# Tag Info

0

I'd write: $$\, p(\theta_1, \theta_2 \mid X,Y)=\frac{p(X,Y \mid \theta_1, \theta_2)\, p(\theta_1, \theta_2)}{p(X,Y)}= \tag{1}$$ $$=\frac{p(X \mid \theta_1, \theta_2) \, p(Y \mid \theta_1, \theta_2)\, p(\theta_1, \theta_2)}{p(X,Y)}= \tag{2}$$ $$=\frac{\, p(\theta_1, \theta_2 \mid X) \, p(Y \mid \theta_1, \theta_2) P(X)}{P(X,Y)}=\tag{3}$$ $$=\frac{\, ... 0 Using algebraic rearrangement$$\begin{align} \frac{\alpha(x)}{1-\alpha(x)} & =\frac{\alpha\; \mathsf E(g(\mu)\mid x)}{\beta\; \mathsf E(h(\mu)\mid x)} \\[1ex] \alpha(x)\;\beta\; \mathsf E(h(\mu)\mid x) & =\alpha\; \mathsf E(g(\mu)\mid x)\;\big(1-\alpha(x)\big) \\[1ex] \alpha (x) \left(\alpha\; \mathsf E(g(\mu)\mid x)+\beta\; \mathsf E(h(\mu)\mid ...

0

If $$\frac \alpha {1-\alpha} = \frac p q,\tag 1$$ then $$\alpha = \dfrac p {p+q}.$$ To see this, first multiply both sides of $(1)$ by $1-\alpha$: $$\alpha = (1-\alpha)\frac p q.$$ Expand: $$\alpha = \frac p q - \alpha \frac p q.$$ Move all $\alpha$s to one side: $$\alpha + \alpha \frac p q = \frac p q.$$ Factor  \alpha \left( 1 + \frac p q ...

1

Your notation of p'(h|s,r) is confusing. Simply use p(h,s,r)= p(h|s,r)*p(s)*p(r) Then $p(r|h,s)$ $= p(h,s,r)/p(h,s) \\ = p(h,s,r)/[p(h,s,r)+p(h,s,!r)] \\ = p(h|s,r)\cdot p(r)\Big/[p(h|s,r)\cdot p(r)+p(h|s,!r)\cdot p(!r)] \\ = 1\cdot 0.01 / [1\cdot 0.01 + 0.7\cdot 0.99] \\ = 0.01422475106685633001422475106686...$ And $p(r|h,!s)$ = p(h|!s,r)\cdot ... 3 I think the calculation should be: \begin{align} P(T\mid R) &= \dfrac{P(R,T)}{P(R)} = \dfrac{P(R,T)}{P(R,T)+P(R,\neg T)} \\ & \\ P(R,T) &= \sum_{L,A,F\in\{True,False\}} P(R,T,A,F) \\ &= \sum_{L,A,F\in\{True,False\}} P(R\mid L)\;P(L\mid A)\;P(L\mid T,F)\;P(T)\;P(F) \\ & \\ &= (0.75)(0.88)(0.5)(0.02)(0.01) & \qquad [L,A,F] \\ ... 0 I think I figured it out. Essentially, using the law of total probability I can expand P(D1) with an integral, and then since\mu$is uniformly distributed,$p(\mu)$is 1. In addition,$p(D_1 | \mu)$is$\mu * (1-\mu)$since$\mu$is the probability of heads. Using those above facts, I can solve for$p(\mu | D)$. The idea I was missing was that$\mu\$ is a ...

2

Counter-example: Throw a fair die and define events: \begin{align} A &= \{1,2,3,4\} \\ B &= \{2,3,4,5\} \\ C &= \{1,2,3,5\}. \\ \end{align} Then \begin{align} & P(ABC)=1/3 \\ & P(AB)=P(AC)=P(BC)=1/2 \\ & P(A)=P(B)=P(C)=2/3 \\ & \\ \text{So}\quad &P(ABC)=P(A)P(BC)=P(B)P(AC)=P(C)P(AB) \\ \text{But}\quad &P(ABC)=1/3\neq ...

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