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Indeed, your comment points out how it is different from the Beer-Quiche game. Fortunately, though, it's not that different. You can represent it in much the same way by just tinkering with the information sets (as you allude to in the question) as I've done below. In the typical setup, the second-mover must decide without knowing the type of the first ...

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This is not really an answer, but please bear with me. Baye's Rule is about conditional probability, and probability trees. There are a lot of factors which affect the price of oil. Therefore, it seems likely that the Baye's rule won't be practical. If you assume that there are really only a handful of factors: production, consumption, wars in the Middle ...

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You could run your MCMC, and after the burn-in period look at the points in the chain where $d_1,...,d_N$ takes some fixed vector of values, say $\mathbf d$. Then calculate the proportion of $\beta$'s which is greater than zero in this set of points, and you have an estimate for $\mathbb P[\beta > 0 | (d_1,...,d_N)=\mathbf d]$. On a side note, I assume ...

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Use the fact that $\mathbb P[(A \cup B)^c] + \mathbb P[A] + \mathbb P[B] - \mathbb P[A \cap B] = 1$. In particular, $$\mathbb P[A \cap B] = 0.95+0.03+0.04-1= 0.02$$ so you can plug this into your method and you are done.

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$P(A\cap B)=P(A\cup B)-P(A)-P(B)$ then you can the fact that 95% of the goods are not deflected.

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If $n$ random variables $X_1,\ldots,X_n$ are independent, then their joint density $f_{X_1,\ldots,X_n}(x_1,\ldots,x_n)$ is equal to $f_{X_1}(x_1)\cdots f_{X_n}(x_n)$. In this case, for every index $i\in\{1,\ldots,n\}$, and every value of $x>0$, we have $$f_{X_i}(x) = \begin{cases} 1/\theta & \text{if }x<\theta, \\ 0 & \text{if }x>\theta. ... 0 The prior distribution is based on previous experience, data, intuition, whatever. If none of these are available, then a flat or non-informative prior is given. Presumably, you know the distributional form of p(x|\theta) based on your experiment. If possible, it is desirable to pick a prior distribution p(\theta) that is "conjugate" to the likelihood ... 0 These quantities are known as part of the model. p(\theta) is the prior, which you chose (a classic example is the Beta distribution), and p(x|\theta) is the density function of X|\theta, for example the model is such that X|\theta\sim\mathcal{N}(\theta,1). 1 By the Chain Rule, we have: \begin{eqnarray*} P(W,R,E\mid B) &=& P(W\mid R,E,B)\;P(R\mid E,B)\;P(E\mid B) \\ &=& P(W\mid E,B)\;P(R\mid E)\;P(E). \end{eqnarray*} This last equality because: Given E,B then W is conditionally independent of R, so P(W\mid R,E,B) = P(W\mid E,B) Given E then R is conditionally independent of B, so ... 1 The definition of conditional probability gives us:$$P(W\mid A,B,E) = \dfrac{P(W,A\mid B,E)}{P(A\mid B,E)}.$$Given A, we know W is conditionally independent of both B and E, so P(W\mid A,B,E) = P(W\mid A). With this, and re-arranging, we get$$P(A\mid B,E)\cdot P(W\mid A) = P(W,A\mid B,E).

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With 8 Boolean variables, the joint has $2^8-1 = 255$ independent entries. In general, the joint must specify $2^n-1$ numbers (where n is the number of variables). The -1 comes from the fact that the sum of those numbers must add up to 1 because the probability of all possible outcomes must sum to 1.

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One of the general formalism for using ensemble methods for classification was developed by this author and his followers https://en.m.wikipedia.org/wiki/Yuri_Zhuravlev and called algebraic theory of algorithms, but i haven't seen any book or course in english, may be some papers, most information in russian.

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So my confusion is with the fact that it's a Bayesian Network. How does the dependency and Independence come into play? Thanks! In a Bayesian network, the branches indicate dependence of the nodes. So children are dependent on their parents but (conditionally) independent of each other; unless there's some incestuous branching in the network.

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The probability of the conditions being icy and Watson crashing is equal to the prior probability that it is icy times the probability that if it is icy Watson will crash: $P(I \land W) = P(W|I)P(I)$.

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