Let $Y_{(n)}=\max(Y_{1},Y_{2},\ldots,Y_{n})$, where $Y_{1},Y_{2},\ldots,Y_{n}$ is a sample uniform distribution in $(0,\theta)$. Find the MVUE for $\theta$.  My approach:  I know that $U=Y_{(n)}$ is a sufficient statistic for $\theta$. I could prove this using the factorization theorem by Neymann. I know that $\displaystyle h(U)=\left(\frac{n+1}{n}\right)U$ satisfy that $\mathbb{E}[h(U)]=\theta$.  So, $$\left(\frac{n+1}{n}\right)Y_{(n)}$$ is a (or the?) MVUE for $\theta$.  My approach is correct? Can I find other MVUE for $\theta$?
120883
I am having trouble understanding how to compute $\operatorname E[\bar{X}\mid X_{(n)}]$ related to the following premise.  Compute the UMVUE using the Rao–Blackwell Theorem for the following. $X_1,X_2, \ldots ,X_n$ i.i.d. to $\operatorname{Uniform}(0,\theta)$.  I am able to derive that $\hat{\theta}_\text{MM}=2\bar{X}$ and $\hat{\theta}_\text{MLE}=X_{(n)}$.  Since $\hat{\theta}_\text{MM}$ is unbiased and $X_{(n)}$ is a sufficient estimator, I know that $$\operatorname E[2\bar{X}\mid X_{(n)}]$$ must give us the UMVUE.  However, I have no idea how to proceed from here. I appreciate your help.
60321
Let $X_i\sim\mathrm{Uniform}(0,\theta)$ be iid, what is $E[\max\{X_1,\ldots,X_n\}]$? Apparently the answer is $$\frac{n}{n+1}\theta,$$ but I do not see why?  It seems intuitive in that you would "expect" them to be spaced out evenly, hence the maximum would be $\frac{n}{n+1}$-of-the-way along the interval, but how can we prove this mathematically? I feel like it should be simple, but evidently $$E[\max\{X_1,\ldots,X_n\}]\neq\max\{E[X_1],\ldots,E[X_n]\}.$$ Thanks.
108193
I was reading about the german tank problem and they say that in a sample of size $k$, from a population of integers from $1,\ldots,N$ the probability that the sample maximum equals $m$ is:  $$\frac{\binom{m-1}{k-1}}{\binom{N}{k}}$$  This make sense. But then they take expected value of the sample maximum and claim:  $$\mu = \sum_{m=k}^N m \frac{\binom{m-1}{k-1}}{\binom{N}{k}} = \frac{k(N+1)}{k+1}$$  And I don't quite see how to simplify that summation. I can pull out the denominator and a $(k-1)!$ term out and get:  $$\mu = \frac{(k-1)!}{\binom{N}{k}} \sum_{m=k}^N m(m-1) \ldots (m-k+1)$$  But I get stuck there...
51291
Let $\hat{\theta}_n= -\frac{n}{\sum_{i=1}^n \log(X_i)}$, where $X_i$ are i.i.d. samples from distribution with pdf $\theta x^{\theta-1}$ for $x \in (0,1)$. How to prove that $\hat{\theta}_n$ is consistent estimator of $\theta$.
84344
I'm trying to answer the following question from the book :  Let $X_1,X_2,\dots$ be a sequence of sub-gaussian random variables, which are not necessarily independent. Show that   $E\bigg[ \max_i \frac{|X_i|}{\sqrt{1 + \log i}} \bigg] \le CK$,  where $K = \max_i \|X_i\|_{\psi_2}$. Deduce that for ever $N \ge 2$ we have   $E\bigg[ \max_{i \le N} |X_i| \bigg] \le CK \sqrt{\log N}$.  I've tried to figure out what is the distribution of the maximum of Gaussians, but I'm reaching only inequalities that that don't help me answer the question.   I've also seen a similar question .  Does anyone have a clue or something to start with in order to answer this question?  Thanks!
71524
Find the MLE of the unknown parameter $\theta$ when $X_1,X_2,...,X_n$ is a sample from the distribution whose density function is: $$f_X(x) = \frac12e^{-|x-\theta|}, -\infty&lt;x&lt;\infty$$ What I did so far: $$\text{maximize }(\frac12)^ne^{-|x_1-\theta|-|x_2-\theta|-...-|x_n-\theta|} \\ \text{thus we want to minimize }|x_1-\theta|+|x_2-\theta|+...+|x_n-\theta|$$ Since $|x_i-\theta|$ is positive for all $i$, we want to solve for $\theta$ so: $$|x_1-\theta|+|x_2-\theta|+...+|x_n-\theta|=0$$ What do I do from here?
104134
Let $X_1,X_2,...,X_n$ be random sample from a DF $F$, and let $F_n^* (x)$ be the sample distribution function. We have to find $\operatorname{Cov}(F_n^* (x), F_n^* (y))$ for fixed real numbers $x, y$ where $F_n^* (x)$ is a sample distribution. My approach: $$\text{Cov}(F_n^* (x), F_n^* (x)) = \mathbb{E}[(F_n^* (x) - \mathbb{E}[F_n^* (y)])(F_n^* (y) - \mathbb{E}[F_n^* (y)])]$$ $$\text{Cov}(F_n^* (x), F_n^* (x)) = \mathbb{E}[F_n^* (x) .F_n^* (y)] - \mathbb{E}[F_n^* (x)]\mathbb{E}[F_n^* (y)]$$ $$\text{Cov}(F_n^* (x), F_n^* (x)) = \mathbb{E}\bigg[\bigg(F_n^* (\min(x, y))\bigg) \bigg(F_n^* (\min(x, y)) + \int_{\min(x,y)}^{\max(x,y)} f_n^* (x) dx\bigg)\bigg] - \mathbb{E}[F_n^* (x)]\mathbb{E}[F_n^* (y)]$$ where $f_n^*(x)$ is a probability density function of the sample. $$\text{Cov}(F_n^* (x), F_n^* (x)) = \mathbb{E}\bigg[\bigg(F_n^* (\min(x, y))\bigg)^2\bigg]+\mathbb{E}\bigg[ \bigg(F_n^* (\min(x, y)).\int_{\min(x,y)}^{\max(x,y)} f_n^* (x) dx\bigg)\bigg] - \frac{F_n (x)F_n (y)}{n^2}$$ $$\text{Cov}(F_n^* (x), F_n^* (x)) = \frac{(F_n (\min(x, y))^2}{n} - \frac{F_n (x)F_n (y)}{n^2} +\mathbb{E}\bigg[ \bigg(F_n^* (\min(x, y)).\int_{\min(x,y)}^{\max(x,y)} f_n^* (x) dx\bigg)\bigg] $$ How can I proceed from here?
83455
Assume we have $n$ distinct points $x_1,x_2,...,x_n\in \Bbb{R}$, and observations which are independent random variables $y_1,y_2,...,y_n$ depending on formula: $$ y_i=\theta_0+\theta_1x_i^2+\theta_3e^{x_i}+\varepsilon_i $$ where $\varepsilon_i$ are distributed as $N(0,\sigma^2)$. Derive the maximum likelihood estimator for $\theta_{0},\theta_{1},\theta_{2}$ .  I know how to find the MLE for a known distribution like normal or Poisson, but for this question, I don't how I can find the likelihood function, should I consider it as normal distribution? And I don't know what the mean and variance would be since I can't figure out the distribution and likelihood function. Also, should I consider $\varepsilon_i$ as a useful variable, or just treat it as an error in estimation? Thanks for any help.
55461
Let $X$, $Y$ and $Z$ be i.i.d. uniform (0,1) random variables. What is the distribution of $(XY)^Z$? I've tried to solve it via mgfs, and what I've found is:  $$E\left(e^{(XY)^Z}\right)=E\left(E\left(e^{(XY)^Z}|Z\right)\right)$$ $$=E\left(E\left(e^{(XY)}\right)^Z\right)$$ $$=E\left(2\left(\frac{e^t-1}{t}\right)^Z\right)$$ $$=2\int_0^1\left(\frac{e^t-1}{t}\right)^zdz$$ $$=2\int_0^1\exp\left({z\log\left(\frac{e^t-1}{t}\right)}\right)dz$$ $$=2\left[\frac{\exp\left({z\log\left(\frac{e^t-1}{t}\right)}\right)}{\log\left(\frac{e^t-1}{t}\right)}\right]_0^1$$ $$=2\frac{{\frac{e^t-1}{t}}-1}{\log\left(\frac{e^t-1}{t}\right)}$$  But I don't see where to go from here. It doesn't look like the mgf of any distribution that I know. How can I determine the distribution from this, or is there another way?
54625
Let $X_1,\ldots,X_n \sim$ Poisson$(\lambda)$. I wish to find UMVUE of $e^{-\lambda}$. Here $\overline{X}_n$ is complete and sufficient for $\lambda$ (hence for $e^{-\lambda}$??). Define $Y_i=1$ if $X_i=0,$ $Y_i=0$ otherwise. Clearly (I checked!) each $Y_i$ and hence $\overline{Y}_n$ is unbiased for $e^{-\lambda}$. Then I should be able to use the Lehmann Scheffe theorem to deduce the UMVUE. The estimator I get looks like $$\frac{1}{n} E\left\{\#X_i:X_i=0 \mid \sum_{i=1}^n X_i\right\}$$ whereas the correct answer is $$\left(\frac{n-1}{n}\right)^{\sum_{i=1}^n X_i}$$ I'd appreciate if anyone lets me know where I'm going wrong and guides me into the right path. Thank you.
74497
A random sample $X_1,\ldots,X_n$ is taken from a distribution with probability density function which is $B(\theta,2\theta)$. Show that $\prod X_i(1-X_i)^2$ is a sufficient statistic for $\theta$.  Below is the beta distribution with the parameters referred to:  $$f_X(x;\theta)=\frac{\Gamma(3\theta)}{\Gamma(\theta)(\Gamma(2\theta)}x^{\theta-1}(1-x)^{2\theta-1}$$  To be honest, this question is beyond me. I don't know how to start.   I have been reading up about sufficient statistics, and using factorisation theorem, but I don't know if this is the right approach and I don't really know where to begin with this.  Do I need to work out a joint pdf? Do I need to know the likelihood function? Is factorisation theorem even the right thing to use?  Any help (even if it is just directing me to a source for "sufficient statistics for dummies") is much appreciated!
77951
$x_{1}...x_{n}$ 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!
28926
Let $X_1,...X_n$ be a random sample from the Pareto distribution with parameters $\alpha$ and $\theta$, where $\alpha$ is known.   Find the maximum likelihood estimator for $\theta$ and say if it is unbiased, if not find an unbiased estimator  My Approach:  $$f(x;\alpha, \theta) = \alpha \theta^\alpha x^{-(\alpha +1)},\quad x \ge \beta$$  $$L(\theta) = \alpha^n \theta^{\alpha n} \left(\prod_{i=1}^n x_i\right)^{-(\alpha+1)}$$  Taking log for $L(\alpha)$ gives  $$\ln L(\theta) = n \ln(\alpha) + \alpha n \ln(\theta) + \sum_{i=1}^n -(\alpha+1) \ln(x_i)$$  Then since $\ln L(\theta)$ is an increasing function if $\theta$ increases, and for a Pareto distribution we have that $\theta \le x$ we conclude that the maximum likelihood estimator is $\hat\theta=\min {x_i}$ (the first order statistic) Am I right?  Then to prove that it is and unbised statistic we have to prove that $E(\hat{\theta}) = \theta$.  I donot know how to do it I just thought using the p.d.f of the firs order statistic and integrate from $\theta $ to infinite but I'm not sure about this.  Any ideas?
91834
Let $X,Y$ independent random variables with $X,Y\sim \mathcal{N}(0,1)$. Let $Z=\max(X,Y)$. I already showed that $F_Z$ of $Z$ suffices $F_Z(z)=F(z)^2$.  Now I need to find $EZ$.  Should I start like this ? $$EZ=\int_{-\infty}^{\infty}\int_{-\infty}^\infty \max(x,y)\frac{1}{\sqrt{2\pi}}e^{-1/2x^2}\frac{1}{\sqrt{2\pi}}e^{-1/2y^2} dxdy$$
53110
Could anybody help me with solving such problem:  You are given two independent random variables $X$ and $Y$ with continuous uniform distribution. You are to find expected value $E$ of random variable $Z = \max(X,Y)$.
105214
I want to find the expected value of $\text{max}\{X,Y\}$ where $X$ ist $\text{exp}(\lambda)$-distributed and $Y$ ist $\text{exp}(\eta)$-distributed. X and Y are independent. I figured out how to do this for the minimum of $n$ variables, but i struggle with doing it for 2 with the maximum.  (The context in which this was given is waiting for the later of two trains, with their arrival times being exp-distributed).  Thanks!
106599
If I have n samples from an $N(\mu,\sigma)$ distribution, how can I estimate the distribution from a subset of m of these n samples where the order within those n is known.  For example, if n = 100 and m is 3, I might know that one sample is 10 and 3rd highest of the 100, one is 5 and 42nd highest, and the last is 2 and 74th highest.