7

Why do we multiply log likelihood times -2 when conducting MLE? We really don't. The -2 was not about parameter estimation; for that, we'd just use the (negative log-)likelihood. It was about hypothesis testing. Your intuition about negation is correct. Traditionally, in the optimization literature, we minimize functions. It's easy enough to convert a ...


3

If you know the sample size (as opposed to just knowing that the sample size is between 50 and 200), then maximum likelihood might work in this case. Using Mathematica: (* Joint distribution of largest 4 order statistics from a sample size of 100 *) dist = OrderDistribution[{NormalDistribution[0, \[Sigma]], 100}, {97, 98, 99, 100}]; (* Log of the ...


3

I find the literature in MLE a bit fuzzy with nomenclature here, so I might have some stuff off, and I will try to stick to the nomenclature you introduced. We have the observed Fisher information: $$\left[\mathcal {J}(\theta)\right]_{ij} = -\left(\frac{\partial^2 \log f}{\partial \theta_i \partial \theta_j}\right)$$ And the empirical Fisher information: $$\...


1

The formulas you have given can be used to construct moment estimators. But you can also use maximum likelihood estimation in the marginal negative binomial distribution. The variance of the gamma distribution is a function of the negbin parameters, so just use the same function on the maximum likelihood estimates of $r,p$ to get the maximum likelihood ...


1

Comment: This the germ of an idea, not yet a solution. [I am assuming the mean $\mu$ is also unknown.] Suppose $n = 100$ observations from a standard normal distribution. You might find the distance $D$ between order statistics 97 and 100. Then $1/D$ should estimate $1.$ Several runs give various answers. In R, set.seed(123) diff(sort(rnorm(100))[c(97,100)]) ...


1

Simultaneous estimation will surely produce different output than stepwise estimation. The conditional mean model is estimated assuming a GARCH-type conditional variance in the simultaneous case but a constant variance in the stepwise case, yielding different optima. The conditional variance model is estimated in line with the first optimum in the ...


1

I would say that in 98% of cases, this works fine. If you are interested in the 2% cases where it doesn't, read on. This presupposes that the 95% prediction interval is a central one, consisting of the 2.5% and the 97.5% quantile predictions. This is often an unspoken assumption, but of course, you could also have a 95% prediction interval that consists of ...


1

The prediction interval is $$ \hat{y}_{h} \pm t_{(1-\alpha / 2, n-2)} \times \sqrt{M S E \times\left(1+\frac{1}{n}+\frac{\left(x_{h}-\bar{x}\right)^{2}}{\sum\left(x_{i}-\bar{x}\right)^{2}}\right)}$$ for a simple linear regression. The mean of this quantity is $\hat{y}_h$. Thus, you can simply take the average of the prediction interval points and recover ...


1

The distribution you are using here is the generalised error distribution (also called the generalised normal distribution) with unit scale and shape parameter $a>0$. I take it from your specification that the scale $a$ is a fixed value in this analysis, and is therefore not subject to estimation. I will proceed on that basis. For the data $\mathbf{x} = ...


1

Try things in the univariate case. I'll suppose $f(x; \theta)$ is the density and we're estimating $\theta$ with $\hat\theta$ which is unbiased but otherwise unknown. I'm changing to $\theta$ from $k$ just to be more consistent with the usual notation. I'll use $\nu$ as the dominating measure for these densities and $\mathcal X$ as the support. We have$\...


1

I think that your problem is that you considered the model as one equation. However, as you wrote yourself, there are two equation, not one. The problem you faced is called the simultaneous equations. Your equations are given in the structural form. Try to rewrite them in reduced form and then apply the logic to showed in your attempt.


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