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Let $q_i=P(Y_i=1)$. The variables $q_i$ are called nuisance parameters and to deal with them we have to put a prior on them and then average out over them. To be more precise, the model is the following: we have parameters $p_1,\dots,p_N$ and $q_1,\dots,q_N$. We know the $p_i$, but we don't know the $q_i$ and so must put a prior distribution on them, say ...

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It's not Bayes theorem; it's just the definition of conditional probability. The desired prob, by definition of conditional probability, is $${P(\mbox{child is heterozygote & child has brown eyes & parents have brown eyes})\over P(\mbox{child has brown eyes & parents have brown eyes})}.$$ But the numerator simplifies to $$P(\mbox{child is ... 0 Choosing priors can be subjective when you don't know all the details of the problem, but this is what I would do: Normally ignorance priors for Bernoulli parameters are chosen to be Beta distributions, with common choices being \mathrm{Beta}(0,0), \mathrm{Beta}(1/2,1/2), and \mathrm{Beta}(1,1) (see wikipedia). Here you are going to have to condition ... 1 The way you have written it, "\theta = probability that the psychic has ESP", \theta essentially is your prior distribution. There are only two possibilities, ESP and not-ESP, so the full statement of the prior is (I write e for ESP and \neg for negation): P(e) = \theta P(\neg e) = 1 - \theta Writing d for the observed data (3 out of 5 ... 1 First, some basic calculations. let p be the probability of guessing a card correctly. Then the probability of getting exactly 3 correct is \binom 53 p^3(1-p)^2. If p=.2 this is .0512, if p=.5 this is .3125 Let's say your prior is \theta_0. That is, before you test anything, you estimate that the "ESP probability" is \theta_0. Then, ... 2 Your computation doesn't make any sense. Assuming that n of the tosses were rigged doesn't mean that you're assigning a prior probability of n/6 to "not fair". If you're saying the only possibilities are "fair" and "not fair" and "not fair" means a 100% of Clinton winning all 6 tosses (which is what your computation of "P(6H)" implies), then that ... 0 Let E_1 and E_2 be two events defined as E_1: boy is born E_2: girl is born Let say that initially there were n girls. B: boy picked up by nurse G: girl picked up by nurse ... 0 Take a covariance weighted average. (Also, your notation here needs fixing: you need to define the x as samples from a distribution) 0 For the decision rule formula to hold, x_i needs to be binary and P(i | y) = P(x_i = 1 | y) (think about the formula in the case where x_i = 1 and the case where x_i = 0). So i is the event x_i = 1. 1 If the 40 W includes the number 20 W and F, then$$P(F|W) = \frac{P(FW)}{P(W)} = \frac{20/100}{40/100} = \frac{20}{40} = \frac{1}{2}.$$If 40 W does not include the 20 W and F, then notice that$$P(W) = P(W\bar F)+P(WF) = \frac{40}{100}+\frac{20}{100} = \frac{60}{100}.$$So,$$P(F|W) = \frac{P(FW)}{P(W)}= \frac{20/100}{60/100} = \frac{20}{60} = ...

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2) Let $A = \{ \text{exactly 3 were damaged}\}$ and $B = \{\text{neither of the insured were damaged}\}$ \begin{align} &P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{\binom{4}{3}0.1^30.9^3}{0.9^2} = \binom{4}{3}0.1^30.9 \end{align}

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I am not 100% sure I understand your question, but consider a parameter space $\Theta$ with prior probability $f(\theta)$ and a probability distribution $p(x|\theta)$ over a space $X$ indexed by this parameter. The conjugate prior is simply obtained by computing Bayes rule: \begin{equation*} p(\theta|x) = \dfrac{f(\theta) p(x|\theta)}{\int_{\theta' \in ...

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Should we instead be viewing this as taking an average of gaussian random variables? Yes. See Doucet & Johansen 2008 - A tutorial on Particle Filtering and Smoothing section 3.1; they make that explicit in their discussion of MC methods and use notation that stops this confusion from arising in the first place.

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$p(D',\theta | D) = p(D' | \theta,D)p(\theta | D)$ is from Bayes rules, provided we have densities. Now integrate out the nuisance variable $\theta$ on both sides. Your formula also appears to have a Markov-type assumption $p(D'|\theta,D)=p(D'|\theta)$.

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To show this one can follow a somewhat standard argument. In what follows, for notational convenience, I have replaced your "$D$"s with "$S$"s. By the law of total expectation (in terms of conditional expectation) and Fubini's theorem, applied to any bounded measurable function $f$ defined on the relevant sample space $\Omega$, we observe that  \eqalign{ ...

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Let's be more explicit and things will be clearer. There is a probability distribution over outcomes on $t$ flips. For each possible outcome of the first $\tau$ flips, $x$, there is a set of outcomes on $t$ flips whose first $\tau$ flips are $x$. This set is an event, which you may denote $s^\tau$ (which depends on $x$). For any outcome on $t$ flips, $y$, ...

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