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7

The one-sided version of Chebyshev that you are trying to prove is known as Cantelli's Inequality. See Theorem 2.2 of link. It's been around for 104 years.


5

Dedimensionalizing all this to get rid of the constraints on $E(X^2)$ and $z$ is cumbersome but it can be done and it shows that the inequality to prove is equivalent to the following statement: For every nonnegative integrable random variable $X$ and every $t$ in $(0,1)$, $$P(X\geqslant tE(X))\geqslant\frac{(1-t)(E(X))^2}{E(X^2)}.\tag{$\ast$}$$ For ...


4

No--Assume that $X_n=Y$ for every $n$ with $P(Y=1)=P(Y=2)=\frac12$ and that $X=3-Y$. Then $|X_n|\leqslant Y$ almost surely, for every $n$, and the distributions of $X$ and $Y$ coincide hence $X_n\to X$ in distribution, but $[|X|\leqslant Y]=[Y=2]$ has probability $\frac12$ only. Extending this to $k$ values instead of $2$ allows the probability of ...


4

No, this implication does not hold. Roughly speaking: Convergence in distribution does not preserve pointwise properties of random variables. Consider $([0,1],\mathcal{B}([0,1]))$ endowed with the Lebesgue measure and define $$\begin{align*} X(x) &:= \begin{cases} 0 & x < \frac{1}{2} \\ 1 & x \geq \frac{1}{2} \end{cases} \\ Y(x) &:= ...


4

\begin{align*} e^{-\lambda-\mu}\sum_x\frac{\lambda^x\mu^{w-x}}{x!(w-x)!}&=e^{-(\lambda+\mu)}\sum_x\frac{w!}{x!(w-x)!}\frac{\lambda^x\mu^{w-x}}{w!}=\\ &=e^{-(\lambda+\mu)}\sum_x\dbinom{w}{x}\frac{\lambda^x\mu^{w-x}}{w!}=\\ &=\frac{e^{-(\lambda+\mu)}}{w!}\sum_x\dbinom{w}{x}\lambda^x\mu^{w-x} \\ ...


3

Your number is right. For another way of computing, let us find the probability that the maximum is $\le 5$. That's the probability of getting three numbers that are $5$ or below. The probability of that is $\left(\frac{5}{6}\right)^3$. So the probability the maximum is $6$ is $1-\left(\frac{5}{6}\right)^3$. Remark: Here is a more complicated problem. ...


3

Let $$ Z=\frac{Y-1}{2}. $$ $Z$ is distributed normally as $Y$ is a normal random variable and $Z$ is a linear combination of $Y$. Then $$ E\big[Z\big]=E\big[\frac{Y-1}{2}\big]=\frac{E[Y]-1}{2}=0, $$ and $$ \sigma_Z=E\big[\big(Z-E[Z]\big)\big(Z-E[Z]\big)\big]=E\big[Z^2\big] =E\Big[ \frac{Y^2-2Y+1}{4} \Big] =\frac{E[Y^2]-2E[Y]+1}{4} =\frac{4 ...


3

Hints: $\mathbb E(U+V)=\mathbb E(U)+\mathbb E(V)$. $\mathbb E(cU)=c\mathbb E(U)$ where $c$ denotes a constant.


3

One way to proceed is to generate a point uniformly on the sphere, apply the mapping $f : (x,y,z) \mapsto (x'=ax,y'=by,z'=cz)$ and then correct the distortion created by the map by discarding the point randomly with some probability $p(x,y,z)$ (after discarding you restart the whole thing). When we apply $f$, a small area $dS$ around some point $P(x,y,z)$ ...


3

Corrigendum $$\mathbb{var}[\overline{x}]=\mathbb{var}\left[ \frac{1}{N}\sum_{i=1}^Nx_i \right]\stackrel{\text{iid}}{=} \frac{1}{N^2}\sum_{i=1}^N\mathbb{var}\left[x_i \right]=N \cdot \frac{m_{c_2}}{N^2}=\frac{m_{c_2}}{N}$$ $Cov({x_i},{x_j}) = E\left[ {{x_i}{x_j}} \right] - E\left[ {{x_i}} \right]E\left[ {{x_j}} \right] = 0$ Since $\{x_1,\ldots, x_N\}$ iid : ...


3

Let $Y=X^2$. We presumably want to find the density function of $Y$. First we (sort of) find the cumulative distribution function $F_Y(y)$ of $Y$. Note that $Y$ is never negative. For $y\gt 0$, we have $$F_Y(y)=\Pr(Y\le y)=\Pr(X^2\le y)=\Pr(-\sqrt{y}\le X\le \sqrt{y}).\tag{1}$$ It follows that ...


3

In lotteries the numbers $(X_k)_{1\leqslant k\leqslant n}$ are distinct and selected uniformly, thus, for every $k$, $$E(X_k)=\frac1N\sum\limits_{i=1}^Ni=\frac{N+1}2,\quad E(X_k^2)=\frac1N\sum\limits_{i=1}^Ni^2=\frac{(N+1)(2N+1)}6,$$ and, for every $k\ne\ell$, $$E(X_kX_\ell)=\frac1N\sum_{i=1}^Ni\frac1{N-1}\sum_{j\ne ...


3

You can use the Cantelli inequality, which states that $$P(X-E[X]\ge a)\le\frac{\sigma^2}{\sigma^2+a^2}$$ Taking $a=t\sigma$ yields directly the one-sided Chebyshev inequality, since $$P(X-E[X]\ge t\sigma)\le\frac{\sigma^2}{\sigma^2+t^2\sigma^2}=\frac{1}{1+t^2}$$ (but I am not sure if that is what you wanted, or if you wanted also a proof of the Cantelli ...


3

Hint: Let $X_n$ equal $0$ with probability $1-1/n^2$, and otherwise be very large. Argue using Borel-Cantelli that $\sum_n X_n$ converges almost surely. On the other hand, you can arrange that $\mathrm{Var} X_n = 1$, say. Note that the $X_n$ would not be uniformly bounded.


3

For every $s\gt0$, let $\nu(s)=\mu(s)/s+P(X\gt s)$, then $\nu(s)=E(u_s(X))$ where, for every $x\geqslant0$, $u_s(x)=\min\{1,x/s\}$. Assume that $P(X=0)\ne1$. Then, for every $x\geqslant0$, $u_s(x)\leqslant1$, $u_s(x)\leqslant u_t(x)$ for every $s\geqslant t$, $u_s(x)\to0$ when $s\to+\infty$, and, for every $x\gt0$, $u_s(x)\to P(X\gt0)$ when $s\to0$. Thus, ...


2

You do not follow the elementary indications given to you in comments hence it is a bit difficult to know what could help you (and the bounty is no substitute for that). Anyway, steps towards the solution could be as follows. The correct statement of the exercise involves $$Y_n = \frac{S_n/\tau_n - \lambda ...


2

By symmetry, there can be no other probability other than $\frac12$. The probability of the two numbers being equal is $0$, and the probability of the first bein bigger is the same as the probability of the first being smaller.


2

Let $X$ and $Y$ be iid random variables. Then $$P(X<Y)=P(Y<X)$$ Next to that we have: $$1=P(X<Y)+P(X=Y)+P(Y<X)$$ This together leads to: $$P(X<Y)=P(Y<X)=\frac{1}{2}(1-P(X=Y))$$ In your case $P(X=Y)=0$ so we end up with $$P(X<Y)=P(Y<X)=\frac{1}{2}$$


2

Let $X=(X_i)$ and $C=\mathrm{Cov}(X,X)$, then $X$ is centered hence its Laplace transform is such that $$E(\mathrm e^{s'X})=\mathrm e^{s'Cs/2},$$ for every $s=(s_i)$. The homogenous terms of degree $4$ with respect to $s$ from both sides coincide, hence $$E((s'X)^4)=\frac{4!}{2!2^2}(s'Cs)^2=3(s'Cs)^2.$$ Note that ...


2

In a comment it is specified that we are working in a finite field. Here is a simple counterexample. Consider the field of $3$ elements, and the polynomial $p(x)=x^2$. If we evaluate at points in the field chosen at random, we get that $p(x)=0$ with probability $\frac{1}{3}$, and $p(x)=1$ with probability $\frac{2}{3}$.


2

If $E[Y^2]=\infty$ then the statement holds trivially. Otherwise, Markov's inequality yields $$P(\left|Y-E[Y]\right|\ge E[Y^2]) \le \frac{\text{Var}(Y)}{E[Y^2]}$$ Now since $Y$ is nonnegative and integer-valued $$P(Y=0)\le P(\left|Y-E[Y]\right|\ge E[Y^2])$$ (see explanation below) so that $$P(Y=0)\le\frac{\text{Var}(Y)}{E[Y^2]}$$ The last equation can be ...


2

Define $M_n = \max\{U_k, 1 \leq k \leq n\}$, then $P(M_n < 1 - \epsilon) = (1-\epsilon)^n$, so $$\sum_{n=1}^{+\infty} P(M_n < 1- \epsilon) = \dfrac{1-\epsilon}{\epsilon} < +\infty$$. This means $E(\sum_{n=1}^{+\infty}1_{M_n < 1-\epsilon} )< +\infty$, i.e. almost surely only finite many $M_n$ are smaller than $1-\epsilon$. So we have $M_n \to ...


2

Let $X$ and $Y$ be iid Bernoulli(1/2), i.e., coin flips $P(X=0)=P(X=1)=1/2$. Let $Z=(X+Y)\mod 2$. Then $Z$ is independent on $X$ and $Y$ but not $(X,Y)$. So, the answer is no.


2

The point is that in order to verify that $X^{-1}(B)\in\mathscr{F}$ for all Borel sets $B\in\mathcal{B}(\mathbb{R})$, it is enough to verify that $X^{-1}(B)\in\mathscr{F}$ for all $B\in\mathcal{A}$, where $\mathcal{A}$ generates $\mathcal{B}(\mathbb{R})$. To see why this is true, just note that if $\sigma(\mathcal{A})=\mathcal{B}(\mathbb{R})$, i.e. ...


2

Hint: $X$ has the binomial distribution


2

Well, generating it the usual way ensures that the distribution is "as close" to a uniform distribution as it possibly could be. So, you can be sure of that method, assuming your psuedorandom number generator is behaving itself. If we wanted to be really formal about it, we could note that, as $b$ goes to $\infty$, the probability of the generated number ...


2

If you are content with the weaker inequality $$ \mathrm{Pr} (X-\mu \ge t\sigma) \le \frac{2}{1+t^2}, $$ then you are already done. Apply your Chebyshev-like inequality when $t\ge 1$ (so that $\frac{1}{t^2}\le \frac{2}{1+t^2}$) and the trivial bound $\mathrm{Pr}(X-\mu \ge t\sigma)\le 1$ when $0\le t< 1$.


2

In general, 3 does not imply 2 Suppose $g$ is $\mathcal{F}$-measruable and $h=g$ almost surely, it does not imply $h$ is also $\mathcal{F}$-measurable. Take a set $A$ which is contained in $B \in \mathcal{F}$ with $P(B) = 1$, but $A$ is not itself in $\mathcal{F}$. By definition, $h = 1_{A}$ is alomst surely equal to the constant variable $g \equiv 0$. ...


1

Hint: The probability that $M\leq \theta/2$ is the same as the probability that all $X_i$s are less than $\theta/2$.


1

If $Y$ is a linear combination of characteristic functions, then $$a\mathbb P(|X|>a)\leqslant a\mathbb P(|X-Y|+|Y|>a)\leqslant a\mathbb P(\max\{|X-Y|, |Y|\}>a/2),$$ hence $$a\mathbb P(|X|>a)\leqslant a\mathbb P(|X-Y|>a/2)+a\mathbb P(|Y|>a/2).$$ Using Markov's inequality, we get $$a\mathbb P(|X|>a)\leqslant 2\mathbb E|X-Y|+a\mathbb ...



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