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It seems like you are using your intuition for what it takes to maximize the function $g.$ But that has nothing to do with maximizing the likelihood function $L(\theta;x)$. They are completely differe …

Remember the likelihood function is a function of $a.$ The likelihood function on data $x_1,\ldots x_n$ is $$L(a) = \left\{\begin{array}{ll}\left(\frac{53}{2a^{53}}\right)^n(x_1x_2\ldots x_n)^{52} & a …

Per our discussion in the comments, MLE does not seem like a pragmatic choice for estimating these parameters. Because it works using the whole probability distribution, the MLE must hone in on the fa …

I think you've done what's intended. The 'data' is that you picked a silver and the 'hypotheses' are that the candy is a nougat / a licorice (though this is a questionable use of 'hypothesis' in my op …

Here the CDF is the thing you are estimating. You can think of its values as an infinite number of parameters (in a constrained space that says they need to comprise a right-continuous, nondecreasing …

Your reasoning for the $U(0,\theta)$ case is wrong, so is interfering with the $U(-\theta,0)$ case. In the $U0,\theta)$ case the likelihood (which is a function of $\theta$ !) is $\frac{1}{\theta^n}$ …

The likelihood function you're maximizing is a function of $\theta$ so the $k!$ is just a multiplicative constant. It has no effect on what value of $\theta$ maximizes the function. (One way to see th …

The density function is not $\frac{2x}{\alpha^2}.$ It is $\frac{2x}{\alpha^2}$ for $0<x<\alpha$ and zero otherwise. This means that the joint density for $n$ variables is zero if any of the $x_i$ are …

The likelihood on this data is $1/n$ if $n\ge 100$ and zero otherwise (since a draw of $100$ is impossible if $n<100$). This is maximized for $n=100,$ so the MLE is $100.$

If I'm interpreting you correctly you are given $20$ independent samples that are Poisson-distributed with mean $\theta t$ (where $t$ is the same for each one - representing the duration of some inter …

HINT Do the mirror image problem of $U(0,\theta)$ and find what $\rho$ makes the estimator $\rho x_{(n)}$ have lowest expected mean squared error. The answer to this question will be the same $\rho$ …

We have $$\begin{eqnarray}\sum_{i=1}^n(x_i-\bar x)(y_i-\bar y) &=&\sum_{i=1}^n x_iy_i- \bar x\sum_{i=1}^ny_i - \bar{y}\sum_{i=1}^nx_i + \bar x\bar y \sum_{i=1}^n (1) \\&=& \sum_{i=1}^nx_iy_i-n\bar x\b …

The second derivative needs to be less than zero at $\theta^*,$ so the first term of the prefactor is zero and what’s left, $-n$, is clearly negative. On a side note, it’s easier here, and a good id …

In short, MLE is one way of fitting a model to data. This is the real life application, and it is in fact used in a lot of real-life fitting procedures in practice. (For instance, it is one standard w …

The EM algorithm has the two steps. First you find the expected log likelihood (where the expectation is taken under the current parameters and conditional on whatever data you can see) and then you a …