Addressing only the following, as an extended comment: The documentation claims (and it is in line with what we know about LASSO) that $l_1$ regularisation leads to a sparse vector of coefficients. ...

In my view, the assumption of a deterministic relationship between labels and inputs made in chapter 2 of that book can be interpreted, but it's a bit artificial because it's not probabilistic machine ...

Suppose one has a sample of $n$ chi-squared random variables. Then $\mathbb{E}[X_i]=1$. So, $\mathbb{E}[\overline{X}_n] = \frac{n}{n} = 1$ Now the sample has a chi-squared($n$) distribution so it's ...

To supplement Dikran Marsupial's answer, the following is an articulation of a process I use personally. This is from the perspective of coding machine learning algorithms for research, rather than ...

The issue emerges in the evaluation of the second term in line $(3)$ and $(4)$ of your derivation. Note that $$\nabla_{\theta} Z(\theta)^{-1} = \nabla_{\theta} \frac{1}{\int_x \exp(-E_{\theta}(x))\, ... View answer Accepted answer 6 votes There are a number of quantitative finite-sample results, and also asymptotic arguments, in support of using the heuristic k = \sqrt{n}, where n is the sample size. However, in practice, it would ... View answer 1 votes The right hand side of the equality you have quoted is a geometric series. Identifying the first term a=1 and the common ratio r=1-\alpha, the series converges to a/(1-r) = 1 / \alpha if and ... View answer Accepted answer 1 votes In the context of asymptotic results involving the maximum likelihood estimator, the use of the subscript \theta_0 in the expectations operator means the following,$$\mathbb{E}_{\theta_0}[l(\theta; ...

Fix $z \in \mathbb{R}$ to be some value. Now for illustrative purposes, define corresponding i.i.d. random variables $Y_i = g(Z_i) = 1\{Z_i \leq z\}$. The following prompts should hopefully assist in ...

The chapters of the monograph that are being listed are from a draft book by Michael Jordan that to the best of my knowledge was never published, but continues to be used in many probabilistic ...

\begin{align} p(t \vert \mathbf{x}, \mathcal{D}) &= \sum^L_{i=1}p(t, \mathcal{M}_i \vert \mathbf{x}, \mathcal{D}) \tag{marginalisation} \\ &= \sum^L_{i=1} p(t \vert \mathbf{x}, \mathcal{M_i}, \...

To supplement the other answers by Taylor and mhdadk, I find it has never led me astray to have absolute clarity on the probability distributions with respect to which one is computing expectations. ...

The derivation relies on a result known as the matrix inversion lemma, or Woodbury matrix identity. From wikipedia: $$(A + UCV)^{-1} = A^{-1} - A^{-1}U(C^{-1} + VA^{-1}U)^{-1}VA^{-1}$$ Identifying $... View answer Accepted answer 3 votes You are correct in stating that a Bayesian network allows us to answer probability queries. However, the way you have framed the question suggests as if somehow you expected them to provide you with ... View answer 1 votes The class of methods devised to solve this generic problem are known in the optimisation literature as projected Newton methods. The following is extracted from Bertsekas and Gafni (1983), with some ... View answer Accepted answer 2 votes The source of the issue in my view comes from a confusion concerning dimensionality, and because the Hessian departs from the usual context in that there is sub-partitioning going on. Other than a ... View answer Accepted answer 1 votes This answer assumes that you are interested in computing$\nabla_{\theta} \log Z$rather than$\nabla_{x} \log Z$. Without foreclosing the possibility that there may exist other results relevant to ... View answer Accepted answer 1 votes Only in response to: I'm looking for a reference that talks about the relationship between the variance of the gradient and convergence of SGD. Whilst the Robbins-Monro stochastic approximation ... View answer Accepted answer 4 votes I'm trying to learn some machine learning theory, in particular maximum likelihood estimation. At risk of being pedantic, but for the avoidance of doubt, I have found it useful to be clear on which ... View answer Accepted answer 0 votes Solution. Using the results in here linked by StubbornAtom. The loss function you have specified can be interpreted as a generalisation of squared error loss, known as weighted squared error loss. ... View answer Accepted answer 1 votes I think it would be helpful in this instance to draw a distinction between: 1. How the hyperparametrisation encodes prior beliefs about the topic vectors$\phi_k$sampled from the Dirichlet prior$p(\...

Some references. To supplement the comments of @Dave Harris, here are a cluster of references you might consider using to start further formalising what you've done already. Depending on your ...

Here is my attempt. 1.Show that $f(x; \varphi)$ is in the one-parameter exponential family. Assume that $\varphi > 0$. Rewriting $f(x; \varphi)$, we have that \begin{align*} f(x; \varphi) &= \...

Proof strategy. Show that $T = \sum Y^2_i$ is minimal sufficient. Show that $U = (\sum Y_i, \sum Y^2_i)$ is sufficient. Use the following theorem to show that $T = g(U)$ for some function $g$. ...

In that book, $n$ random variables $X_1, \dots, X_n$, constituting the "data set", are represented as $X^n$. Whereas realisations of those $n$ random variables $x_1, x_2, \dots , x_n$ are ...

It depends on the nature of the assumptions on your $n$ random variables $Y_1, ..., Y_n$, which in turn will influence the kind of parametric statistical model that is specified. Here is a coin ...

In response to: So how do I find $E(x^2_i)$? Assuming there are no issues with the steps in the derivation up to that point, you can use the following standard result relating expectation, variance ...

In response to I did not fully understand why we are multiplying the intermediate conditionals i.e.: $...Pr(Burglar|Storm=T)xPr(Cat|Storm=T)...$ and Is there a logical explanation for the ...

To be clear, the learning algorithm $\mathcal{A}$ refers to the procedure of selecting a hypothesis $h$ from the restricted hypothesis class $\mathcal{H}$, so if you are selecting $h$ by say ...
The scipy-stats package is useful if you are using Python. Here is a code-snippet to get you started - it generates one realisation of the random variable $Y$ using $n = 50$ iid copies of $X$. Have ...