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Consider the Theorem of Bayes for two competing events $A, B$ (with "$c$" for $complement$): $P(B|A) = \dfrac{P(A|B) \cdot P(B)}{P(A|B) \cdot P(B) + P(A|B^c) \cdot P(B^c)}$ In the case of the Bayesian network problem described above [P(B|A and D)] we have: $P(B|A \cap D) = \dfrac{P(A \cap D|B) \cdot P(B)}{P(A \cap D|B) \cdot P(B) + P(A \cap D|B^c) \cdot ... 1 What is the parameter$C$? Is it considered to be a random variable if it has probability density function$f_{C}$? Yes,$C$is a random variable with PDF$f_C$(when$C$has a density). If so, what is its relationship to$(\Omega,\mathcal{A},\mu)$? Does it mean that$C:\Omega\to\mathbb{R}$? Yes,$C$is defined on$\Omega$, if only to be able to ... 0 http://en.wikipedia.org/wiki/Bayesian_network: "A Bayesian network, Bayes network, belief network, Bayes(ian) model or probabilistic directed acyclic graphical model is a probabilistic graphical model (a type of statistical model) that represents a set of random variables and their conditional dependencies via a directed acyclic graph (DAG)." A DBN consist ... 0 No,$P(C|A,B)$can't be determined solely from the marginal distributions$P(A)$and$P(B)$. For example, say$C$perfectly determines$A$and$B$, via one of two mechanisms: If$C=1$, then$A=1$and$B=1$(and the marginal distribution of$C$is the same as$A$and$B$:$P(C=1) = 0.1$). In this case,$P(C=1|A=1, B=1)=1$. If$C=1$, then$A=0$and$B=0$... 0 Non-Factorised: Say we have$3$random variables for simplicity and that$r=2$: each variable having values$T$or$F$. To know the full joint probability distribution, we need to know the actual probability value of each row in this table: $$\begin{array}{ccc|c} x_1 & x_2 & x_3 & P(X_1=x_1,\; X_2=x_2,\; X_3=x_3) \\ \hline T & T & T ... 1 You have a prior distribution$$\pi_0(\theta) = \phi\left(\dfrac{\theta - \eta}{\sigma_0}\right)$$and two likelihoods proportional to \phi\left(\dfrac{s_1 - \theta}{\sigma_1}\right) and \phi\left(\dfrac{s_2 - \theta}{\sigma_2}\right) so you want$$\pi(\theta|s_1,s_2)=\dfrac{\phi\left(\dfrac{\theta - \eta}{\sigma_0}\right)\phi\left(\dfrac{s_1 - ... 0 The key mistake is incorrectly writing the Binomial distribution pdf as$Pr(x_i|p_i,n_i)={n_i \choose x_i} x_i^{p_i} (n_i-x_i)^{1-p_i}$, instead of$Pr(x_i|p_i,n_i)={n_i \choose x_i} p_i^{x_i} (1-p_i)^{n_i-x_i}$. From here we can see our integrand is now of the form of a Beta distribution and thus integrates to$B(x_i + \alpha, n_i - x_i +\beta)$. ... 1 One thing I would change in the way you phrased the problem is to be sure you distinguish the event of getting black ball on the first draw from the event of getting a black ball on the second draw. These are two completely separate (though not independent) events. Clearer notation could help. You have to distinguish such things as "selected bin$A$for the ... 2 Let$b_1$be the event drawing a black ball the first draw; and$b_2$the event of drawing a black ball the second draw. We want$P(b_2|b_1)$. That is, we want$\frac{P(b_2\cap b_1)}{P(b_1)}$. We have$P(b_1)=\frac12\cdot\frac12+\frac12\cdot\frac34=\frac58$. To figure out$P(b_1\cap b_2)$we note that there are four ways this can happen: (i) Bin A, ... 0 Firstly, one should distinguish between the capital$X$used to denote the random variable and the lower-case$x$used as the argument to the density function or the cumulative distribution function, etc. This makes it possible to understand an expression like$\Pr(X\le x)=F(x)$(with lower-case$x$in two places and capital$X$in one). Then$x\mapsto ...
If they would be conditionally independent, it would be possible to decompose $$P(A,B|C)=P(A|C)P(B|C)\textrm{.}$$ However, as you correctly write $$P(A,B|C) = \frac{P(A)P(B)P(C|A,B)}{P(C)}\textrm{.}$$ Thus, whether $A$ and $B$ are conditionally dependent depends on the decomposition of $P(C|A,B)=Q_1(A,C)Q_2(B,C)$ where $Q_1$ and $Q_2$ are some nonnegative ...