35
votes
Why are we using a biased and misleading standard deviation formula for $\sigma$ of a normal distribution?
For the more restricted question
Why is a biased standard deviation formula typically used?
the simple answer
Because the associated variance estimator is unbiased. There is no real ...
- 13.1k
16
votes
Accepted
Why are we using a biased and misleading standard deviation formula for $\sigma$ of a normal distribution?
The sample standard deviation $S=\sqrt{\frac{\sum (X - \bar{X})^2}{n-1}}$ is complete and sufficient for $\sigma$ so the set of unbiased estimators of $\sigma^k$ given by
$$
\frac{(n-1)^\frac{k}{2}}{2^...
- 31.2k
9
votes
Why are we using a biased and misleading standard deviation formula for $\sigma$ of a normal distribution?
Q2: Would someone please explain to me why we are using SD anyway as it is clearly biased and misleading?
This came up as an aside in comments, but I think it bears repeating because it's the crux ...
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9
votes
UMVUE for normal distribution $\sigma$
Although the question was posted almost 4 years ago, I would like to answer this question. English is not my mother tongue and I am learning it so please don't mind my awkward sentences.
To solve ...
- 873
9
votes
Accepted
Find UMVUE of $\frac{1}{\theta}$ where $f_X(x\mid\theta) =\theta(1 +x)^{−(1+\theta)}I_{(0,\infty)}(x)$
Your reasoning is mostly correct.
The joint density of the sample $(X_1,X_2,\ldots,X_n)$ is
\begin{align}
f_{\theta}(x_1,x_2,\ldots,x_n)&=\frac{\theta^n}{\left(\prod_{i=1}^n (1+x_i)\right)^{1+\...
- 11.4k
7
votes
Accepted
Finding UMVUE for a function of a Bernoulli parameter
Except when $k=1$, given a finite sequence of i.i.d. Bernoulli
$\mathcal B(θ)$ random variables $X_1,X_2,\ldots,X_m$, there exists no
unbiased estimator of $(1−θ)^{1/k}$, when $k$ is a positive ...
- 107k
7
votes
Accepted
Find best unbiased estimator for $\theta$ when $X_i\sim U(-\theta,\theta)$
Note that likelihood function depends on the sample $X_1,\ldots,X_n$. Therefore, there can be no $x$ in the argument.
$$f(X_1,\ldots,X_n\mid \theta)=\prod_{i=1}^n\left[\left(\frac{1}{2\theta}\right) I(...
- 616
7
votes
Accepted
UMVUE for $g(\theta)=\theta^2$ of Poisson random variables
Yes, that's correct. To make it rigorous, you can apply Lehmann–Scheffé Theorem with $Y = T(X)$ and $\varphi(Y) = \frac{1}{n^2}(Y^2 - Y)$.
After a closer look at the Wikipedia link above, it seems ...
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6
votes
Find the joint distribution of $X_1$ and $\sum_{i=1}^n X_i$
Correct me if I am wrong, but I don't think one needs to find the conditional distribution to find the conditional expectation for the UMVUE. We can find the conditional mean using well-known ...
- 11.4k
6
votes
Accepted
Unbiased Estimator for the CDF of a Normal Distribution
As a comment suggested, an unbiased estimator is (one minus) the empirical distribution function
$$\hat P(X_1 > 0) = 1-\hat F_X(0) = 1-\frac 1n \sum_{i=1}^n I\{x_i \leq 0\}$$
where $I\{\}$ is the ...
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6
votes
Accepted
UMVUE of $e^{-\lambda}$ from poisson distribution
Yes, Lehman Scheffe gives you the sufficiency of $\bar{X}$ for $e^{-\lambda}$. The problem is that $\bar{Y}$ does not actually condition on the sufficient statistic. Although the $Y$ are a function of ...
- 64.1k
6
votes
Accepted
UMVU estimator for non-linear transformation of a parameter
Your final answer is not quite right. The conclusion due to the sample mean $\bar X$ being only sufficient for $\mu$ also looks faulty.
Recall that $T(X_1,X_2,\cdots,X_n)=\sum_{i=1}^n X_i$ is a ...
- 11.4k
5
votes
Complete statistic for $\sigma^2$ in a $N(\mu,\sigma^2)$
I think I solved my own question. Comments about this answer and new answers are welcome.
If $x_1,\ldots,x_n$ are observations in a $N(\mu,\sigma^2)$ population and $\mu$ is unknown, then $$f(x_1,\...
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5
votes
The pdf of $\frac{X_1-\bar{X}}{S}$
I'd like to suggest this way to get the pdf of Z by directly calculating the MVUE of $P(X\leq c)$ using Bayes' theorem although it's handful and complex.
Since $E[I_{(-\infty,c)}(X_1)]=P(X_1\leq c)$ ...
- 873
5
votes
Accepted
Finding UMVUE of $\theta e^{-\theta}$ where $X_i\sim\text{Pois}(\theta)$
The Poisson distribution is a one-parameter exponential family distribution, with natural sufficient statistic given by the sample total $T(\mathbf{x}) = \sum_{i=1}^n x_i$. The canonical form is:
$$...
- 130k
5
votes
Accepted
Find UMVUE of $\theta$ where $f_X(x\mid\theta) =\theta(1 +x)^{−(1+\theta)}I_{(0,\infty)}(x)$
From your previous question, you already have the complete sufficient statistic:
$$T(\mathbf{X}) = \sum_{i=1}^n \ln(1+X_i).$$
The simplest way to find the UMVUE estimator for $\theta$ is to appeal ...
- 130k
5
votes
Accepted
Find the UMVUE of $\frac{\mu^2}{\sigma}$ where $X_i\sim\mathsf N(\mu,\sigma^2)$
I have skipped some details in the following calculations and would ask you to verify them.
As usual, we have the statistics $$\overline X=\frac{1}{4}\sum_{i=1}^4 X_i\qquad,\qquad S^2=\frac{1}{3}\...
- 11.4k
5
votes
Does UMVUE of $\frac{\theta_{x}}{\theta_{y}}$ exist? X $\sim$ exp($\theta_{x}$), Y $\sim$ exp($\theta_{y}$)
If $\bar X$ and $\bar Y$ are independent, since $\bar Y$ is an unbiased estimator of $\theta_y^{-1}$, the question boils down to whether or not there exists an UMVUE of $𝜃_x$ which calls first for ...
- 107k
5
votes
Accepted
UMVUE of parameter from Zero Truncated Poission distribution
First of all you could rewrite the distribution to fit the general form of the exponential family.
$$ \frac{1}{x!}e^{x\log\theta -\theta -\log(1-e^{-\theta})} $$
so the sufficient statistics is $T:=x$,...
- 1,033
5
votes
Accepted
How do I find the UMVUE of $\sqrt{\alpha}$ here?
You are going down the correct path—when you are looking for the UMVUE in a parametric problem, the simplest method in most cases is to use the Lehmann–Scheffé theorem, which says that if you can form ...
- 130k
4
votes
What is the necessary condition for a unbiased estimator to be UMVUE?
On Uniformly Minimum Variance Unbiased Estimation when no Complete Sufficient Statistics Exist by L. Bondesson gives some examples of UMVUEs which are not complete sufficient statistics, including the ...
- 176
4
votes
Accepted
Is the OLS estimator the UMVUE (assuming Normality)?
Under the assumptions
$$
\begin{align}
&\mathbf{y} = \mathbf{X} \mathbf{b} + \mathbf{e}, \;\mathbf X \;\text{full column rank},\\
&\mathbf e \mid \mathbf{X} \sim \mathop{\mathcal{N}}\left(\...
- 6,355
4
votes
UMVUE of distribution function $F$ when $X_i\sim F$ are i.i.d random variables
Sufficiency: The ordered data can be recovered from the ECDF, so the sufficiency of the former implies the sufficiency of the latter.
Completeness: As you have shown, for any fixed value $x$ we have $...
- 130k
4
votes
Accepted
Find UMVUE of $p^3$
Actually, this problem is a classic example of the Lehmann–Scheffé theorem. The theorem states
If a statistic that is unbiased, complete and sufficient for some parameter $\theta$, then it is the ...
- 1,797
4
votes
UMVUE of $\cos\theta$ when $X_i\sim U(0,\theta)$
Assuming you have a sample of $n$ observations.
The density of the complete sufficient statistic $X_{(n)}$ is
$$f_{X_{(n)}}(t)=\frac{nt^{n-1}}{\theta^n}\mathbf1_{0<t<\theta}$$
Any function ...
- 11.4k
4
votes
Accepted
UMVUE of the probability a Poisson R.V is odd?
We have the exact expression, verifiable using the power series expansion of $e^{\lambda}$: $$\sum_{k=0}^\infty \frac{\lambda^{2k+1}}{(2k+1)!}=\frac{1}{2}(e^{\lambda}-e^{-\lambda})$$
So that reduces ...
- 11.4k
4
votes
Accepted
Estimator with variance equal to Cramér-Rao lower bound in $N(x_i\theta,1)$-distribution
The complete sufficient statistic here is actually $\sum_i x_i Y_i$ and not $\sum_i Y_i$. You can see this by writing out the joint distribution in the exponential family form. As it forms a full-rank ...
- 2,544
4
votes
Accepted
UMVUE for probability of cutoff
$\newcommand{\v}{\operatorname{var}}\newcommand{\c}{\operatorname{cov}}\newcommand{\e}{\operatorname{E}}$
Lemma: Suppose $U,V$ are normally distributed, but also jointly normally distributed, i.e. ...
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4
votes
Can a Bayesian estimator perform better than an MVUE?
First of all the equation is $MSE = Bias^2 + Variance$ and not Bias.
Now in MVUE estimators, the Bias is zero and the variance is equal to the CRLB (Cramer-Rao Lower Bound) calculated as follows:
CRLB(...
- 161
4
votes
Can a minimum variance unbiased estimator be inconsistent?
One counterexample is when there's no consistent estimator. Suppose $X_i\sim N(\mu_i,1)$ where $\mu_i$ are all distinct unknown parameters. The UMVUE of any $\mu_i$ is $X_i$, but it's not consistent.
...
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