Homework Question  Hello, I have a simple homework question that asks:  We know that when we sample, with equal probability from a finite population {x1, x2,..., xN} without replacement, we obtain a simple random sample instead of an (iid) random sample. Let X1,...,Xn (n &lt; N) be the sample obtained. It is claimed in class that X1,..., Xn, though no longer independent, are still identically distributed according to the pmf:  f(x) = 1/N, if x = xi, for some 1 &lt;= i &lt;= N;        0, for all other x.  The first choice X1 obviously follows this distribution. Show that the second choice X2 is distributed according to f(x) too.   My work: I did some research on exchangeable random variables. My thinking is that if I can show the sequence of random variables X1,..., Xn is exchangeable, then it follows that any of the Xi's can be the first random variable in the sequence. Therefore, each Xi will follow f(x).  My question: I am unsure if I properly understand what being exchangeable means. Also, I am not sure how you would properly prove that this sequence is exchangeable. I would really appreciate any tips on going about it.  Thank you.
122071
I see the concept of 'exchangeability' being used in different contexts (e.g., bayesian models) but I have never understood the term very well.    What does this concept mean? Under what circumstances is this concept invoked and why?
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Hy, I hope everyone is doing fine. Lately I have been studying the topic of probability, I am aiming to improve on it, recently i came across this hard problem.   Let $(u_{n})$ a sequence of random independent variables identically distributed following Rademacher distribution. And let  $f(x)=\sum_{n \geq 0} u_{n} x^{n}$.  Prove that : $$ f(x)~~\text{almost surely has infinitely many zeros  in}~~ [0,1].$$   I have been stuck on it for weeks now, but I did find out a hint to solve it on a forum by searching about it, Here is the hint:  Construct an increasing sequence $(x_{k})$ such that the event $A_{k}=\{f(x_{0}),...,f(x_{k}) \text{are not zero and have the same sign} \}$ are such that $p(A_{k+1})\leq \frac{6}{7} p(A_{k})$.      I find it difficult to construct by induction such sequence in a way that may permit us to solve the problem. Any proposition is welcome.
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Let $X_1(x),X_2(x),\ldots ,X_n(x)$ be $n$ iid random variables. I know that  $$\mathbf{P}(\lim_n X_n(x)=X(x))=1$$ for each $x\in \mathbb{R}$. We also know that the limit random variable $X$ is a.s. constant and considering $X\colon \mathbb{R}\to\mathbb{R}$, it is continuous. I want to show that  $$\mathbf{P}(\lim_nX_n(x)=X(x)\ \text{ for all }x\in \mathbb{R})=1.$$ I thought about using countable additivity of probability on $\mathbb{Q}$ and then applying continuouty to extend the result to the whole $\mathbb{R}$. Since $$\mathbf{P}(\lim_nX_n(x)\neq X(x)\ \text{ for all }x\in \mathbb{Q})=\sum_{x\in \mathbb{Q}}\mathbb{P}(X_n(x)\neq X(x))=0+0+\ldots.$$ Is there anything else I should add to this explanation? Ideas?
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For a random sample $X_1,X_2,\ldots,X_n$ from a $\operatorname{Uniform}[0,\theta]$ distribution, with probability density function $$ f(x;\theta) = \begin{cases} 1/\theta,  &amp; 0 \le x \le \theta \\ 0, &amp; \text{otherwise} \end{cases} $$  Let $X_{\max} = \max(X_1,X_2,\ldots,X_n).$ What is the value of k such that $\hat \theta  = kX_{\max}$ is an unbiased estimator of $\theta$ ?    I'm not sure if there is more to this question, because my intuitive answer answer is just  $k=1$. This is because if you order the sample like $$x_{(1)} \le x_{(2)} \le \cdots \le x_{(n)}$$ such that $x_{(n)} = E[X_{\max}]$.  and the fact that the distribution is uniform, the estimator of $\theta$ should just be $X_{\max}$.  Unbiased estimator -> $E\left[\widehat{\theta\,}\right] = kE[X_{\max}] = \theta$  Is my logic wrong here?
61811
Show by a counterexample that for a family $(X_i)_{i\in I}$ of random variables the independence of all pairs $(X_i,X_j)$ with $i,j\in I, i\neq j$ does not imply the independence of the family (It is enough to consider $\mbox{card}(I)=3$.).      Hello, I am searching for a simple example. The professor gave the hint that one might consider random variables which only can take two different values, but I do not know exactly what is meant.  By the way:  We call the family $(X_i)_{i\in I}$ of random variables independent, if $(\sigma(X_i))_{i\in I}$ is independent, whereat we call a family $(\mathcal{E}_i)_{i\in I}$ with $\mathcal{E}_i\subset\mathcal{A}$ independent, if for any finite subset $I_0\subset I$ and any choice $E_i\in\mathcal{E}_i, i\in I_0$, it is $$ \mathbb{P}(\bigcap_{i\in I_0}E_i)=\prod_{i\in I_0}\mathbb{P}(E_i). $$
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Let $X\sim N(0,1)$ and let $Z$ be a random variable independent of $X$ such that:  \begin{equation} \Pr(Z=z) = \begin{cases} \frac{1}{2} &amp; \mbox{if $z = -1$ or $z=1$}, \\ \\ 0 &amp; \mbox{otherwise}. \end{cases} \end{equation}   Define $Y = ZX$. Show that $Y \sim N(0,1)$. Determine $\operatorname{Cor}(X,Y)$. Are $X$ and $Y$ independent?   Progress:  I'm basically stuck at the beginning. I got as far as $P(ZX &lt; y) = P(Z &lt; y/X)$, which I'm not even sure is correct. I'm also not clear on what it means by defining $Y = ZX$ - does it mean I need to state the probability density function or something? And how do I show that $Y \sim N(0,1)$?
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$x_{1}...x_{n}$ are independent continuous random variables with common distribution function $F(x)$,consider the order statistics $(x_{(1)},...,x_{(n)})$, compute $E(F(x_{(n)})-F(x_{(1)}))$  I have no idea to this problem, anyone could help me? Thanks!
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I was implementing a simple scheme of Bernoulli distribution sampler. $ X \sim B(p) $. I have a function that generates a uniform random number $r \in (0,1)$. Then, I set $ X = 1 $  if $p &gt; r $, and $X =0$ otherwise. Is this correct?
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I have a question concerning about one step in the Metropolis algorithm. The algorithm ,   Generate a proposed new sample value  from the jumping distribution $Q(x'|x_t)$ Calculate the acceptance ratio $a = \frac{P(x')}{P(x_t)}$ If $a\geq 1$, accept $x'$ by setting $x_{t+1}= x'$ Else, accept $x'$ with probability $a$. That is, pick a uniformly distributed random number $r$ between 0 and 1, and if $r\leq a$ set $x_{t+1}=x'$, else set $x_{t+1} = x_t$   I feel unclear about step 4, in step 4 if we have $a&lt;1$, which means $P(x_t)&gt;P(x')$, $x'$ is further away from the large mass of the density than $x_t$. However, we are not dumping $x'$ directly, we leave it to a 'gamble' to decide its go or stay. This gamble is the uniform distribution, a random draw from uniform distribution then compare it with $a$. I do not understand what's happening here. Why won't we just dump $x'$, why we have to give it a chance to 'nature'?
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Let $X \sim Bin(n,p)$   Part A.  Show that the argument      "Then $X/n \sim Bin()$ with $E[X/n]=p$and $Var[X/n]=pq/n$"    is false.   My book says we can prove this result using a moment generating function. But to me it makes no intuitive sense: We have Bin-distributed RV $X$ and if we rescale it with $1/n$ of course it will be Bin-distributed.  And we know from standard formulas that $E[aX]=aE[X]$ and $Var[aX] = a^2Var[X]$ so I think the argument is correct.   (For full disclosure, the argument we should falsify according to the book is "Let X be a binomial random variable with mean np and variance np(1−p). Show that the ratio X/n also has a binomial distribution with mean p and variance p(1−p)/n." But I think the argument is correct. )    Ok, now I get part A. Here is Part B. Prove it using the moment generating function of $X/n$.   Attempt: I get $$m_X(t)=E[\exp(tX/n)]=(m_X(t))^{1/n}$$
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Let $X_{1},X_{2},....,X_{n}$ be independent $N(0,1)$ random variables and set $\bar{X}=\frac{1}{n}\sum_{i=1}^{n}X_{i}$ show that $\bigg(\bar{X},\frac{1}{n}\sum_{i=1}^{n}(X_{i}-\bar{X})^{2}\bigg)$ and $\bigg({X}/\sqrt{n},\frac{1}{n}\sum_{i=1}^{n-1}(X_{i})^{2}\bigg)$ have the same distribution. This shows that the sample mean and sample variance are independent in this model.  Thoughts I tried showing that they have the same characteristic function but couldn't get it to work that way. I think maybe I could do something involving the CLT but I'm not sure how.
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We have $X_1,...$ indepdent random variables with common distribution $F$ and $N$ is a geometric random variable independent of $X_i$'s. Let $M = \max(X_1,...X_N)$. Im trying to convince myself that   $$ P( M \leq x \mid N &gt; 1 ) = F(x) P(M \leq x ) $$  I understand that since $X_1$ is independent of all $X_i's$ and from $N$, when we can pull it out of the probability and we have   $$ P(X_1 \leq x) P( \max(X_2,....,X_N ) \leq x \mid N &gt; 1 ) $$  And then we have  $$ F(x)  P( \max( X_1,...,X_{N-1} \leq x \mid N &gt; 1 ) $$  How does    $$ P( \max( X_1,...,X_{N-1} \leq x \mid N &gt; 1 )  = P( M \leq x ) $$   ??
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Let $T_n$ be a random variable with $T_n=X_1+...+X_n$ where the $X_i$'s are iid. Further we set $N(t)=max\{ n: T_n\leq n\}$ with the property $\Pr(N(t)&lt;\infty)=1$.  I want to prove that $\mathbb{E}[N(t)]&lt;\infty$.   I have done this so far. Let $F_n$ be the distribution function of $T_n$. $$\mathbb{E}[N(t)]=\sum_{n=1}^{\infty} n P(N(t)=n)=\sum_{n=1}^{\infty} n (F_n(t)-F_{n+1}(t))=\sum_{n=1}^{\infty} F_n(t)$$  Further I know that $0 \leq F_n \leq 1$. But does this implies $\mathbb{E}[N(t)]&lt;\infty$ ? I don't think so... I don't know what I can do further...   Can anybody help me how I can proof it?   Thank you for any help!!!
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I am reading an article in which the author states that given two i.i.d random variables $X,Y\sim\mathcal{N}(0,1)$, we have $\mathbb{P}(\frac{X}{Y}\le t)=\mathbb{P}(\frac{X}{|Y|}\le t)$ since the random varaibles $\frac{X}{Y}$ and $\frac{X}{|Y|}$ are identically distributed by the symmetry of the standard normal distribution.  But I don't really understand why the random variables $\frac{X}{Y}$ and $\frac{X}{|Y|}$ are identically distributed. Can you provide a reason for why these two random variables are identically distributed?
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I know there's no uniform distribution for a countably infinite set, but I'm wondering if there's still a way to determine the probability of picking from a subset of a countably infinite set.  For example, what's the probability that I pick an odd number from the set of naturals, assuming I'm picking randomly? Is this even a coherent question? If so, is there a textbook approach to this problem in measure theory/probability theory/probability measure?  Thanks!
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I am trying to understand the relationship of a Random Variable to a population and random samples. I understand that a Random Variable is a function that maps each event in a sample space to a number. In this context, what would be a sample space and mapping? Would each sample be governed by the probability mass function?  Do we think of each sample as a Random Variable that takes on values from a population (sample space) randomly?   Or do we think of a population as a Random Variable and our sample as a random collection of those points?  I would appreciate if you can explain this to someone who is new to mathematical statistics. Thank you!
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I tried to do the following task:  Let $X_1,\ldots,X_n$ be a random sample with the pdf $$f_{X_i}(x|\theta)=\begin{cases}e^{i\theta-x},&amp;&amp;x\ge i\theta\\0,&amp;&amp;x&lt;i\theta\end{cases}$$ Show that $T=\min_i \frac{X_i}{i}$ is a sufficient statistic for $\theta$.  First of all, I don't know what the notation $f_{X_i}(x|\theta)$ should mean. Does it mean that $f(x_1,\ldots,x_n|\theta)=\prod_{i=1}^{n}e^{i\theta-x_i}1_{x_i\ge i\theta ,\forall i \in\{1,\ldots,n)\}}$? (I plan to use the factorization theorem)