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Refers to an estimator of a population parameter that "hits the true value" on average. That is, a function of the observed data $\hat{\theta}$ is an unbiased estimator of a parameter $\theta$ if $E(\hat{\theta}) = \theta$. The simplest example of an unbiased estimator is the sample mean as an estimator of the population mean.
1
vote
Unbiased estimator of p in geometric distribution
Assuma as StubbornAtom in a comment that we have only one observation and that $P(X=j)=p(1-p)^j\mathbf1_{j\in\{0,1,2,\ldots\}}$ and $T(X) = \mathcal{I}_{\{X=0\}}$. Then just calculate the expectatio …
1
vote
How to find an unbiased estimator for reciprocal of scale parameter given an iid exponential...
So far this is more of a long comment. You are right that $\bar{x}$ have an Erlang distribution, that is, a Gamma distribution. (Refer to Wikipedia). I find the following density for $\bar{x}$
$$
f( …
6
votes
Accepted
Where does linear regression fit into the bias-variance tradeoff?
OLS is an unbiased estimator assuming the model is true, which is to say,
Effects are exactly linear
All variables with non-zero effects are included
All interactions are included
no non-linear effe …
1
vote
Do we need an unbiased estimator of the variance?
Depends on what your end goal is. It the end goal is inference on regression coefficients (t-tests) then the unbiased variance estimator is just a stepping stone, and unimportant in itself.
But for …
12
votes
Questions about unbiased sample variance
Many misunderstandings here. It would be easier to answer if you defined your terms and included some formulas. But:
1) NO, it is not correct to say that an unbiased estimator is necessarily close …
4
votes
Is an WLS estimator unbiased, when wrong weights are used?
It is unbiased, let's see: Let the linear model be $Y=X\beta +e$, in matrix form, with $E e=0$ and the variance-covariance matrix of the errors $e$ be $\Omega$. We use for weights the matrix $W$. Th …
5
votes
Accepted
Is the sample mean an unbiased estimator of population mean in the presence of autocorrelation?
Yes, autocorrelation (or spatial correlation or ...) do not destroy the unbiasedness of the sample mean as an estimator of population mean.
Expectation is a linear operator, so when you calculate the …
5
votes
Ratio of Unbiased Estimators
No, it will not be unbiased (unless the estimator of the denominator have zero variance.) And it will not help if the numerator and denominator are independent. In general, if $\hat{\theta}$ is an un …
11
votes
Best estimator of the mean of a normal distribution based only on box-plot statistics
An exact answer will be difficult, so first I will look at asymptotic theory. Answers from that could be tested by simulation, comparing it to a maximum likelihood estimator computed by maximizing an …
1
vote
Accepted
Good parameter estimates vs good computed moment estimates
You have data from some distribution family $f(y; \theta)$ and some estimator $\hat{\theta}$ of $\theta$ with "good properties". But you are interested in some function of $\theta$, say $g(\theta)$ ( …
2
votes
Unbiased estimators of the log odds
For the case $g(p)=1/p$, a more rigorous proof for nonexistence of unbiased estimators is For the binomial distribution, why does no unbiased estimator exist for $1/p$?. With $g(p)=\log\frac{p}{1-p}$ …
3
votes
Accepted
What is an "unbiased forecast"?
We can write your predictor for $Y_h$ as
$$
\widehat{Y_h}=T\left(W(x,y),X_h\right)
$$
Then $\widehat{Y_h}$ is unbiased as a predictor for $Y_h$ if
$$ \DeclareMathOperator{\E}{\mathbb{E}}
\E \{\wideh …
0
votes
Complete sufficient statistic and unbiased estimator
No, there are examples where there is a complete sufficient statistic but there is some function of the parameter that does not admit an unbiased estimator. One binomial example, noted in comments, is …
7
votes
Accepted
Variance of unbiased estimator for the shape parameter of Pareto distribution
I will write the standard Pareto distribution with density
$$
f(x;\alpha,x_m)=\frac{\alpha x_m^\alpha}{x^{\alpha+1}}\cdot I(x > x_m),
$$ for some $\alpha>0, x_m>0$. Then the loglikelihood function …
2
votes
Accepted
What is the problem in the Neyman-Scott problem?
Partially answered in comments:
The problem has nothing to do with the fact that the bias can be corrected. It's that a particular procedure--namely, the Maximum Likelihood estimator--does not enjoy a …