microhaus
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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 ...

This conference paper from NeurIPs 2020 may contain what you are looking for - it contains some theoretical guarantees on using mini-batch stochastic gradient descent in context of Gaussian processes.

The issue emerges in the evaluation of the second term in line $(3)$ and $(4)$ of your derivation. Note that $$\nabla_{\theta} Z(\theta)^{-1} = \nabla_{\theta} \frac{1}{\int_x \exp(-E_{\theta}(x))\, ... View answer Accepted answer 6 votes There are a number of quantitative finite-sample results, and also asymptotic arguments, in support of using the heuristic k = \sqrt{n}, where n is the sample size. However, in practice, it would ... View answer Accepted answer 6 votes It means expectation with respect to q_{\phi}(\mathbf{z} | \mathbf{x}^{(i)}). So:$$\mathbb{E}_{q_{\phi}(\mathbf{z} | \mathbf{x}^{(i)})}[\log p_{\theta}(\mathbf{x}^{(i)} | \mathbf{z})] = \int_{\...

Joint distribution. Using the graphical model you provided, we get the following joint distribution over all variables of interest, conditioning on model parameters. $$p(\Theta, \mathbf{v} | a_0, b_0, ... View answer Accepted answer 5 votes There are some minor issues with your understanding, and I think it would help to clarify exactly what is being maximised, together with what is not being maximised. Loss function. I am going through ... 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 4 votes 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. ... View answer 3 votes 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 ... 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 Accepted answer 3 votes 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 ... View answer 3 votes Here is why further prompts on your question have been made by commenters. It is to avoid the following vague answer, which I suspect is not what you want: Which is the 'known data', X or \beta? ... View answer Accepted answer 3 votes I think what is going on here is the matrix-vector product \Sigma^{-1} \mathbf{1} and quadratic form \mathbf{1}^T\Sigma^{-1} \mathbf{1} are being used as a convenient way of specifying summation. ... View answer Accepted answer 3 votes What I understand Andrew Ng is saying is the following: We want to derive estimates of the parameters by solving for \phi, \mu, \Sigma. We do this by maximising l(\phi, \mu, \Sigma) i.e. the log ... View answer Accepted answer 3 votes The correct expansion of \oplus_{i \in S}(\log \Psi_t(j, i, x_t) + \log \alpha_{t-1}(i)) you are looking for is (A). Taking the logs of the forward recursion in equation (4.6), we have:$$\begin{...

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. ...

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 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 ...

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 ...

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 ...

At the most basic level, here is what the issue is when you have a DAG with a cycle. Let $X$ and $Y$ be random variables, that is, nodes in a DAG. Case 1. Consider the factorised joint distribution ...

From what I understand, your query concerns whether it is "correct" to compare the Bayes estimators, belonging to a "Bayesian paradigm", with other frequentist estimators such as ...

Empirical Bayes is a means of using the observed data to compute point estimates of the hyperparameters parametrising your priors. Which only makes sense in context of a hierarchical Bayesian model, ...

A surprisingly powerful check on whether there is an issue with your mathematical derivation of the Baum-Welch algorithm; or whether there is a bug in your implementation, is whether the log-...

Your question raises some important points which I also struggled with during self-study. I am by no means an expert, but I will attempt to clarify the various usages of the term "parameter" ...

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 ...

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 ...

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; ...