New answers tagged

0

The way that I would think about it is the following: First I would try to define the sample space of that problem, i.e the space that will contain all the possible results. In order to define that I would consider what values the events $A$ and $B$ can take. $A$ which is the number of goals can take the following value $\left \{ 0,1,2,... \right \}$, while $...


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This equation is an application of the law of total probability conditioning on events $U=u$, integrated over all values $u$ in the support of $U$. Here is a more explicit version of the same working: $$\begin{align} p(X(U) = k|X(0) = 1) &= \int \limits_\mathbb{R} p(X(U) = k|X(0) = 1, U=u) \ g_U(u) \ du \\[6pt] &= \int \limits_0^1 p(X(u) = k|X(0) = ...


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It's not. For example, let $X$ and $Y$ be independent, $Y$ can be any distribution.


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Pollard argues that disintegrations, which give conditional probability distributions under fairly general conditions, are the right way to think about the factorisation theorem and sufficiency in the continuous case.


14

I like this example: $$ \DeclareMathOperator{\P}{\mathbb{P}} \P(\text{Dead} \mid \text{Was hanged}) \qquad \text{is high}\\ \P(\text{Was hanged} \mid \text{Dead}) \qquad \text{is low} $$ and that should answer your question.


1

Let's review the conditional probability of both $P(x|y)$ and $P(y|x)$: $$P(x|y) = \frac{P(x \cap y)}{P(y)}$$ and $$P(y|x) = \frac{P(x \cap y)}{P(x)}$$ The common thing in both is their joint probability. So, their discrepancy depends on $P(x)$ and $P(y)$. If these two probabilities are equal or close to each other then according to formulas, the conditional ...


8

$P(x|y)$ and $P(y|x)$ are different things and do not have to be related. Say that you have an “algorithm” that predicts the weather. It is very simple, it always predicts “it is going to rain”. So P(predicted rain|rain) = 1, in all cases it rained, it also predicted rain. On another hand, say that in your area it rains only 5% of the time, so P(rain|...


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This situation seems to amenable to the analysis of competing risks provided by the R survival package. You have a waitlist state, from which a transition is possible to exactly 1 of 3 terminal states: ineligible, transplant, and death. (As you note, one might further model a transition from transplant to death but we'll stay with the simple model.) That is ...


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I would categorize both methods as "correct". This feels like the mindset of the teacher is "this is the answer in the book, therefore it is the ONLY answer". Or "this is in the section with formula so you have to use the formula". This question I think would be geared more towards the multiplicative rule: $P(A \cap B) = P(B|A)P(...


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The question you linked to is by someone curious about an incorrect formula they found. The answer corrects it. The corrected formula does include $A$ in the denominator. In your notation, the formula from the answer includes $A$ and is: $$ \Pr(A \mid B, C) = \frac{ \Pr(B \mid A) \Pr(C \mid A) \Pr(A) }{ \Pr(B \mid A) \Pr(C \mid A) \Pr(A) + \Pr(B \mid \...


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I've upvoted a couple answers that already provide many of the ingredients to the answer. I'll provide what I view as a more direct answer. Suppose you find a dataset with observations on 2 fields: x (fertilizer) and y (yield) but you don't know exactly how this dataset was obtained. You think of Fisher's experiments and realize that this is probably ...


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Since this is self-study (please tag and research!) I won't give you the answer, but the simple formula you are missing is the law of total variance: $$\text{var}(X) = E(\text{var}(X|W)) + \text{var}(E(X|W))$$


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The problem with your reasoning is that you don't take into account the probability of the second card to be a heart or not (p = 0.5). P(A|B) = (12/51)x0.5 + (13/51)x0.5 = 25/102.


3

Using intuition, knowing that the first card was red doesn't change the probability of choosing a heart that much, so it seems wrong that it could be almost double. $P[A|B \text{ or } C]$ is not necessarily equal to $P[A|B]+P[A|C]$, even if $B$ and $C$ are disjoint. Suppose $B$ and $C$ are disjoint, like they are in the example where $B$ means the first card ...


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Thank you to Arya. $P(A) = 0.15$ $P(B) = 0.6$ $P(B|A) = 0.8$ $P(A|B) = \frac{0.15*0.8}{0.6} $ $P(A|B) = 0.2$


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I'll refresh future readers on the variables you defined. $A$ is whether the callee becomes a new customer. $B$ is whether the callee has used a rival company in the previous year. The information you've been given is: $P(A) = 0.15$ $P(B \mid A) = 0.8$ $P(B) = 0.6%$ You tried to compute $P(B \mid A)$ by multiplying $P(B \mid A) \times P(A)$. That's not ...


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All the information about lifetimes (whether Weibull distributed or not) is contained in the survival function. So let's start by deriving the survival function given the subject is age $t_0$. Let $T$ denote the (random) lifetime of the subject. Recall that the survival function $S(t)$ is really just notation for $P(T > t)$. We'd like the survival ...


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Thanks to Arya McCarthy for the tip to use $a \times b + c \times d$, where $a$ is the probability of choosing a paperback fiction = $\frac{59}{95}$ $c$ is the probability of choosing a hardcover fiction = $\frac{13}{95}$ given the first book is paperback, $b$ is then the probability of choosing a hardcover without replacement = $\frac{28}{94}$ given the ...


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From Wikipedia: A Bayesian network (also known as a Bayes network, belief network, or decision network) is a probabilistic graphical model that represents a set of variables and their conditional dependencies via a directed acyclic graph (DAG). Bayes' rule is used for inference in Bayesian networks, as will be shown below. A better name for a Bayesian ...


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In response to I did not fully understand why we are multiplying the intermediate conditionals i.e.: $...Pr(Burglar|Storm=T)xPr(Cat|Storm=T)...$ and Is there a logical explanation for the multiplication of the intermediate probabilities - or is it just a trick? Denoting each random variable with their first letter, the model represented using the ...


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In Bayesian Networks, there are three types of junctions, https://en.wikipedia.org/wiki/Bayesian_network. In your case, you have a Fork junction, $Storm \rightarrow Burglar \ \& \ Storm \rightarrow Cat.$ Each junction case has some conditional or unconditional independence properties. For your case, the Fork junction has $$Burglar \perp Cat \ | \ Storm $...


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