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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 ...


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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 ...


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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 ...


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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, ...


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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 ...


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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 ...


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Take a covariance weighted average. (Also, your notation here needs fixing: you need to define the x as samples from a distribution)


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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 $x_i = 1$.


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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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