New answers tagged

1 vote

A Kinder Egg Problem

I want to give you a simple approach, while Ben's approach is the proper one. I reuse some of Ben's variable names. If i interpret your data correctly, you aggregated data for three time periods with ...
user avatar
0 votes

MLE vs MAP estimation, when to use which?

Theoretically, if you have the information about the prior probability, use MAP; otherwise MLE. However, as the amount of data increases, the leading role of prior assumptions (which used by MAP) on ...
user avatar
4 votes

A Kinder Egg Problem

You could potentially approach this as an inference from an occupancy problem, depending on the sampling method. For simplicity, let's assume that there are $N$ types of toys and your sampling method ...
user avatar
  • 97.2k
3 votes

How to construct a confidence interval for the coefficients of a multivariate regression with dependence between dependent variables?

Although your question asks about a model where the error terms are independent, I'm going to generalise to drop this assumption in order to give a slightly broader answer. This will allow me to show ...
user avatar
  • 97.2k
1 vote

What is an unbiased estimate of population R-square?

We made some progress regarding this. Thus here an updated answer. Assumptions All existing comparisons for this (thus also the results summarized by Jeromy, and all claims I make dependent on all ...
user avatar
  • 1,600
0 votes

Using the Median to Estimate a Parameter

(I've deleted my previous answer as it was a bit sloppy.) Because in a comment you've mentioned the desire for other estimators, here is an implementation of a maximum likelihood estimator. Note that ...
user avatar
  • 2,208
0 votes
Accepted

MLE for distribution having most general form

I don't know how you tried to differentiate the log-likelihood, but I tried to proceed in following way: The pmf is given by, $f(x_i;\theta)=(\frac{x_i}{\theta})^{\theta A'(\theta)}e^{A(\theta)+C(x_i)}...
user avatar
1 vote

Using the Median to Estimate a Parameter

The median of this distribution is at $a/9$, since $$P[X<\frac{a}9]=48.6\%$$ $$P[X=\frac{a}9]=\ \ 3.4\%$$ $$P[X>\frac{a}9]=48.0\%$$ So you can estimate $a$ as $9$ times the median.
user avatar
  • 2,047
0 votes

Iterative proportional fitting with constraints

$ \def\LR#1{\left(#1\right)} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\m#1{\left[\begin{array}{r}#1\end{array}\right]} \def\c#1{\color{red}{#1}} \def\CLR#1{\c{\LR{#1}}} \...
user avatar
  • 236
1 vote
Accepted

Sampling distribution of GBM Maximum-Likelihood estimator

It is common knowledge in asymptotic statistics that the asymptotic distribution of MLE in an exponential family is the normal distribution with mean being the MLE, and variance being the inverse ...
user avatar
0 votes

MLE estimation of Autoregressive Conditional Poisson model

I use your code with slight housekeeping and it works for me! Please see here- ...
user avatar

Top 50 recent answers are included