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The usual method to solve such questions for Bayesian networks is to use the Chain Rule for probability and use the rules regarding independence and conditional independence implied by the lines joining the nodes. Firstly, $P(\lnot A, C, D) = P(\lnot A, C, D, B) + P(\lnot A, C, D, \lnot B)$, since either $B$ must occur or not occur. We can calculate each ...


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Other than the normal rules on conditional probability, the key thing specific to working with Bayesian networks (BN) is the rules regarding independence and conditional independence of the events represented by the nodes. Without using the independence inferred by the particular network you can only get so far. There are lots of notes around if you google ...


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In general, yes, your interpretation is correct. Bayesian models support a generative interpretation, unlike classical frequentist models (a grey area is mixed-effects modeling).


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Do I understand your problem correctly if I state you want to solve the following problem (assumption $A(x)=Id$) $ \hat U = \arg \min \{ \| U - Y \|^2_2 + \lambda \mathcal R( U) \} $ where the data $Y$ is given on a $N \times N $ grid and $\mathcal R(.)$ is a convex regularization functional? Or is your data given only at certain points on the grid?


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Hint: Use the following facts: (1)$P(A\cap B|C)=\frac{P(A\cap B\cap C)}{P(C)}$ (2)$P(A\cap B\cap C)=P(A|B\cap C)P(B\cap C)$


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In the example, there are three protocols. In the Bayesian viewpoint, the idea is that the number of events of interest is a binomial random variable whose probability of "success" is itself a random variable $p_i$. Specific realizations of the data may be drawn from realizations of the underlying random variables in the model, but that does not mean that ...


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They did use Baye's formula imprecisely. Their values assigned to the claim's premise are correct.   $P(A) = 0.000001$ is the assumed measure of the claim, and $P(B\mid A)=0.9$ is the measure of accuracy given the claim. However, $0.5$ is not the probability measure of the evidence, $P(B),$ as they gave.   It is the measure of the evidence if ...


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Bayes' formula can never give $P(A|B) > 1$ when used correctly. The main trouble with using it in applications is that the prior probabilities are subjective. Suppose, for simplicity, that there are just two alternatives: ($A_1$) The guesser has probability $0.9$ of guessing any given coin-flip correctly (whether this comes from magical powers or ...


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Suppose after some draws you've seen $k$ different colors. Furthermore, suppose you haven't seen any new colors in the previous $t$ draws. If there are more than $k$ colors, the probability that you see a new color within $t$ draws is at least $1-\left(\frac{k}{k+1}\right)^t$. By letting $t$ (the number of draws since the last ($k^{\text{th}}$) new color ...


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If you draw $m$ marbles, the chance that you not have seen a particular color is $\left(\frac {N-1}N \right)^m$, so the chance you have seen it is $1-\left(\frac {N-1}N \right)^m$. If we make the incorrect assumption that the colors are independent, the chance that you have seen them all is $\left(1-\left(\frac {N-1}N \right)^m\right)^N$ This will be very ...


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The (a-priori) pdf of $p$ is $f(p)=1,\; 0\leq p \leq 1$. Let $f(p|G)$ denote the (a-posteriori) pdf of $p$ conditioned on the event $G$. It holds \begin{align} f(p|G) &= \frac{\mathbb{P}(G|p) f(p)}{\int_0^1 \mathbb{P}(G|p) f(p) dp} \\ &= \frac{p}{\int_0^1 p dp} \\ &=2p, \; 0\leq p \leq 1. \end{align} Therefore, $\mathbb{P}(p\geq ...


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Thanks very much - both answers were very helpful. Once I constructed a 2x2 table of the possibilities it became more apparent that I had all the information I needed. I could write all the known probabilities down as the result of a set of simultaneous equations and then solve these to get my unknown probabilities. Using the notation I had in my original ...


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One possibility to organize your information is to put it in a two-by-two table where rows are A wins, B wins; and columns are A scores first, B scores first. (You could also put your info in a 2-circle Venn diagram). Then see where it leads.


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It is not as simple as saying that $A_1 = C_1 * D_1$ (different teams may have different probabilities of winning given that they won the first game.) For this problem your space consists of four events. AA = Team A scores first and wins. AB = Team A scores first and loses. BA = Team B scores first and loses. BB = Team B scores first and wins. Write ...



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