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Hot answers tagged maximum-likelihood

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You observe $\text{“}{+}\text{''}$ with probability $p=\Phi\left( \dfrac{0 - \mu} 1 \right)$ and $\text{“}{-}\text{''}$ with probability $1-p=1-\Phi\left( \dfrac{0-\mu}1 \right),$ where $\Phi$ is the c.d.f. of the standard normal distribution. Let $y$ be the number of $\text{“}{+}\text{''s}$ you observe, so that $n-y$ is the number of $\text{“}{-}\text{''s}.$...

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Yes, the likelihood is the likelihood. You sometimes see likelihood defined only up to a multiplicative constant (as Fisher did) but that doesn't harm either of those applications if you are consistent in how you deal with it. Unfortunately, by asking a yes-or-no question to which the answer is "yes" there's not much more to say. If the answer had been no, ...

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Unbiased estimator has an expected value that equals to the true parameter. An efficient and unbiased estimator has the lowest possible variance among all unbiased estimators. There exists no trade off between efficiency and bias. What I think you might be referring to is the trade off between variance and bias. There are some biased estimators that have a ...

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First, note that the CDF of the minimum of $n$ iid observations from a continuous variable is \begin{align*} F_{X_{1:n}}(x) &= Pr(\min(X_1, \dots,X_n) \leq x)\\ &= (1 - Pr(\min(X_1, \dots,X_n) > x))\\ &= (1 - \prod_{i=1}^n Pr(X_i > x))\\ &= 1 - (1 - F_{X}(x))^n. \end{align*} By differentiating this with respect to $x$, we ...

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As Whuber points out it depends on the assumption. So here is an example of some assumptions that could be made. First stack the individual equations $$y_i = x_i^\top \beta + u_i$$ to get multivariate version $$y = X\beta + u$$ then make some assumptions $$\mathbb E[u] = 0$$ $$\mathbb Var(u) = \mathbb E[uu^\top] = \Omega$$ $$det(\Omega) \not = 0$$ $$... 1 k-NN just measures the distances between observations and may suffer the curse of dimensionality as well as other algorithms. It also does not try finding the distribution of the variables, just makes local approximations. So it is hard to compare to the two other methods you mention. Logistic regression (same applies to linear regression) makes the ... 1 Let's first write down a density for a scale-shape parameterization for a Gamma and then shift it. Taking the density from Wikipedia (which has it correct), but making the variable z rather than x:$$f(z;\alpha ,\theta )=\frac{z^{\alpha -1}e^{-z/\theta }}{\theta ^{\alpha }\Gamma (\alpha )}\quad {\text{ for }}z>0,\quad \alpha ,\theta >0$$... 1 I'm not sure what you mean by "legitimate", but you are correct - perhaps that's just as good, haha. I'll point out why this is the case in general. Let's say we have a random variable with probability mass function f, which has a parameter \theta. We observe x_1, ..., x_n, which are drawn IID from this distribution. As I'll point out below, each of ... 1 The likelihood function is:$$p(x|\theta)=\prod_{i=1}^n p(x_i|\theta)= \prod_{i=1}^n I(\theta < x_i<\theta+1) The maximum value of this function is $1$, and the $\theta$ satisfying this maximum value is not unique , i.e. $\theta\in (x_{max}-1,x_{min}+1)$, if the interval is not empty, which can happen when the model is wrong, or there is noise.

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Your question is very broad, so this is more of a comment. One possible problem is that the (generalized) likelihood ratio test might be suboptimal in some cases, this paper points to some such examples. You say Indeed, many, if not all, of the above mentioned tests can be shown to be equivalent to a LRT, but I was referring to LRTs directly constructed by ...

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It takes the partial derivative, not the total derivative, although seemingly doesn't say it explicitly. Partial derivative assumes that the parameters other than the one being differentiated are constant. Intuitively, the reparametrization is an axis change. Now the log-likelihood is plotted against $(x,y,z)=(\sigma_g^2,\delta,\beta)$ instead of \$(x,y,z)=(\...

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I wrote a little Python helper to help with this problem (see here). You can use the fit.get_vcov() function to get the standard errors of the parameters. It uses automatic differentiation to compute the Hessian and uses that to compute the standard errors of the best-fit parameters. Background You can look through the slides here, but I will explain it as ...

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This is of course not a rigorous answer to your question 1, but since you asked the question in general, evidence for a counterexample already indicates that the answer is no. So here is a little simulation study using exact ML estimation from arima0 to argue that there is at least one case where there is bias: reps <- 10000 n <- 30 true.ar1.coef &...

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