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1

That probability is: $$\mathsf P(X=x) = \dfrac{{4\choose x}{48\choose 13-x}}{52\choose 13} = \dfrac{4!48!13!35!}{x!(4-x)!(13-x)!(35+x)!52!}$$ However, for this you should use the Linearity of Expectation, and the most useful tool of indicator variables. Let $X_i\in\{0,1\}$ be the indicator that card $i$ is an ace for $i\in\{1,,,13\}$. So ...


0

Note that $X = X_1 + \ldots + X_{10}$, where $X_i=0$ if the $i$th flip is tails, and $1$ if the $i$th flip is heads. Since the $X_i$'s are independent, we have $Var(X) = Var(X_1 + \ldots + X_{10}) = Var(X_1) + \ldots + Var(X_{10})$. Finally, $Var(X_i) = E(X_i^2) - E(X_i)^2 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$, so that $Var(X_1) + \ldots + ...


0

I just had to work out the exact same problem for homework, here's my solution: V(x) = E(x^2) - E(x) As you stated, you know that E(x) = 5, which is true. I'll add the solution for that below. For E(x^2) we need to take the value of each outcome and square it, then multiply it by its probability. Finally, they're all summed. You need to do a binomial ...


0

A coupling of random variables $X$ and $Y$ is just a pair of r.v. whose marginals distributions are the distrbution of $X$ and the distribution of $Y$. So the statement in question looks a bit tautological, but is clearly true.


0

Let us find the probability of $0$ red, of $1$ red, or $2$ red. We will not use the "hypergeometric formula," though we do in a remark at the end. For $0$ red, imagine picking one at a time. The probability the first is non-red is $\frac{4}{6}$. Given this happened, the probability the second is non-red is $\frac{3}{5}$, so the probability of $0$ red is ...


1

Since $Y\sim Geom(1/n)$ you have that $$P(Y=k)=\left(\dfrac{1}{n}\right)\left(1-\dfrac{1}{n}\right)^{k-1}$$ for $y \in \{1, 2, 3, \ldots \}$. Thus $$P(X=k)=P\left(\dfrac{Y}{n}=k\right)=P(Y=nk)=\left(\dfrac{1}{n}\right)\left(1-\dfrac{1}{n}\right)^{nk-1}$$ for $k\in \{0/n, 1/n, 2/n, 3/n, \ldots\}$. Now ...


-1

For (i): $$1=c\sum_{x=0}^{\infty}\sum_{y=0}^{\infty}\dfrac{2^{x+y}}{x!y!}=c\sum_{x=0}^{\infty}\dfrac{2^{x}}{x!}\sum_{y=0}^{\infty}\dfrac{2^{y}}{y!}=ce^{2}e^{2}=ce^4$$ Thus $c=e^{-4}$. For (ii): Note that since $c=e^{-4}=e^{-2}e^{-2}$ you can conclude that $X,Y$ are independently Poisson$(\lambda=2)$ distributed, since ...


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 $P(x,y,z)$ ...


0

Hints: If $a\ne p\ne b$, use the law of large numbers for sums of i.i.d. random variables $X_k$ with distribution $P(X_k=1)=p$, $P(X_k=0)=1-p$ If $a=p$ or $b=p$, use the central limit theorem applied to the same i.i.d. sequence Final answer: The limit is $1$ if $a\lt p\lt b$, $\frac12$ if $a=p\lt b$ or if $a\lt p=b$, and $0$ if $a\lt b\lt p$ or if ...


0

Hints: If the random variables are independent and identically distributed then what is the probability of one of them being greater than another one? $\mathsf P(X>Y)=?$ versus $\mathsf P(Y>X)=?$ What is the probability that one of them in particular is greater than both the others? $\mathsf P(X>Y, X>Z)=?$ versus $\mathsf P(Y>X, Y>Z)=?$ ...


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. ...


1

If you have some information about the value of $Y$, it gives you information about the value of $X$. By definition, it means they are dependent. Regarding $I$ and $Y$: It depends on the expected value of $X$; If it is $0$, then knowing the value of $I$ doesn't change the distribution of $Y$, since the density function of $X$ is even. Otherwise, fixing a ...


1

To show that two random variables $X,Y$ are not independent, it suffices to show that the there exist sets $S, T$ such that $$P(Y\in T)\neq P(Y\in T|X \in S)$$ It is easy to find such sets in the above exercise. For example $$P(Y>8|0<X<3)=0$$ whereas $$P(Y>8)>0$$ so that $Y$ and $X$ are not independent.


0

$a)$ the probability that a computer is correct is $0.8$, hence the products of that day will pass the quality test if at most one of the computers is out of service. Hence: $Pr(passing\space the\space quality\space test)=(0.8)^{12}+C(12,1)(0.8)^{11}(0.2)$, in which $C(12,1)=\frac{12!}{1!\times (12-1)!}$ $b)$ In this case, the first testing among 12 ...


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 ...


0

The joint density $f_{X,Y}$ is defined on $\mathbb R^2$ by $$f_{X,Y}(x,y)=\frac1{2x^2y}\mathbf 1_{x\gt1}\mathbf 1_{1/x\lt y\lt x}=\frac1{2x^2y}\mathbf 1_{y\gt0}\mathbf 1_{x\gt\max(y,1/y)},$$ hence $$f_X(x)=\mathbf 1_{x\gt1}\int_{1/x}^x\frac1{2x^2y}\mathrm dy,\qquad f_Y(y)=\mathbf 1_{y\gt0}\int_{\max(y,1/y)}^\infty\frac1{2x^2y}\mathrm dx.$$ Surely you can ...


0

The C.F of N is the following: $\varphi_N = E[e^{itN}]=\sum_{k=1}^ne^{itk}P(N=k)$ Where $P(N=k)=pq^{k-1}$. Which can be deduced from taking P(N=1), P(N=2), P(N=3) and so on in the following manner: $P(N=1)= P(X_1=0) = \frac{1}{2}$ $P(N=2)= P(X_1\neq 0,X_2=0) = P(X_1 \neq 0)P(X_2=0)= \frac{1}{2}\frac{1}{2}$ $P(N=3)= P(X_1\neq 0,X_2 \neq 0,X_3 = 0) = ...


0

Hint: The marginal pdf of $X$ is given by $$f_X(x)=\int_Y f_{XY}(x,y)dy=\int_{0}^{1-x}f_{XY}(x,y)dy$$ That is because you have the restriction $0<x+y<1$ i.e. $$-x<y<1-x \implies 0<y<1-x$$ since $-x<0$ and it is already stipulated that $y>0$.


1

Try of a second proof: For fixed $A \subseteq \Sigma$ with $\mu(A)=0$ and for all $t \in X \setminus A$ we have that $$|x(t)+y(t)| \leq |x(t)| + |y(t)| \leq \sup_{t\in X \setminus A } |x(t)|+ \sup_{t\in X \setminus A } |y(t)|. $$ Taking the supremum on both sides leads to $$\sup_{t\in X \setminus A } |x(t)+y(t)| \leq \sup_{t\in X \setminus A } |x(t)|+ ...


0

Because of $$\left\lbrace t \in X \colon |x(t) +y(t)| \geq u+v \right\rbrace \subseteq \left\lbrace t \in X \colon |x(t)| +|y(t)| \geq u+v \right\rbrace \subseteq \\ \subseteq \left\lbrace t \in X \colon |x(t)| \geq u \ \vee \ |y(t)| \geq v \right\rbrace $$ it follows that $$\mu \left(\left\lbrace t \in X \colon |x(t) +y(t)| \geq u+v \right\rbrace ...


0

Fix $u\gt\|x\|_\infty$ and $v\gt\|y\|_\infty$, then $\mu(\{t\in X\mid|x(t)|\geqslant u\})=\mu(\{t\in X\mid|y(t)|\geqslant v\})=0$, what does this tell you about $\mu(\{t\in X\mid|x(t)+y(t)|\geqslant u+v\})$? ...Thus, $\|x+y\|_\infty\leqslant u+v$, hence...


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

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 ...


1

If you are yourself chosen uniformly randomly from the class (a feature which we will assume but that you do not explain), $[X\geqslant i]$ means that you are the tallest person from a uniform sample of size $i+1$ (you and your $i$ first "opponents"). By symmetry the probability of that event is $1/(i+1)$ hence $P(X\geqslant i)=1/(i+1)$ for every $0\leqslant ...


4

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 : ...


1

$$ 𝔼[\frac{1}{N^2}\sum_{i = 1}^n\sum_{j \neq i}x_ix_j] = \frac{1}{N^2}\sum_{i = 1}^n\sum_{j \neq i}𝔼[x_i]𝔼[x_j] $$ And this is not equal to zero.


1

The variance may even be infinite. If we want finite very large variance, let $X$ for example have density functions of the form $$\frac{k\alpha}{(1+kx)^{\alpha+1}}$$ for $x\ge 0$. Here $k$ and $\alpha$ are positive parameters. For the variance of $X$ to exist, we need $\alpha\gt 3$. Given a target positive mean $\mu$, and $\alpha\gt 2$, we can find a ...


1

If $a$ and $b$ are constants, then for a random variable $X$, $$Var(a+bX)=b^2Var(X)$$ So here $$Var(Z)=Var(4+9X)=9^2Var(X)$$ And you already know $Var(X)$ from your given information, since the standard deviation is the square root of variance.


3

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


1

You arrive at: $$P\left(X=k\right)=\sum_{n=0}^{\infty}P\left(X=k\mid N=n\right)P\left(N=n\right)$$ which is okay, but we will have $P\left(X=k\mid N=n\right)=0$ if $k\notin\left\{ 0,1,\dots,n+1\right\} $. For a positive $P\left(X=k\mid N=n\right)$ we need $k\leq n+1$ resulting in: $$P\left(X=k\right)=\sum_{n=k-1}^{\infty}P\left(X=k\mid ...


0

This is simply geometry... Rotate the $(x,y)$ domain you called the "shaded area", by $45°$ clockwise, now the abscissa axis corresponds to $z=x+y$. For every $|z|\leqslant1$, the height of the segment where the vertical line of abscissa $z$ intersects the domain is proportional to $f_Z(z)$. This segment is proportional to $[-|z|,+|z|]$, whose length is ...


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, ...


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 ...


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 ...


1

If $E[Y] \gt 0$ then $\Pr(Y \ne 0) \gt 0$ and $E[Y] = E[Y\mid Y \not = 0] \times \Pr(Y \not = 0) \gt 0$, so $E[Y]^2 = E[Y\mid Y \not = 0]^2 \times \Pr(Y \not = 0)^2 \gt 0$ and finite; and similarly $E[Y^2] = E[Y^2\mid Y \not = 0] \times \Pr(Y \not = 0) \gt 0$ possibly infinite. $Var(Y\mid Y \not = 0) = E[Y^2\mid Y \not = 0] - E[Y\mid Y \not = ...


1

So, following my earlier comment and your work so far, we have that our hypothesis is $$\frac{(x_1 + 2x_2 + 3x_3 + \cdots)^2}{x_1 + 4x_2 + 9x_3 + \cdots}\leq x_1 + x_2 + x_3 + \cdots$$ multiplying both sides by the denominator (it is non-negative so the sign doesn't flip) $$(x_1 + 2x_2 + 3x_3 + \cdots)^2 \leq (x_1 + x_2 + x_3 + \cdots)(x_1 + 4x_2 + 9x_3 + ...


1

use the fact that $\sum_{k=1}^\infty \frac{1}{k^3}=\zeta(3)$ This means $\sum_{k=1}^\infty \frac{1}{\zeta(3)k^3}=1$ would be a great probability distribution. Let $P(X=k)=\frac{1}{\zeta(3)k^3}$ on $k=1,2,\dots$ Now $E[X]=\sum_{k=1}^\infty kP(X=k)=\sum_{k=1}^\infty \frac{1}{\zeta(3)k^2}=\frac{\zeta(2)}{\zeta(3)}<\infty$ But $E[X^2]=\sum_{k=1}^\infty ...


1

Hint: Try density $f(x) = C/x^p$ for $x \ge 1$ for suitable $C$ and $p$ (or if you prefer a discrete case, $P(X = n) = C/n^p$).


1

Hint: Apply the Borel-Cantelli lemma to the sequence of events $$A_n := \{\omega \in \Omega; X_n(\omega) \neq -1\}, \qquad n \in \mathbb{N}.$$


1

If $X_1, \ldots, X_n$ are the numbers selected, $S_n = \sum_{k=1}^n X_k$ so $\text{Var}(S_n) = \sum_{i=1}^n \sum_{j=1}^n \text{Cov}(X_i, X_j)$. There are just two cases to consider, $i=j$ (which occurs $n$ times) and $i \ne j$ ($n^2 - n$ times), so $\text{Var}(S_n) = n \text{Var}(X_1) + (n^2-n) \text{Cov}(X_1,X_2)$. $X_1$ is equally likely to be any of ...


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 ...


0

If they are chosen independently of each other, then the variance of the sum is the sum of the variances, so it would be $n$ times the variance of the number chosen randomly from among $1,\ldots,N$. In fact, that is the reason why standard deviations are used as a measure of dispersion, rather than the seemingly simpler and more obvious mean distances. If ...


0

Consider two borelian $>0$ functions $f,g$. $$ \begin{align} E(f(X)g(Y)) &= \frac 12\left(\int _{-1}^0 f(-1) g(-x) dx + \int_0^1 f(1)g(x)dx\right) \\&= \frac 12\left(\int _{0}^1 f(-1) g(x) dx + \int_0^1 f(1)g(x)dx\right) \\&= \frac 12\left(f(-1) + f(1)\right) \int _{0}^1 g(x)dx \\&= \left[\frac 12\left( \int_{-1}^0 f(-1) dx + \int_0^1 ...


0

Random variables $X : \Omega_X \rightarrow \mathbb R$ and $Y : \Omega_Y \rightarrow \mathbb R$ are independent random variables if events ${X \in [a,b]}$ and $Y \in [c,d]; \forall a,b,c,d \in \mathbb R$ are independent events. We can see that $Y$ equals $|Z|$, while $X = \frac{Z}{|Z|}$. Now try to understand whether $Y$ distribution is altered if we know the ...


0

Let $X \sim Geometric(p)$ and $Y \sim Geometric(q)$ where $X$ and $Y$ are strictly positive independent random variables. Then $(X,Y)$ have joint pmf $f(x,y)$: Part 1: You seek: where I am using the Expect function from the mathStatica package for Mathematica to do the nitty-gritties. Here is a plot of the solution $E[max(X,Y)]$, as parameters $p$ ...


1

Hint: Let $Z = \max(X,Y)$. Then $$P[Z \leq z] = P[\max(X,Y) \leq z] = P[X \leq z,Y\leq z]$$ This is because $\max(a,b) \leq z \iff a \leq z, b\leq z$ As for $E[X|X \leq Y]$,consider first $E[X|X \leq y]$. You have to compute the conditional pmf and work with that. Let me know if you have difficulties with this in comments. If so, I will update my answer.


0

To win the game you must find the two 'good' squares before the one 'bad' square, out of seven uncovered one at a time. Call the other four squares 'okay'. Suppose it takes $k$ turns to until you win, where $k$ is some number between $2$ and $6$.   Then to have won you must have selected $k-2$ 'okay' squares and $1$ 'good' square out of all the ways ...


1

Hint: consider the sequence of random variables $X_n = X\wedge n$ for $n\in\mathbf{N}$, where $a\wedge b = \min\{a,b\}$.


1

Suppose $E[X_i]<\infty$, then for all $c>0$, $$ E[X/c] = \int_0^\infty P(X/c>t)\,dt\ge \sum_{k\ge 1}P(X/c>k)= \sum_{k\ge 1}P(X/k>c) $$ Thus, since $\sum P(X/k>c)$ is finite, by first Borel-Cantelli, $P(X/k>c \,\,\,i.o.)=0.$ This holds for all $c>0$, proving $A=0$ a.s. Argument is similar for $E[X_i]=\infty$, but there you use an upper ...


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} \\ ...



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