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Questions tagged [normalizing-constant]

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Regarding samples gotten from MCMC

In one article explaining MCMC, I once read the following statement. The idea of sampling methods is the following. Let’s assume first that we have a way (MCMC) to draw samples from a probability ...
user3125's user avatar
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Do the following normalizing constants cancel out in Reversible Jump ratio?

We know that a Strauss Point Process has density $$p(x_{1}, x_{2},..., x_{K})\propto \prod_{i=1}^{K}\phi(x_{i};\theta)\prod_{1\leq i\leq j \leq K}a^{1(\left | x_{i}-x_{j} \right |\leq \delta)}$$ where ...
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Exponential family admissibility of base measure, sufficient statistic and log partition function

Let $$ f(y | \eta) = h(y) \exp\left( \eta^\top T(y) + A(\eta) \right)$$ be the exponential family with base density/pmf $h$, sufficient statistic $T$, log partition function $A$ and natural parameter $...
Student's user avatar
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Monte Carlo Approximation of a Normalizing Constant [duplicate]

I know that one can approximate expectations of a function with respect to a pdf as such $$ \mathbb{E}_{p(x)}[\phi(x)] = \int \phi(x) p(x) dx \approx \frac{1}{N}\sum_{i=1}^N \phi(x^{(i)}) \qquad\qquad ...
Physics_Student's user avatar
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Why can't we use Monte carlo approximation for normalising constant in Baye's theorem

The normalising constant $Z$ in baye's theorem is the probability that the model generates the data $D$. $$\begin{align}P(D) &= \int P(D|\theta)P(\theta)d\theta \\ &= E_{\theta \sim p(\theta)}[...
calveeen's user avatar
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15 votes
4 answers
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Why is the normalisation constant in Bayesian not a marginal distribution

The formula for Baye's rule is as follows $$p(\theta |D) = \frac{p(D|\theta)p(\theta)}{\int p(D|\theta)p(\theta)d\theta}$$ where $\int p(D|\theta)p(\theta)d\theta$ is the normalising constant $z$. ...
calveeen's user avatar
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3 votes
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Simplifying modified Bessel function of the first kind

The modified Bessel function of the first kind shows up in the normalizing constant of a lot of random variables (e.g. the normal product distribution, the noncentral chi-square distribution, the ...
Taylor's user avatar
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1 vote
1 answer
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How do I compute the closed form normalizing constant for this distribution?

The funnel distribution for random variable $X = (x_1,x_2,..,x_D)$is $$P(X) = N(x_1|0,9)\prod_{d=2}^D N(x_d | 0,exp(x_1))$$ The closed form normalizing constant for normal distribution $N(x|0,\...
ElleryL's user avatar
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weighted sum of posterior Dirichlet distributions

I have the following distribution: $q(\vec\theta) = \frac{\sum_k \alpha_k}{\sum_k \beta_k \alpha_k} (\sum_k\theta_k\beta_k) \frac{\Gamma(\sum_k \alpha_k)}{\prod_k\Gamma(\alpha_k)} \prod_k \theta_{k}^{...
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What is the relationship between the normalization constants of the normal distribution and the (Inverse-)Wishart distribution?

I was looking at the probability density function of the multivariate normal distribution and that of the inverse-Wishart distribution: $$ \begin{array}{rccll} p_{\mathcal{N}}\left(\mathbf{x}; \...
Xiubo Zhang's user avatar
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1 answer
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Bayesian inference, finding unknown function of parameter

I'm completely new to Baayesian statistics and am having some questioning about the following: Given a probability distribution defined by the pdf $\pi(x|\theta) = C(\theta)x^2 e^{-\theta x^{3}}$ ...
Knut's user avatar
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2 answers
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Deriving the normalizing constant for the multivariate Gaussian

I am trying to derive the normalizing constant for the multivariate Gaussian. The book I'm following suggests diagonalizing the covariance matrix and then using a change of variables. So, we consider ...
cangrejo's user avatar
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3 votes
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Dropping the normalization constant in Bayesian inference [duplicate]

Could anyone please explain to me (or provide a situation) where one would drop the normalising constant (or marginal probability term) from Bayes rule when performing Bayesian Inference? New to ...
Maeve90's user avatar
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1 answer
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Sampling a random binary matrix with "Gaussian" probability distribution

Let $A_{ij}$ be a $n\times n$ random binary matrix with probability mass function $P(A)$ given by $$ \log P(A)=-\frac 12 \mathrm{tr}\left[\left(A-M\right)^TV\left(A-M\right)\right] + C, $$ where $M$ ...
Till Hoffmann's user avatar