New answers tagged

2

It looks like the author intends for the set $(\theta_0-a, \theta_0+a)$ to be on the interior of $\Omega$. Strictly speaking, what the proof should have done is to choose a value $a>0$ such that: $$(\theta_0-a, \theta_0+a) \subseteq \text{int }\Omega,$$ which then implies $\theta \pm a \in \Omega$. Bear in mind here that we only need to have the ...


1

I am not sure that I understand your concern, but sounds like it is that possibly $\theta-a$ or $\theta+a$ might be on the boundary of the parameter space, with one being either the maximum or the minimum of $\Omega$. I agree that you need to assume both $\theta-a$ and $\theta+a$ are interior points in the parameter space. It is just a fact that even when ...


2

One of the problems with "Expected Risk Mimimization" is that the distribution, $p(y׳|x_i;\phi)$, is unknown, and thus it is not clear how to minimize it. You could theoretically argue, that we can try to find this distribution, during the minimization process, but as you said, in this case it will not approximate the true distribution, because the ...


0

A continuous random variable has the Probability Density Function (PDF) or simply Density Function that helps us to calculate its probability for a certain range or interval. When we substitute x in the Density Function with a value from a random variable's range, then we just get a value of the Density Function which is not the probability. To calculate the ...


4

Firstly, your equation for the log-likelihood function for the negative binomial distribution looks wrong to me, and it's not clear how your $\beta$ enters into the parameterisation of the distribution. In any case, I will show you how to do this kind of problem using the standard parameterisation of the negative binomial distribution. If $X_1,...,X_n \sim ...


1

These are quite complicated, but Mathematica is up to the task: \begin{align*} \frac{\partial}{\partial\theta}\,\Gamma\!\left(\frac{\theta x^2}{\beta},\theta\right) &=\frac{x^2\left(G_{2,3}^{3,0}\!\left(\theta\left|\begin{array}{c}1,1\\0,0,\frac{\theta x^2}{\beta}\\\end{array}\right.\right)+\log(\theta)\,\Gamma\!\left(\frac{\theta x^2}{\beta},\theta\...


2

Let's call the MLEs $\theta_1$ (best fit to Jan-Nov), $\theta_2$ (best fit to December), $\theta_{12}$ (best fit to full year). Your option 1 is indeed the classic formulation of the likelihood ratio test. The restricted model (Model 1) $\{\theta_{12}, \theta_{12}\}$ (i.e. using the same parameters for the whole year) is nested within the full model (Model 2)...


0

Given the maximum likelihoods estimates of mean and variance for a Gaussian distribution are \begin{align*} \hat{\mu} &= \frac{1}{N} \sum_{i=1}^N x_i \\ \hat{\sigma}^2 &= \frac{1}{N} \sum_{i=1}^N (x_i - \hat{\mu})^2 \\ \end{align*} Suppose the real mean and variance for the Gaussian distribution is $\mu$ and $\sigma^2$. First, we show $\hat{\mu}$ is ...


2

I assume that the random variables $X_{1}, X_{2},..., X_{6}$ are independent draws from $Unif(\theta-2,\theta+6)$ with $\theta>0$ Then the joint pdf of $X_{1}, X_{2},..., X_{6}$ can be written as $$f(X_{1},X_{2},...,X_{6};\theta) = \prod_{i=1}^{6}f(X_{i}) = \prod_{i=1}^{6}\frac{1}{8}\mathbb{I}_{X_{i}\in(\theta-2,\theta+6)}$$ where $\mathbb{I}_{X_{i}\in(\...


2

Before you get to the likelihood function, the first issue here is to confirm that this is a valid cumulative distribution function. If you examine the requirements for this, you will see that you must have $a=1$ for validity of the form, so your distribution will reduce to the two-parameter Weibull distribution. Once you have narrowed down the CDF, you ...


1

You can get easily get the PDF. If your values are in the range $0$ to $+\infty$ then you should have $a=1$; this because CDF goes to $1$ as $x$ goes to infinity. The PDF is simply the derivative of the CDF, it is $f(x)=\frac{a*c}b*\left({\frac{x}b}\right)^\left(c-1\right)*e^{-\left(\frac{x}b\right)^c}$


1

If you are certain that one wrote the beginning and the other wrote the end, you can annotate each sentence in the first paragraph as author A and the rest as author B. Train a supervised machine learning model or several (e.g. an ensemble of SVM, NB, LR, RF) on this annotation, cross-validate and note the accuracy of the model. Since words indicative of ...


0

The first term would become the probability that $y_j \ge y_L$, so I think you are right. This would be $\Phi \left(\frac{X_j \beta - y_L}{\sigma} \right)$. I guess the case where the objective is truncated to both ends is equivalent to probit (I did not prove it formally. There could be some difference).


3

The density is $$ f(x \vert \theta) = \dfrac{x}{s} \exp\left( -\dfrac{x^2}{2s}\right)$$ Here $s = \theta^2$. We will take derivatives with respect to $s$ for simplicity. The first derivative of the log likelihood is $$ \dfrac{\partial \ell}{\partial s} = \sum_i \dfrac{2s - x_i^2}{2s^2} $$ and the second derivative is $$ \dfrac{\partial ^2 \ell }{\partial s^...


2

I think your derivatives look okay. We don't have a basis to judge your implementation though. What steps have you taken to check it? However, if we're trying to estimate $\theta^2$, why not do the derivatives directly in terms of that, and maybe save a step, since you're then iterating directly to an estimate of the desired quantity. I think it goes like ...


0

Since the MLE is the first order statistic it has the pdf: $n \theta a^{n\theta} x^{-(n\theta+1)} \boldsymbol 1_{(a,\infty)}(x) $. Naturally then $\displaystyle \mathbb{E}[X_{(1)}] = \frac{a n\theta}{n\theta-1 } $, so that $\displaystyle \frac{(n\theta-1)X_{(1)} }{n\theta } = T(x)$ is the UMVUE by Lehmann-Scheffé. $$\implies \frac{Var(T(x)) }{Var(X_{1})} = \...


0

MLE of $a$ is indeed the first order statistic $X_{(1)}=\min\limits_{1\le i\le n}X_i$ because the likelihood is non-decreasing in $a$ subject to the restriction $a<X_{(1)}$. Because the population distribution is Pareto, you can verify that $X_{(1)}$ also has a Pareto distribution from which you can get its exact variance. UMVUE of $a$ however depends on ...


3

You will not get the correct variance of the standard deviation by working out the variance of the variance and taking its square root $-$ that's the population standard deviation of the sample variance not the population variance of the sample standard deviation. You also have to be careful about what $\hat{\sigma}$ is! At the normal, the MLE of $\sigma$ is ...


4

Casella and Berger (2002) explain this by saying that, "In many cases, this simple version of the invariance of MLEs is not useful because many of the functions we are interested in are not one-to-one" (p. 320). This is the essence of their motivation for extending the concept of the likelihood function. If you have an existing sampling density ...


1

It means that in the first experiment, you observed 1 blue ball. In the second experiment, you observed 3 blue balls. In the third and fourth experiments, you observed 2 blue balls.


0

Suppose that the model does not involve any parameter other than $\boldsymbol{\mu}$. You can regard $\mu$ as a random quantity in a Bayesian approach. Then, if the prior on $\boldsymbol{\mu}$ is Gaussian, the posterior is also Gaussian and is given by the last step of the Kalman Filter. The filtered mean $\widehat{\boldsymbol{\mu}}_{T\vert T}$ for the last ...


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