Search Results

Finding the MLE: You should be more careful with your sigma-notation in your working, by bracketing the terms in the sum. The simplest way to express the score log-likelihood here is in terms of the …

Your answer is not correct. From your (correct) specification of the density for the minimum value, the log-likelihood is: $$\ell_{\boldsymbol{x}} (b) = \sum_{i=1}^n \ln(b - x_i) - 2n \ln(b) \quad \ …

Consider the simpler case of $X_1,...,X_n \sim \text{IID InvN}(\mu, \lambda)$, which gives you the sampling density: $$f(\mathbf{x} | \mu, \lambda) = \prod_{i=1}^n \Big( \frac{\lambda}{2 \pi x_i^3} \B …

Maximum-likelihood estimators are invariant under functional transformations, and so if $\boldsymbol{\alpha} = \boldsymbol{\beta} \boldsymbol{A}$ is a new parameterisation for your regression then $\h …

There is no need to use asymptotic approximations here, since we can get the exact distribution of the MLE. The distribution you didn't recognise is $Y = X^2 \sim \text{Gamma}(\text{Shape} =\tfrac{3} …

Your answer is correct, and it is simple to verify using ordinary calculus methods. It is usually simpler to express this with the log-likelihood rather than the likelihood, since it is easier to dif …

As your question stands you have independent observations $Y_i \sim \text{N}(\beta_i, \sigma^2)$. There is no relevance to the contextual statement that these are angles of a triangle, since you have …

The approach you have used is fine, and what you are essentially doing is getting MLE equations for each parameter in terms of the other one, and then solving these as simultaneous equations to get th …

In problems like these you can still use ordinary calculus methods; you have a monotonic likelihood function with the MLE occurring at a boundary point. This gives rise to a biased estimator of the p …

The book you have referenced uses some general theory about zero-inflated distributions (i.e., the application of some results that are not specific to the Poisson case). I have been unable to replic …

Since you have stated that $y = ax+b$, it is unclear in your question where the errors come into your model. However, if you are saying that there is some error in the measurement of $x$ and $y$ then …

The conditional MLE for an auto-regressive model can be solved using standard linear regression theory. It is easy to do this in matrix form and use the standard formula for the coefficient estimator …

In this case your log-likelihood function is: $$\ell (b) = - \sum_{i=1}^n | y_i - b x_i |.$$ Since your $x_i$ values are either negative or positive one, we can let $\mathscr{X} \equiv \{ i = 1,..., …

In the case where you observe $\bar{x}_n \leqslant 1$ you have test statistic $t = 0$ and the corresponding p-value for your test is $p=1$. This means that there is no evidence to reject the null (i. …

This cannot be true as a general theorem, since there are situations where the MLE is not unbiased, but other unbiased estimators exist. For example, this occurs when estimating the parameter $\theta …