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Considering $P(E \vert A)$, you might first think about whether $E$ is dependent on $A$ at all. If this is not the case, it holds that $P(E \vert A) = P(E)$ And considering $P(A,B)$: Then $P(A,B) = P(A)P(B)$, if $A$ and $B$ are independent of each other. If you understood how d-separation works, then you should be able to answer these questions yourself.

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In general, the conditional pdf of $X$ given that $X \leq a$ is just $$f_{X \mid \{X \leq a\}}(x) = \begin{cases} \displaystyle \frac{f_{X}(x)}{P\{X \leq a\}}, & x \leq a,\\0, &x > a,\end{cases}$$ that is, it is just the pdf of $X$ scaled to have total area $1$ (as all pdfs must have) in the region of the conditioning event, and $0$ in the ...

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The formula on that slide was a straw man and not intended to make sense. The point was that moment matching does not make sense on an individual likelihood term in isolation. This is illustrated further on the next slides. I have actually seen this bad approach used in papers, so I thought it was worth pointing out. This is one of those cases where "you ...

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If a random vector $X$ has density $f$, and $A=\{\omega:X(\omega)\in B_0\}$, for some fixed Borel set $B_0$, supposing that $P(A)>0$, we can define a conditional density $f(\;\cdot\mid A)$ as a function that satisfies $$P(X\in B\mid A) =\int_B f(x\mid A)\,dx \, , \qquad\qquad (*)$$ for every Borel set $B$. It is easy to prove that $$f(x\mid A) = ... 0 So the answer is yes this is correct, there is another way to get b) without the 1-p formula like p(A or B)= p(A)+ p(B) - p(A and B), but I wanted to see the conditionals and use it. Thanks. 2 Wom Man Mar x y Sing w z You are mixing up the concepts of joint and conditional probability. Problem: x = P(M,W) = 0.06 x+w = P(W) = 0.3 P(M|W) = ? Solution: P(M|W) = P(M,W) / P(W)  2 You should be able to draw and use something like the diagram below to lay out the information (write it in the spaces and margins) and then you may be able to see how to do the problem. You would write all of the values on the diagram and from them fill in probabilities for every subregion and colored region(/margin). Then you should have a clearer idea ... 2 That is the correct "long" way to do it. Essentially any combinatorics problem could be solved this way with enough ink and paper or computing power. This seems a bit like homework, so I'll get you started in the right direction without giving the answer away completely. You've got a bunch of conditional probabilities P(A|B) that you need to calculate. Use ... 0 Marginalising out x_c means that you can forget about the last row and column of the covariance and \mu_c in the mean vector. http://en.wikipedia.org/wiki/Multivariate_normal_distribution#Conditional_distributions$$p(x_a|x_b)=\mathcal{N}\left(\mu_{a|b},\Sigma_{a|b}\right) Where $\mu_{a|b},\Sigma_{a|b}$ is given in the above link.

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You can solve this question analytically. Denote $P(R_m)=0.5$ as the probability that it will rain in the morning, $P(R_e)=0.5$ as the probability that it will rain in the evening and $P(U)$ as the probability sally brought her umbrella to school. You are interested in figuring out $P(\bar{U}, R_e)$, i.e. the probability that she has no umbrella and it rains ...

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