12

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.


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|...


3

You have 19 sequential and independent trials, and need to be successful in all of them. The first trial is a die roll, with a success probability of $\frac{1}{6}$. The second trial is a coin toss, with a success probability of $\frac{1}{2}$. The third trial is a die roll, with a success probability of $\frac{1}{6}$. And so forth. Now, since all trials are ...


3

It appears your confusion arises from couple arithmetic errors that come before it. I assumed that you have obtained that $Pr(M|C) = \frac{0.05}{0.0525}$, and calculated that $Pr(C) = 0.0525$. Thus, by the correct definition of conditional probability $Pr(M|C) = \frac{Pr(M \cap C)}{Pr(C)}$, you are trying to reason that $Pr(M \cap C) = 0.05$, and wonder why ...


3

This question is really about Stan rather than pystan, so I will demonstrate how one might be able to use the posterior samples to make predictions. Say you have a very simple Bayesian model for linear regression. $$ y \sim \mathcal{N} (\alpha +X \beta, \sigma^2) $$ $$ \beta \sim \mathcal{N}(0, 1) $$ $$ \sigma \sim \operatorname{Half-Cauchy}(0, 1) $$ A Stan ...


2

Apart from the usual mathematical techniques of checking your derivation, checking for consistency (e.g., differentiating the CDF to see whether it equals the PDF), and considering whether your answer makes sense, whenever there is a question of determining the distribution of a random variable (or collection of variables) you almost always have available ...


2

Covariance and correlation that is just its normalized version, measure the linear relationship between two variables. A linear relationship is just one of the many possible relationships, so they cannot be thought of as a general measure of how strongly the two variables are related, though it's a popular proxy for measuring it. If you have two events $X$ ...


2

How many sequences of coin flips preserve the deck? Call this number $p(n)$ for $n=1, 2, 3,$ and so on. Since there are $2^n$ equiprobable sequences, your answer to the question (about probability) will be $2^{-n}\,p(n).$ Obviously $p(1)=2,$ because no matter what the outcome of that one flip is, the deck is the same afterwards. Consider $p(n+1)$ for $n\ge ...


2

If we assume that each outcome is from a fair die then the outcomes are IID uniform random variables on the $m=6$ possible outcomes. Suppose we roll $n$ dice and let $K_n$ be the number of distinct outcomes (called the "occupancy number") from all dice. It is well-known that this random variable is distributed according to the classical occupancy ...


2

By combinatorics: $$(6/6)(5/6)(4/6)(3/6)(2/6) = {}_6P_5/6^5 = 6!/6^5 = 0.09259259.$$ But you should also investigate the connection with the coupon-collector's problem. By simulation in R, a million iterations gives $0.092627\pm 0.0006.$ die = 1:6 set.seed(2021) nr.faces = replicate(10^6, length(unique(sample(die,5,rep=T)))) mean(nr.faces==5) [1] 0.092627 ...


2

All answers seem to focus on the nature of random while the question concerns truly the nature of knowledge. What is it to know something? You seem to implicitly allude to an unattainable and impractical notion of knowledge, some God-like insight into matters. We're human, for us knowledge is not an absolute state of clarity and vision. We know that it's ...


1

Instead of flipping a coin in between, you can roll a die with twelve sides. (for instance: construct the die so that there are light and dark sides numbered 1..6) The coinflip is not needed after the last roll so you roll your new die nine times and finally one roll with the old die; this gives you the probability $$ \bigg(\frac{1}{12}\bigg)^9\cdot\frac{1}{...


1

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


1

First you have to specify what distribution you are drawing the the variable from, e.g. $\mathcal{N}(0, 1)$. Then, denote the variable by a letter, typically a capital one, like $X$. Finally, you can say $X \sim \mathcal{N}(0, 1)$. PS: your question deals with the concept of "realization" of random variable. Referring to this[1], you can say "$...


1

It's important to realize that probability, by its definition/construction as a mathematical concept, is only countably additive and not "uncountably additive." Generally in mathematics there is no such concept as uncountably additive (there are a few contexts where uncountable sums have been defined, but they aren't applicable to interpreting ...


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 ...


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