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0

It seems that I have not pushed the calculation to the end. In fact: \begin{align} \sum_{i=1,x_i=0}^T \log(1 - p) &= \log(1 - p) \left( T - \| x \|_0 \right) \\ &= \log(1 - p) T - \log(1 - p) \| x \|_0 \\ &= \log(1 - p) T + \left[ p + o(p) \right] \| x \|_0 \quad \text{for } p \rightarrow 0 \end{align} The first term doesn't depend on $ x ...


0

The prior $ f(\theta) $ where $ \theta = \{ \tau_j \}_{j \in \{1, \cdots, n\}} $ are a Poisson process with rate $ \mu $ is given by: \begin{align} f(\theta) &= f_{\tau_{(1)}, \cdots, \tau_{(n)}}(t_1, \cdots, t_n) \\ & = f_{\tau_{(1)}, \cdots, \tau_{(n)} \mid N(T) = n}(t_1, \cdots, t_n) \times \mathbb{P}(N(T) = n) \\ & = \frac{n!}{t^n} ...


2

I preassume that $AB$ is an abbreviation of $A\cap B$ here, and that we are dealing with disjoint events $B_1,B_2,\dots$ that cover the whole space. Also I just don't know what is meant by marginalization and conditioning, but here is a direct route to prove the statement. We have the general rule: $$P(S|R)P(R)=P(SR)\tag1$$ leading to: ...


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ii) Is actually correct. iii) Is simply $\frac{1}{3}.\frac{4}{5}$ so $P(G\cap G1)$.


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The observed data implies $r^2\ge3^2+1^2=10$, so $r\ge\sqrt{10}$. Thus the hypothesis $r\le2$ has already been refuted. The likelihood function for $r$ is now $f(r)=\frac{\sqrt{10}}{r^2}$ for $r\ge\sqrt{10}$ and $0$ otherwise.


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1. Let $X$ be an event that the probabilistic program does not crash, $H$ be an event that the hardware is ok. From the question it is given that $P(X|H) = \frac{97}{100}$, $P(H) = \frac{3}{5}$ and $P(X|H^{c}) = \frac{1}{2}$, where $H^{c}$ is the compement of $H$, that is an event that the hardware is buggy. a) We would like to find $P(X)$ and for that we ...


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$P(\text{hot})$ isn't one. All you know is that the event "hot" occurred; this doesn't imply that it has probability 1. You need to calculate using Bayes' rule. Using $H$ to denote the event "hot", we see: $$P(A\mid H)={P(H\mid A)P(A)\over P(H)}={P(H\mid A)P(A)\over P(H\mid A)P(A) + P(H|B)P(B)}.$$ Now plug in $P(A)=0.4$, $P(B)=0.6$, $P(H\mid A)=0.9$, ...


2

Your second approach is the right idea, but what you were after is: $$\mathsf P(A_2\mid A_1) ~=~ \mathsf P(A_2\mid P_{\rm arasite})~\mathsf P(P_{\rm arasite}\mid A_1)+\mathsf P(A_2\mid P_{\rm arasite}^\complement)~\mathsf P(P_{\rm arasite}^\complement\mid A_1)$$ Now you are looking for $\mathsf P(P_{\rm arasite}\mid A_1,A_2)$ using Bayes' Rule, and the ...


1

If your endgame is to get the posterior density for y, I'm not understanding why you need an approximation when you can get the exact posterior density for any particular set of data. Here's how to do it using Mathematica: (* Data *) n = 10; p = {2/10, 3/10, 222/1000, 15/100, 19/100, 4/10, 44/100, 21/100, 34/100, 1/10}; x = {1, 1, 0, 0, 1, 1, 1, 1, 0, 1}; ...


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It depends on the original distribution, and the strength of the prior. For example, if the original distribution is $\mathcal{N}(0,1)$, and the prior is $\mathcal{N}(100,.1)$, then almost surely you will never converge to the true distribution.


1

Proposition: Let $a,b,c,d$ be positive real numbers such that $ad-bc<0$. Then $$\frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$$ First, notice that the requirement is equivalent to $a/b<c/d$, so we can use either as the supposition in the proposition. $$ad-bc=ad+ac-(ab+bc)<0\Rightarrow\frac{a}{b}<\frac{a+c}{b+d}$$ by dividing both sides by ...


1

$\dfrac{\alpha+y}{\alpha+n+\beta}=\dfrac{\alpha+\beta}{\alpha+n+\beta}\cdot\dfrac{\alpha}{\alpha+\beta}+\dfrac{n}{\alpha+n+\beta}\cdot\dfrac{y}{n}$ Take $\gamma=\dfrac{n}{\alpha+n+\beta}\in(0,1)$ then $\dfrac{\alpha+y}{\alpha+n+\beta}=(1-\gamma)\dfrac{\alpha}{\alpha+\beta}+\gamma\dfrac{y}{n}$. It is a convex combination of $\dfrac{\alpha}{\alpha+\beta}$ ...


0

$$P(R)=1/5$$ For each forecast $i=1,2,3$ $$P(H_i = R|R)= 3/4$$ and $$P(H_i=R|\bar{R})=1/4$$ We need to find $$P(R|H_1=R, H_2= R, H_3 = \bar{R}) = \frac{P(R) P(H_1=R, H_2=R, H_3=\bar{R}|R)}{P(H_1=R, H_2=R, H_3=\bar{R})} = \frac{P(R) P(H_1=R, H_2=R, H_3=\bar{R}|R)}{P(R) P(H_1=R, H_2=R, H_3=\bar{R}|R)+P(\bar{R}) P(H_1=R, H_2=R, H_3=\bar{R}|\bar{R})} ...


2

Using Bayes' rule, for each value $x$ you have $$P(x | data) \propto P(data | x) prior(x)$$ Therefore, $P(x|data)$ is zero if $x \neq 0$ and so $P(0|data) = 1$, so your posterior is the same as the prior. No, the posterior is the same as the prior. Using this prior says that you have decided that the answer is zero and no evidence can convince you ...



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