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2 votes
1 answer
31 views

Prove that $T$ is a complete statistic and find a UMVUE for $p$

While preparing for my prelims, I came across this problem: Let $X_1, X_2,\cdots, X_n$ be a sequence of Bernoulli trials, $n \geq 4.$ It is given that, $X_1,X_2,X_3 \stackrel{\text{i.i.d.}}{\sim} Ber(\...
Wrik's user avatar
  • 21
1 vote
0 answers
100 views

Finding UMVUE of a parameter in form of probability of discrete random variables

We have $X$ and $Y$ as independent discrete random variables both in ${1, 2, ...}$. Their pmf's are: $f(x|\alpha)=P(X=x)=\alpha(1-\alpha)^{x-1}, x=1, 2, ...$ $f(y|\beta)=P(Y=y)=-\frac{1}{\log\beta}\...
AlgoManiac's user avatar
0 votes
1 answer
294 views

UMVUE for P(X > k) in exponential distribution [duplicate]

I have to find UMVUE for $exp(-k*a)$ where X ~ Exponential(a); k is a positive real number. I tried it using Lehmann-Scheffe theorem. Since, T = $sum(xi) (i = 1,..,n)$ is complete sufficient statistic ...
Kcd's user avatar
  • 107
0 votes
0 answers
338 views

UMVUE of Bernoulli random variables

Let $X_1, X_2..... X_n$ be a random sample from a Bernoulli population with parameter $p$. A sufficient statistic is $\sum_{i=1}^{n}X_i$. If we define $$ U(X_1,X_2,\ldots,X_n)= \begin{cases}1/2n &\...
Muskaan Madan's user avatar
1 vote
1 answer
386 views

Unbiased Estimator based on Sufficient Statistic

suppose $X_1, ... , X_n$ are iid with pdf $f(x|\beta) = e^{-(x-\beta))}I_{(\beta, \infty)}(x)$ and the pdf of ( the smallest order statistic) $X_{(1)}$ is given by $f_{X_1}(x)$ = n $ *$ $e^{n(\...
Pedros's user avatar
  • 213
2 votes
1 answer
187 views

Proving the MVUE is the following

I am stuck on the following question and I was wondering if can get some help. Let $f(x;\theta) = g(\theta)h(x),\ a(\theta) \leqslant x \leqslant b(\theta)$ with $a(\theta)$ decreases and $b(\theta)$...
user_1512314's user avatar
1 vote
0 answers
49 views

Sufficient statistic for the mean of a generic distribution?

Is there such thing as a "sufficient statistics for the expectation?" From what I understand, a sufficient statistic is defined only when there is a family of distributions parametrized by $\theta$ ...
diadochos's user avatar
  • 175
4 votes
0 answers
563 views

The relationship between UMVUE and complete sufficient statistic [duplicate]

Let $X_1,...X_n$ $U(-\theta , \theta)$ I want to find the UMVUE of $\theta$ if it is exists. My answer is , there is no UMVUE in this case. Because there is no complete sufficient statistic that ...
student_R123's user avatar
2 votes
2 answers
1k views

UMVUE for $\theta$ where $X \sim Unif\{1 ,\ldots, \theta\}$

Say we have $X \sim Unif\{1, \ldots , \theta\}$ and we want to find the uniformly minimum variance unbiased estimator for $\theta$. My first assumption was $X_{(n)}$. Which I managed to show is ...
rannoudanames's user avatar
1 vote
1 answer
4k views

Finding sufficient statistic for Weibull density function

I am given the follow problem and am having trouble finding the sufficient statistic. Suppose that Y$_1$, Y$_2$, ..., Y$_n$ denote a Weibull density function, given by: f ( y | $\theta$ ) = Let $...
jmoore00's user avatar
  • 389
1 vote
1 answer
99 views

Find the UMVUE of $U(n_1,θ)$ where $θ>n_1$

Let $X_1,X_2,\dots ,X_n$ follows $U(n_1,θ)$ where $θ>n_1$ and $n_1\le n$. Find the UMVUE of $θ$. My answer is: $$\frac{(n+1)X_n}n - \frac{n_1}n$$ Is that correct?
Halima.Khatun's user avatar
4 votes
0 answers
120 views

Relating sufficient statistics to parameters

I'm studying sufficient statistics and I came across this problem: A dataset consists of independent triples $(W_i,Y_i,Z_i)$ of independent random variables with distributions as follows, $$ W_i \sim ...
ogustavo's user avatar
  • 676
2 votes
1 answer
508 views

UMVUE of a function of sufficient statistic

Let $X_1,\ldots, X_n$ be a random sample from the following distribution: $$f(x\mid\theta)= \frac{4}{\theta}x^3e^{-\frac{x^4}{\theta}}$$ I know that $S = \sum X_i^4$ is a complete and sufficient ...
arina's user avatar
  • 95
2 votes
1 answer
280 views

How does one can guarantee that any unbiased estimator is MVUE due to it containing a minimal sufficient statistic?

Sufficiency is okay. But I don't really get it why the fact guarantees it has minimal variance? Can anyone explain to me somewhat intuitively?
Jason Park's user avatar
10 votes
2 answers
2k views

Find the unique MVUE

This question is from Robert Hogg's Introduction to Mathematical Statistics 6th Version problem 7.4.9 at page 388. Let $X_1,...,X_n$ be iid with pdf $f(x;\theta)=1/3\theta,-\theta<x<2\theta,$ ...
Deep North's user avatar
  • 4,776

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