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This is not true. Say $\Omega=\{(0,0),(0,1),(1,0),(1,1)\}$. Say each point of $\Omega$ has probability $1/4$. Let $A=\{(0,0),(1,0)\}$, $B=\{(0,0),(0,1)\}$, and $C=\{(0,0),(1,1)\}$.

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Thanks to Tomas' comment, I went back to my studies and generalized the formula. Having a random variable with two parameters $p_1$ and $p_2$, the Bayes estimator of $p_1$ (and $p_2$ in a similar manner) is simply \begin{align*} &\hat{p_1}=E[p_1|x]=\int p_1\frac{f(x|p_1)\pi(p_1)}{f(x)}dp_1=\int p_1\frac{\int f(x|p_1,p_2)dp_2\pi(p_1)}{\int\int ...

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So it appears the best methods I can find are detailed in http://researcher.ibm.com/researcher/files/us-phaas/jasa3rj.pdf

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We will consider the general case of $m$ marbles placed in the bag, $n$ independent draws with replacement, and the observed number of white marbles $x$; and substitute $m = n = x = 10$ to obtain the desired probability. Let $X$ be the random variable that counts the number of observed white marbles among $n$ independent draws. Because the marbles are ...

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With trial answers to some of the questions in my own comment, suppose we begin the time series at $t = 1$ with $X_1 = \mu = 50$ and that the SD of the normal distribution is $1/X_{t-1}$. Then the iteration seems to converge quickly to $Norm(\mu = 50,\; \sigma = 1/\mu = 1/50).$ Similarly for $\mu=5$. A According to a Kolmogorov-Smirnov goodness-of-fit ...

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Probability $p$ of having $n$ white balls inside the box is: $$p(n) = \frac{\binom{10}n}{2^{10}}$$ With $n$ white balls in the box, the probability $q$ of selecting a white ball all $10$ times is: $$q(n) = \left(\frac{n}{10}\right)^{10}$$ So the probability of getting 10 balls AND the box having $n$ white ones is: $$p(n) q(n)$$ With this, the ...

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Your formulation of independence seems to ignore the crucial conditional relationships. Also your interpretation concentrates on continuous distributions, whereas some applications use a mixture of discrete and continuous distributions. Furthermore, your statement of Bayes' theorem ignores the usual practice of specifying the likelihood function only up to a ...

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You have a lot of learning to do. To begin, you can estimate $P(A|B)$ as the fraction of elements where B is true that also have A true. For example, $P(age=41|happiness=high)$ is approximately the ratio of the count of all members with age 41 and happiness level of high over the count of all those with happiness level high.

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Unknown parameters are indeed random variables according to Bayesians. You can either specify a proper prior for them (non-negative real-valued density that integrates to 1), or use an "improper prior" such as pretending that the probability of every single real number value for the parameter is equal, even though there is no actual density that does this. ...

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