Questions tagged [factorisation-theorem]

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Why is the weak likelihood principle not a theorem?

The weak likelihood principle (WLP) has been summarized as: If a sufficient statistic computed on two different samples has the same value on each sample, then the two samples contain the same ...
virtuolie's user avatar
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How do I proceed from here for factorization of likelihood / joint normal density for finding sufficient statistics?

I'm trying to show that the statistic $\left(\sum_{i = 1}^n Y_i, \sum_{i = 1}^n Y_i^2 \right)$ is sufficient for $\mu$ where $(Y_1, \dots, Y_n)$ is a random sample from $N(\mu, \mu)$ for $\mu > 0$. ...
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Order Statistics - simple random sampling without replacement

I am studying the Watson-Royall theorem and I have a question in item 2: "under simple random sampling, the vector of order statistics is sufficient ..." I wish to use the factorisation ...
Alisson Silva's user avatar
5 votes
2 answers
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Finding UMVUE for a function of a Bernoulli parameter

Given $m$ i.i.d. Bernoulli( $\theta$ ) r.v.s $X_{1}, X_{2}, \ldots, X_{m},$ I'm interested in finding the UMVUE of $(1-\theta)^{1/k}$, when $k$ is a positive integer. . I know $\sum X_{i}$ is a ...
wanderer's user avatar
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Problem on sufficient statistics

Let the distribution of $X_1,X_2,...X_n$ depend on two parameters $a, b$ such that there exists a single sufficient statistic, for either parameter when the other is fixed/known. Show that there is ...
The Three Muskets's user avatar
3 votes
1 answer
676 views

Showing the sample mean is a sufficient statistics from an exponential distribution

Suppose that the lifelengths ( in thousands of hours) of light bulbs are distributed Exponential($\theta$), where $\theta>0$ is unknown. If we observe $\overline x = 5.2$ for a sample of $20$ light ...
hkj447's user avatar
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Sufficiency of $|X|$ when $X\sim N(0,\sigma^2)$ without using Factorization theorem [duplicate]

Question: Given, $X\sim N(0,\sigma^2)$. By means of conditional approach show that $|X|$ is a sufficient estimator for $\sigma^2$. My Attempt: This problem is very easy if we use Fisher–Neyman ...
RATNODEEP BAIN's user avatar
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1 answer
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When a function of sufficient statistic is itself sufficient?

I'm following notes at onlinecourses and I got confused on transformation of sufficient statistics. For example, if $X$ is a sufficient statistic for $\mu$, why $Y=X^2$ is not a sufficient statistic ...
AlexMe's user avatar
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7 votes
2 answers
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Puzzled by definition of sufficient statistics

I am learning about sufficient statistic from Mood, Graybill, and Boes's Introduction to the Theory of Statistics. I am slightly confused by the book's definition of a sufficient statistic for ...
Noppawee Apichonpongpan's user avatar
-1 votes
1 answer
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Proving sufficiency by showing ratio of statistic pdf to sample pdf is independent of unknown parameter

Let $X_1,...,X_n$ be iid random variables with densities given by $$ f_{x_i}(x|\theta)=e^{i\theta - x}\mathbb{I}_{(i\theta,\infty)}(x), $$ when $x>i\theta $ and $x=0$ otherwise. Let $T$ be the ...
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1 vote
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Minimal Sufficient Statistic for Gaussians with different means

I have the following problems on my Statistics course (using Casella and Berger's book) problem set: 1) Let $Y_{i} = X_{i}'\theta + U_{i}$ where $\theta \in \mathbb{R}^k$ and $U_{i}$ are iid $N(0,...
Raul Guarini Riva's user avatar
1 vote
1 answer
1k views

Showing sufficiency using the Fisher-Neyman factorization theorem

I have derived a likelihood function for $\theta$ as follows: $$L(\theta)=(2\pi\theta)^{-n/2} \exp\left(\frac{ns}{2\theta}\right)$$ Where $\theta$ is an unknown parameter, $n$ is the sample size, ...
David Parks's user avatar
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Is an injective statistic always sufficient?

New to the concept of sufficient statistics. Does it follow, in general, from the factorization theorem that any injective statistic is necessarily a sufficient statistic? I cannot find anything ...
Rachel's user avatar
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2 votes
2 answers
469 views

Can the Fisher factorization theorem be understood as a product of densities?

Let $T$ be some random variable on a probability space $\Omega$. Then we have, for $x\in\Omega$: $$P(x) = P(x|T=T(x))P(T = T(x))$$ This equation is nonsense in an arbitrary probability space but ...
Jack M's user avatar
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1 answer
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Manually computing fisher test with Big Numbers in Cells

Hi all I have a 2x2 contingency table like this: ...
jbest's user avatar
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3 votes
1 answer
896 views

Sufficient statistics for a continuous distribution

Below is a discussion of sufficient statistics for a continuous distribution, taken from the third edition of Lehmann's Testing Statistical Hypotheses. I understand the discussion until the underlined ...
Typhode's user avatar
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5 votes
2 answers
870 views

Does Fisher's factorization theorem provide the pdf of the sufficient statistic?

From Wikipedia Fisher's factorization theorem or factorization criterion provides a convenient characterization of a sufficient statistic. If the probability density function is $ƒ_θ(x)$, then $T$ ...
Tim's user avatar
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