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Consider the situation in which you have $n = 20$ observations of a binary (2-coutcome) process. Often the two possible outcomes on each trial are called Success and Failure. Frequentist confidence interval. Suppose you observe $x = 15$ successes in the $n = 20$ trials. View the number $X$ of Successes as a random variable $X \sim \mathsf{Binom}(n=20; p),$ ...


5

Your interpretation is correct. In my opinion that particular passage in the Wikipedia article obfuscates a simple concept with opaque technical language. The initial passage is much clearer: "is an interval within which an unobserved parameter value falls with a particular subjective probability". The technical term "random variable" is misleading, ...


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Simple answer No, you cannot. Since in your problem setting, $\Sigma | \vec{\mathbf{y}}$ and $\theta | \vec{\mathbf{y}}$ are by no means independent. Thus we cannot simply marginalize it. Your try Here the problem setting is follows: $$ \begin{aligned} \mathbf{y}_1, \cdots, \mathbf{y}_n | \Sigma &\sim \mathcal{N}\left( \theta, \Sigma \right) \\[...


3

Priors are not chosen for convenience but for reflecting one's own beliefs or absence thereof about the parameters of the model. There is thus no foundational reason for always choosing conjugate priors, which main justification is computational. Furthermore, Conjugate priors only exist for exponential family models. Outside these models, there is no ...


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The absolute value of the likelihood, whether it is $10^{100}$ or $10^{-200}$, has no relevance for Bayesian inference. (Think of the extreme case when the likelihood function is constant in the parameter $\boldsymbol\theta$.) What matters is how this likelihood function varies with the parameter $\theta$, since the posterior density is a probability density ...


2

My interpretation of the credible interval was that it encapsulated our own uncertainty about the true value of the estimated parameter but that the estimated parameter itself did have some kind of 'true' value. This is slightly different to saying that the estimated parameter is a 'random variable'. Am I wrong? Although you say that you interpret the ...


1

1&2) You are working on a prediction problem. You need to solve $$\pi(k\ge450|4\text{ success of 10 tries})=1516927277253024\sum_{k=450}^{1000}\int_0^1\binom{1000}{k}p^k(1-p)^{1000-k}(1-p)^{25}p^{23}\mathrm{d}p$$ That is your posterior probability that the next 1000 tosses will result in 450 or more successes. You should gamble no more than the prize ...


1

Very interesting question. I doubt there is a definite answer to this, but let me give an extended comment. Imagine that you've done an experiment on some phenomenon that was never studied by anyone before. During your analysis, as a Bayesian, you would choose some prior for analyzing this data. Since you have no prior knowledge, you choose some "...


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Note that for each customer, the only information you have is the total number of visits ("has_visited" and "repeat_customer" are both functions of the total number of visits). Since you don't have any other distinguishing information for each customer, it stands to reason that you will have to make the prediction in the same way for each customer! You can ...


1

Take the Bayesian perspective, supposing that $E[m(y)\mid\theta] = \theta$ (posterior mean is unbiased). Then $$E[\theta m(y)] = E[E[\theta m(y) \mid \theta]] = E[\theta E[m(y)\mid\theta]] = E[\theta^2].$$ But also $$E[\theta m(y)] = E[E[\theta m(y) \mid y]] = E[m(y) E[\theta\mid y]] = E[m(y)^2].$$ So $$E[(m(y)-\theta)^2] = E[m(y)^2] + E[\theta^2] - 2E[\...


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In theory, your problem is easy: Compute what you need conditionnaly upon you missing data p(what I need| My Data, The missing Data) Then marginalize the missing Data p(what I need| My Data) = Sum_The missing Data p(what I need, The missing Data| My Data) = Sum_The missing Data p(what I need| My Data, The missing Data) p(The missing Data) In ...


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