So we can calculate this probability by dividing the number of ways given BB by the total number of ways: What he meant is that probability is a device for describing uncertainty from the perspective of an observer with limited knowledge; it has no objective reality. Otherwise they are the same as before. Suppose there are two globes, one for Earth and one for Mars. Which of the following statements corresponds to the expression: $$\Pr(\mathrm{Monday} | \mathrm{rain})$$? This is called the marginal likelihood, and to calculate it, we need to take the probability of each possible globe and multiply it by the conditional probability of seeing land given that globe; we then add up every such product: So it can be interpreted (repeating all the previous work) as the probability of rain, given that it is Monday. $\Pr(B) = 0.5$, Next, let’s calculate the marginal probability of twins on the first birth (using the formula on page 37): $\Pr(\mathrm{twins} | B) = 0.2$ $\Pr(\mathrm{single}|B) = 1 – \Pr(\mathrm{twins}|B) = 1 – 0.2 = 0.8$ We can use the same formulas as before; we just need to update the numbers: $\Pr(\mathrm{BB})=\frac{\mathrm{BB}}{\mathrm{BB+BW+BW}}=\frac{2+2}{2+1+0+2}=\frac{4}{5}$ Software. So the final answer is 0.2307692, which indeed rounds to 0.23. After we already know marriage rate, what additional value is there in also knowing age at marriage? The probability that the female is from species A, given that her first birth was twins, is 1/3 or 0.33. $\Pr(\mathrm{rain}, \mathrm{Monday})/\Pr(\mathrm{rain})=\Pr(\mathrm{Monday}|\mathrm{rain})$ Using the test information only, we go back to the idea that the species are equally likely. So it becomes immediately intuitive that the probability of test saying A but it actually is B just means the probability of test being wrong about B. Option 4 would be $$\Pr(\mathrm{Monday}, \mathrm{rain})$$. Last updated on May 12, 2020 22 min read Notes, R, Statistical Rethinking. $\Pr(A) = \frac{1}{3}$ The Mars globe is 100% land. I do […], Here I work through the practice questions in Chapter 2, “Small Worlds and Large Worlds,” of Statistical Rethinking (McElreath, 2016). $\Pr(A | \mathrm{single}) = \frac{\Pr(\mathrm{single}|A)\Pr(A)}{\Pr(\mathrm{single})} = \frac{0.9(1/3)}{5/6} = 0.36$. The purpose of this paper is to shed light on several misconceptions that have emerged as a result of the proposed “new guidelines” for PLS-SEM. So there are three total ways to produce the current observation ($$2+1+0=3$$). $\Pr(B | \mathrm{twins}) = \frac{\Pr(\mathrm{twins} | B) \Pr (B)}{\Pr(\mathrm{twins})} = \frac{0.2(0.5)}{0.15} = \frac{2}{3}$, These values can be used as the new $$\Pr(A)$$ and $$\Pr(B)$$ estimates, so now we are in a position to answer the question about the second birth. As before, let’s begin by listing the information provided in the question: $\Pr(\mathrm{twins} | A) = 0.1$ As the hint suggests, let’s fill in the table below by thinking through each possible combination of first and second cards that could produce the observed data. Then redo your calculation, now using the birth data as well. Statistical Rethinking 2019 Lectures Beginning Anew! We can do this using the third formula on page 37. A common boast of Bayesian statisticians is that Bayesian inferences makes it easy to use all of the data, even if the data are of different types. Chapter 1 A Review of Basic Statistical Concepts 5 assembled a dream team of behavioral economists to help him get elected—and then to tackle the economic meltdown. Since BB could produce this result from either side facing up, it has two ways to produce it ($$2$$). Hint: Treat this like the sequence of globe tosses, counting all the ways to see each observation, for each possible first card. First ignore your previous information from the births and compute the posterior probability that your panda is species A. $\frac{\Pr(\mathrm{rain},\mathrm{Monday})}{\Pr(\mathrm{Monday})} = \Pr(\mathrm{rain}|\mathrm{Monday})$. \beta_{R} \sim \text{Normal}(0,1) & [\text{prior for }\beta_{R}] \\ Here is the chapter summary from page 45: This chapter introduced the conceptual mechanics of Bayesian data analysis. His models are re-fit in brms, plots are redone with ggplot2, and the general data wrangling code predominantly follows the tidyverse style. Sort by. So the probability of the other side being black is indeed now 0.5. To use the previous birth information, we can update our priors of the probability of species A and B. If you find any typos or mistakes in my answers, or if you have any relevant questions, please feel free to add a comment below. Assume again the original card problem, with a single card showing a black side face up. Reflecting the need for even minor programming in today's model-based statistics, the book pushes readers to perform step-by … Not for re-distribution, re-sale or use in derivative works. $\Pr(\mathrm{rain}|\mathrm{Monday})\Pr(\mathrm{Monday})/\Pr(\mathrm{rain})=\Pr(\mathrm{rain}, \mathrm{Monday})/\Pr(\mathrm{rain})$ Transition from descriptive to inferential statistics (Chapters 6-7) Inferential Statistics (Chapters 8-18) Statistics Descriptive Statistics (Chapters 2-5) FIGURE 1.1 A general overview of this book. Specifically, if a positive test result is indication of the subject being from species A, P(+|B) should correspond to the false positive scenario where the test shows positive yet the subject is actually from species B. This means counting up the ways that each card could produce the observed data (a black card facing up on the table). Finally, there would be no ways for the first card to have been the second side of BW or either side of WW. The third card has two white sides. $\Pr(+|B) = 0.65$ They look exactly alike and eat the same food, and there is yet no genetic assay capable of telling them apart. This site uses Akismet to reduce spam. To begin, let’s list all the information provided by the question: $\Pr(\mathrm{land} | \mathrm{Earth}) = 1 – 0.7 = 0.3$ So the statement, “the probability of water is 0.7” means that, given our limited knowledge, our estimate of this parameter’s value is 0.7 (but it has some single true value independent of our uncertainty). I'm working through all the examples, both in R and the PyMC3 port to python, but I find the statistics confusing at times and would love to bounce ideas off fellow students. The American Statistician has published 43 papers on "A World Beyond p < 0.05." Let’s simulate an experiment. $\Pr(\mathrm{Earth} | \mathrm{land}) = \frac{0.15}{\Pr(\mathrm{land})}=\frac{0.15}{0.65}$. The second card has one black and one white side. Statistical Rethinking is the only resource I have ever read that could successfully bring non-Bayesians of a lower mathematical maturity into the fold. $\frac{\Pr(\mathrm{rain},\mathrm{Monday})}{\Pr(\mathrm{Monday})} = \frac{\Pr(\mathrm{rain}|\mathrm{Monday})\Pr(\mathrm{Monday})}{\Pr(\mathrm{Monday})}$ \alpha \sim \text{Normal}(10, 10) & [\text{prior for }\alpha] \\ The probability of the other side being black is indeed 2/3. If anyone notices any errors (of which there will inevitably be some), I would be … Hugo. But the test, like all tests, is imperfect. Powered by the So the probability of the first card having black on the other side is indeed 0.75. We just have to calculate the updated marginal probability of twins. This is the information you have about the test: The vet administers the test to your panda and tells you that the test is positive for species A. Here is a super-easy visual guide to setting up and running RStudio Server for Ubuntu 20 on Windows 10. What does it mean to say “the probability of water is 0.7”? Rather, it is named after Stanislaw Ulam (1909–1984). Statistical Thinking By Beth Chance and Allan Rossman. Lectures. The probability that it is Monday and that it is raining. Again calculate the probability that the other side is black. Like the other BB card, it has $$2$$ ways to produce the observed data. Notes on Statistical Rethinking (Chapter 9 - Big Entropy and the Generalized Linear Model) Apr 22, 2018 9 min read StatisticalRethinking Entropy provides one useful principle to guide choice of probability distributions: bet on the distribution with the biggest entropy. Some of the material in these notes will be published by Cambridge University Press as Statistical Machine Learning: A Gentle Primer by Rui M. Castro and Robert D. Nowak. Let’s update our table to include the new card. Let’s convert each statement to an expression: Option 1 would be $$\Pr(\mathrm{rain} | \mathrm{Monday})$$. Statistical Rethinking Chapter 5 Problems John Fox 2016-11-4. Feb. 21, 2019. California Polytechnic State University, San Luis Obispo. Note that this probability increased from 0.33 to 0.36 when it was observed that the second birth was not twins. In each case, assume a uniform prior for $$p$$. The UNDP Human Development Report 2020 explores how human activity, environmental change, and inequality are changing how we work, live and cooperate. best. share. Statistical Rethinking chapter 5 notes. The Bayesian statistician Bruno de Finetti (1906-1985) began his book on probability theory with the declaration: “PROBABILITY DOES NOT EXIST.” The capitals appeared in the original, so I imagine de Finetti wanted us to shout the statement. Below are my attempts to work through the solutions for the exercises of Chapter 2 of Richard McElreath's 'Statistical Rethinking: A Bayesian course with examples in R and Stan'. The probability that it is Monday, given that it is raining. They differ however in family sizes. $\Pr(B) = 0.5$ Of these three ways, only the ways produced by the BB card would allow the other side to also be black. We can use the same approach to update the probability again. The rst chapter is a short introduction to statistics and probability. Imagine that black ink is heavy, and so cards with black sides are heavier than cards with white sides. So the posterior probability of species A (using just the test result) is 0.552. So the total ways for the first card to be BB is $$3+3=6$$. Why things are normal. More mechanically, a Bayesian model is a composite of a likelihood, a choice of parameters, and a prior. I do my […], Here I work through the practice questions in Chapter 3, “Sampling the Imaginary,” of Statistical Rethinking (McElreath, 2016). Statistical Rethinking: Chapter 3. So we can use the same approach and code as before, but we need to update the prior. P(test says B | B) = 0.65. $\Pr(B) = \frac{2}{3}$ Option 2 would be $$\Pr(\mathrm{rain} | \mathrm{Monday})$$. Code from Statistical Rethinking modified by R Pruim is shown below. Statistical Rethinking (Code) Chapter 12 April, 2017. $\Pr(\mathrm{rain}|\mathrm{Monday})\Pr(\mathrm{Monday}) = \Pr(\mathrm{rain}, \mathrm{Monday})$ $\Pr(A) = 0.5$ This reflects the idea that singleton births are more likely in species A than in species B. save hide report. New comments cannot be posted and votes cannot be cast. $\Pr(A | \mathrm{twins}) = \frac{\Pr(\mathrm{twins} | A) \Pr (A)}{\Pr(\mathrm{twins})} = \frac{0.1(0.5)}{0.15} = \frac{1}{3}$. $\Pr(\mathrm{BB})=\frac{\mathrm{BB}}{\mathrm{BB+BW+BW}}=\frac{2}{2+1+0}=\frac{2}{3}$. Let’s suppose 1,000 people flip a coin 16 times. Use the counting method, as before. Option 3 needs to be converted using the formula on page 36: The test says B, given that it is actually B is 0.65. $\Pr(+|A) = 0.8$ Now we can substitute this value into the formula from before to get our answer: Required fields are marked *. $\Pr(A) = 0.36$ Learn how your comment data is processed. Before looking at the other side, we draw another card from the bag and lay it face up on the table. The rules of probability tell us that the logical way to compute the plausibilities, after accounting for the data, is to use Bayes’ theorem. $\Pr(\mathrm{single}|A) = 1 – \Pr(\mathrm{twins}|A) = 1 – 0.1 = 0.9$ So suppose now that a veterinarian comes along who has a new genetic test that she claims can identify the species of our mother panda. 2 Syllabus. Lecture 02 of the Dec 2018 through March 2019 edition of Statistical Rethinking: A Bayesian Course with R and Stan. Richard McElreath (2016) Statistical Rethinking: A Bayesian Course with Examples in R and Stan. Statistical physics is a beautiful subject. In order for the other side of the first card to be black, the first card would have had to be BB. After experimenting a number of times, you conclude that for every way to pull the BB card from the bag, there are 2 ways to pull the BW card and 3 ways to pull the WW card. \end{array} $\Pr(\mathrm{BB})=\frac{\mathrm{BB}}{\mathrm{BB+BW+BW}}=\frac{2}{2+2+0}=\frac{2}{4}=\frac{1}{2}$ Statistical Rethinking: A Bayesian Course with Examples in R and Stan builds readers' knowledge of and confidence in statistical modeling. Again, this time with 5 $W$s and 7 tosses: Now assume a prior for $$p$$ that is equal to zero when $$p<0.5$$ and is a positive constant when $$p\ge0.5$$. $\Pr(A | +) = \frac{\Pr(+ | A) \Pr(A)}{\Pr(+)} = \frac{0.8(0.5)}{0.725} = 0.552$. $\Pr(B) = 1 – \Pr(A) = 1 – 0.36 = 0.64$, Now we just need to do the same process again using the updated values. Academic theme for Notes on Statistical Rethinking (Chapter 8 - Markov Chain Monte Carlo) Apr 19, 2018 33 min read StatisticalRethinking The Stan programming language is not an abbreviation or acronym. $\Pr(\mathrm{twins}) = \Pr(\mathrm{twins} | A) \Pr(A) + \Pr(\mathrm{twins} | B) \Pr(B) = 0.1\bigg(\frac{1}{3}\bigg) + 0.2\bigg(\frac{2}{3}\bigg) = \frac{1}{6}$. You will actually get to practice Bayesian statistics while learning about it and the book is incredibly easy to follow. The correct answers are thus Option 1 and Option 4. Now we just need to count the number of ways each card could produce the observed data (a black card facing up on the table). $$\Pr(\mathrm{rain}|\mathrm{Monday})\Pr(\mathrm{Monday})$$, $$\Pr(\mathrm{rain}|\mathrm{Monday})\Pr(\mathrm{Monday})/\Pr(\mathrm{rain})$$, $$\Pr(\mathrm{Monday}|\mathrm{rain})\Pr(\mathrm{rain})/\Pr(\mathrm{Monday})$$. lecture note include statistical signal processing, digital communications, information theory, and modern con-trol theory. So again assume that there are three cards: BB, BW, and WW. Here I work through the practice questions in Chapter 2, “Small Worlds and Large Worlds,” of Statistical Rethinking (McElreath, 2016). Using the approach from 2E1, we could show that Option 4 is equal to $$\Pr(\mathrm{Monday}|\mathrm{rain})\Pr(\mathrm{rain})$$, but that is not what we want. 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