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Yes! Let $X,Y$ are two real valued random variables whose common density $f_{X,Y}(x,y)$ exists. The marginal densities, then, are $$f_X(x)=\int_{\mathbb R}f_{X,Y}(x,y)\ dx \ \ \text{ and } \ \ f_Y(y)=\int_{\mathbb R}f_{X,Y}(x,y)\ dy.$$ The conditional densities are defined as follows $$f_{X|Y=y}(x)=\frac{f_{X,Y}(x,y)}{f_Y(y)} \ \ \text{ and } \ \ ...


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Define $G$ and $B$ to be event of being in good and bad state, respectively. Let $I$ be event that no news arrives in the interval $[t, t+dt)$. As you state, we are interested in $\Pr(G|I)$, because if news arrives, then we are sure to be in good state. So, we are interested in evolution of our belief of being in good state as time passes without the arrival ...


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$p(A|B)$ means the probability of A being true given the assumed truth of B; “AB” means “A and B”, etc. This basically follows from the fact that “A and B” must always be equivalent to “B and A”.


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Take a look at the German tank problem. It is a really interesting problem and Wikipedia provides both forms of analysis


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Under the assumption that your parameters are independent and that the applied transformation is continuous. $$\hat x= g(x)$$ Parameters variance-covariance matrix is: $$\Sigma = \begin{bmatrix} \sigma_{x_1}^2 & 0 &0 \\ 0& \sigma_{x_2}^2&0\\0&0&\sigma_{x_2}^2\end{bmatrix}$$ Then calculate the Jacobian of $g$: $$J = \begin{bmatrix} ...


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I think you want to update your beliefs about $\alpha$ and $\beta$ jointly, not separately. Then you can just use Bayes' rule to write $p(\alpha,\beta|Y)\propto p(Y|\alpha,\beta)p(\alpha,\beta)$. Assuming that your joint prior for $(\alpha,\beta)$ is just the product of the marginal priors, I think this works out to: $$ p(\alpha,\beta|Y) = \frac{\sum_{j=1}^2 ...


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The naive Bayes classifier, while appearing to be written as a conditional probability, is not in fact a conditional probability because it assumes that the data points are independent and identically distributed (i.i.d.). Bayes' theorem is used to compute the "probabilities" even though it does not really apply. These computed quantities are used as a ...


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Following up on OP's request to turn my loss function comment into an answer. Note that in statistics there is rarely a universally agreed-upon approach to an estimation problem, much less a notion of "correct" estimator. The rudimentary setup for parametric statistics is as follows. We have a family of distributions $\{P_\theta\}_{\theta \in \Theta}$ ...


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You said : I do not understand how a mathematical question can have contradictory answers in such a way. The problem is that these are not "mathematical questions" as most people would like them to be. If I am about to roll a six sided die once and ask you to estimate the result, you might say "5" and someone else might say "4." Which answer is right?


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Common choice of bayesian prior for the Poisson distribution is the Gamma distribution. Which ultimately has a Gamma distribution for the posterior. If you have a large set of data which you believe comes from a Poisson distribution. The assumption is that $Y_i \text{ iid}\sim \text{Poisson}(\lambda)$. $$f(y|\lambda) = \frac{\lambda^ye^{-\lambda}}{y!}$$ ...


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In Kartik Audhkhasi's answer, he concluded that Laplace's smoothing can be achieved using MAP on the Dirichlet posterior, with $\alpha=2$. However I guess this is not a practical solution. As for infrequent bigrams, $C(w_{n-1}w_n)$ could be $0$ or $1$, and using a "pseudocount" (intuitively speaking) of "2" gives too much prior information. Another flaw is, ...


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I think there's a dyslexic typo here. Swap $0.1$'s with $0.4$'s. So for example, $$P(S=T,R=T|C=F)=P(S=T|C=F)P(R=T|C=F)=0.5\cdot 0.2 =0.1,$$ $$P(S=F,R=T|C=F)=P(S=F|C=F)P(R=T|C=F)=0.5\cdot 0.2 =0.1,$$ $$P(S=T,R=F|C=F)=P(S=T|C=F)P(R=F|C=F)=0.5\cdot 0.8 =0.4,$$ $$P(S=F,R=F|C=F)=P(S=F|C=F)P(R=F|C=F)=0.5\cdot 0.8 =0.4.$$


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You need to use the fact that Sprinkler and Rain are conditionally independent given Cloudy (this is implied by the structure of the network). In other words, $P(S,R|C)=P(S|C)P(R|C)$. Now you can work things out: $$P(S,R)=P(S,R|C=T)P(C=T)+P(S,R|C=F)P(C=F)=...$$ can you finish it from here?


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The prior distribution and the likelihood function (based on data) both contain information about a parameter. Bayes' theorem allows these two kinds and sources of information to put together into a posterior distribution. The combined information from the posterior distribution can be used make inferences about the parameter. A couple of examples ...


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Start with an example. Say you have a test that discovers some disease. Even if a person tests positive for the disease, all is not lost, since the test may not be accurate. What are the persons chances of actually having the disease? There are three factors involved: What are the overall chances of contracting the disease: It would be very unlikely for ...


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Draw a Venn diagram to help you understand $P(A|B)=P(A\cap B)/P(B)$. Then use this to relate the quantities $P(A|B)$ and $P(B|A)$ algebraically. Let's discuss the first point. Suppose we have a finite sample space so we can count the number outcomes in each possible "event." To determine $P(A|B)$, we're essentially asking what the probability of getting an ...


1

I think you are correct, with minor quibbles. The observed times must have been $(t_0, t_1, \dots, t_N).$ Your $N$ observed exponential interarrival times are $d_i = t_{i} - t_{i-1}$, for $i = 1, \dots, N$. Their sum is your $\tau$. The likelihood function based on exponential interarrival times is $$\prod_{i=1}^N \lambda \exp(-\lambda d_i) = \lambda^N ...


1

First notice that with the given prior, $\mathop{Beta}(1,1)$, the probability of the coin being exactly fair is zero (because the distribution is continuous). The same is true of our posterior. So if we ask the question "What's the probability that the coin is fair?" we will just get the answer $0$. From this point there are two things we could do: The ...


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For c). It is more convenient to denote the required probability with $p$ (and not $P(C)$) because you are going to use in d). So $$p=0.95\cdot P(E_1\cap E_2)=0.95\cdot10^{-9}$$ For d). The number $X$ of crashes that occur in $10000$ flights is a binomial random variable with parameters $n=100000$ and $p=0.95\cdot 10^{-9}$ (from c). Thus $$P(X\ge ...


2

The first thing is that the probability of a woman who does not have the gene developing the cancer is actually $P(D|\overline{C})=x$, where $P(\overline{C})=1-P(C)=0.99$ Your application of Bayes' rule looks right in terms of the formula, however you need to proceed as follows:- $$\begin{align}P(C|D)&= \frac{P(D|C)*P(C)}{P(D)}\\&= ...


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The topic of uniformly sampling from (integer) lattice points within a rational convex polytope is often considered in the literature in connection with the problem of counting such points (known to be NP-hard, as the corresponding decision problem is NP-complete). See the paper "Sampling lattice points," by Kannan and Vempala (1997), which proposes a ...



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