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New answers tagged likelihood

1

First, use a conditional Bayes' rule where we keep conditioning on $X$, $\theta$, and $\mathcal H_i$, and only swap $\mathbf w$ and $\mathbf y$: $$p(\mathbf w \mid \mathbf y, X, \boldsymbol\theta, \mathcal H_i) = \frac{p(\mathbf y \mid \mathbf w, X, \boldsymbol\theta, \mathcal H_i) \; p(\mathbf w \mid X, \boldsymbol\theta, \mathcal H_i)}{p(\mathbf y \... 2 The equation y = 0.2x + 4 is not a statistical model, but a mathematical equation. It doesn't even have parameters (unless you mean that x is a parameter), so there is nothing to optimize. In case you meant y = \beta_0 + \beta_1 x, this would still be a deterministic function of the variables, that can be solved by pure algebra, that does not need ... 1 This depends on how you are thinking about the likelihood function: If you have the same regression model but two different data sets then yes the likelihood function takes the same mathematical form. See for example (this is a log likelihood which is just the natural log of the likelihood) equation 3 here: https://www.stat.cmu.edu/~cshalizi/mreg/15/lectures/... 0 It's easy to derive P(p_i, p'_i) from the equation of the two formulations you used for LL. Remember that t_i can only be 0 or 1.$$ P(p_i, p'_i) = \begin{cases} p'_i\,^{w'_i}, & t_i = 1\\ (1 - p'_i)^{w'_i} -1, & t_i = 0 \end{cases}  In this terms alone, a function $w'_i= w(p_i, t_i)$ that makes $P(p_i, p'_i)$ independent on $t_i$ is surely ...

4

Yes, the likelihood is the likelihood. You sometimes see likelihood defined only up to a multiplicative constant (as Fisher did) but that doesn't harm either of those applications if you are consistent in how you deal with it. Unfortunately, by asking a yes-or-no question to which the answer is "yes" there's not much more to say. If the answer had been no, ...

1

To put simply, likelihood is "the likelihood of $\theta$ having generated $\mathcal{D}$" and posterior is essentially "the likelihood of $\theta$ having generated $\mathcal{D}$" further multiplied by the prior distribution of $\theta$. If the prior distribution is flat (or non-informative), likelihood is exactly the same as posterior.

2

First, to have a posterior distribution for $\theta$, $\theta$ must be (modeled as) a random variable. For the likelihood function that is not necessary. So this is deeper than the comment (by @gazza89) saying The likelihood is a pdf, it's just normalised w.r.t all possible data outcomes, and the posterior is a pdf, but it's normalised w.r.t all ...

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The whole point of doing Bayesian Inference, is that you stipulate your data was generated by a model with unknown parameters. An example of this would be "the samples were all drawn independently from the same normal distribution, but we don't know the mean or variance of that distribution" We then try to use the data to come up with beliefs about the ...

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