbb E(XY^2)-\mathbb E(XY)^2= \mathbb E(X)^2V(Y) $$


2

Something can be said if $X\ge 0$: By Cauchy-Schwarz, $$ \left(\Bbb E[XY]\right)^2=\left(\Bbb E[\sqrt{X}\cdot\sqrt{X}Y]\right)^2\le\Bbb E[X]\cdot\Bbb E[XY^2]. $$


0

Maybe he was looking for you to give an approximation and solve the problem. If $X$ and $Y$ are independent then the expectations become $E[X]^2 E[Y]^2$ and $E[X]^2 E[Y^2]$ comparing the two we see the latter is larger by Jensen's inequality.


2

I don't think that much can be said about the relationship between the two quantities in general, other than what Michael Hardy has already said. As the following examples show, it's not true that one of the quantities is always larger than the other: For instance, if $X$ is a random variable taking the values $\pm 1$, each with probability $\frac{1}{2}$, ...


1

Certainly not. Unless the $X_i$ are constant, the probability that $\frac{1}{n} \sum \limits_{i = 1}^n 1_A(X_i)$ equals $\Lambda(A)$ for all $A \in \mathcal{B}(R)$ is $0$ for any $n$. You need to rethink what you want to show.



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