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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 $E(X)=cE(Y)=E(X/Y)E(Y)$.


3

Depends on what valid claim numbers are. If the digits are randomly generated then the odds of them putting that one down are one in $100$ million. If the claim number needs to start in $34-$ then it's one in a million. But, I think this is just a couple of nails in a big coffin for your case. Having your company's name on the document is much more ...


3

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 $E(XY) = E(X)E(Y) = 0$.


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 ...


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.


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}\...


2

We assume that \begin{align*} d\left(\! \begin{array}{c} S^1(t)\\ S^2(t) \end{array} \!\right) =\textrm{diag}\left(S^1(t), S^2(t)\right)\bigg[\left(\! \begin{array}{c} r\\ r \end{array} \!\right)dt + \left(\! \begin{array}{cc} \sigma_{1,1} &\sigma_{1,2}\\ \sigma_{2,1} &\sigma_{2,2} \end{array} \!\right)d \left(\! \begin{array}{c} W_t^1\\ W_t^2 \end{...


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 ...


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 ...


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 ...


2

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=...


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 ...


2

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.


2

In addition to the identical distribution independence is used to obtain that $(Z_1,Z_1+Z_2)$ and $(Z_2,Z_1+Z_2)$ are identically distributed which is needed to conclude $\mathbb E(Z_1 \mid Z_1+Z_2) = \mathbb E(Z_2 \mid Z_1+Z_2)$. Edit. The distribution of $(Z_1,Z_1+Z_2)$ is the image (push forward) of $\mathbb P^{(Z_1,Z_2)}= \mathbb P^{Z_1} \otimes \...


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}-\...


2

Yes, we can choose such a $K$, or just notice that $$\lim_{K\to +\infty}K^{1-p}A=0$$ and that the bound of $\mathbb E\left(\left|X\right|;\left|X\right|\gt K\right)$ is uniform with respect to $X\in\mathcal C$. The follows from the inclusion $$\left\{\left|X\right|\gt K+k^{-1}\right\}\subset \left\{\left|X-X_n\right|\gt k^{-1}\right\}.$$ To see this, it ...


2

$\mathcal{C}$ is UI $\Rightarrow$ $(i)$ Let $\epsilon>0$. Then, since $\mathcal{C}$ UI there exists $K>0$ s.t. \begin{equation} E[|X|] = E[|X|1_{|X|>K}+|X|1_{|X|\le K}]\le \epsilon + K \end{equation} for all $X\in\mathcal{C}$. Thus $(i)$. $\mathcal{C}$ is UI $\Rightarrow$ $(ii)$ You want to find for a given $\epsilon>0$ a $\delta>0$ such ...


1

But with you attempt, you do not reach the conclusion. Let $K>0$, then we have \begin{equation} E[|X|1_{F}] = E[|X|1_F1_{|X|>K}+|X|1_F1_{|X|\le K}]\le E[|X|1_{|X|>K}]+E[K1_F]= E[|X|1_{|X|>K}]+KP(F). \end{equation}


1

Let $(Z_1,Z_2)$ be uniformly distributed on $\{(-1,0),(0,1),(1,-1)\}$. Then $Z_1$ and $Z_2$ are identically distributed but dependent. The value of $Z_1+Z_2$ fixes the (different) values of $Z_1$ and $Z_2$, so $\mathbb E[Z_1\mid Z_1+Z_2]\ne\mathbb E[Z_2\mid Z_1+Z_2]$.


1

Both results are actually equivalent. You can prove one from the other. Regarding the first result: Let $\mathcal{C}$ be a class of subsets of $\Omega$ under finite intersections and containing $\Omega$. Let $\mathcal{B}$ be the smallest class containing $\mathcal{C}$ which is closed under increasing limits and by difference. Then $\mathcal{B} = \...


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

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] $


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 $t>\epsilon$. ...


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.


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}\...


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 $B_{j,n} \in \mathcal{B}(X)$....


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 $\...



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