# statistical rethinking notes

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\mu_{i} = \alpha + \beta_{R}R_{i} + \beta_{A}A_{i} & \text{[linear model]}\\ Option 5 is the same as the previous option but with the terms exchanged. $\frac{\Pr(\mathrm{rain},\mathrm{Monday})}{\Pr(\mathrm{Monday})} = \frac{\Pr(\mathrm{rain}|\mathrm{Monday})\Pr(\mathrm{Monday})}{\Pr(\mathrm{Monday})}$ First ignore your previous information from the births and compute the posterior probability that your panda is species A. P(test says A | B) = 1 – P (test says B | B) = 1 – 0.65 = 0.35, And for the posterior calculation, you would have to use Now we can substitute this value into the formula from before to get our answer: Someone reaches into the bag and pulls out a card and places it flat on a table. 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. Rethinking P-Values: Is "Statistical Significance" Useless? 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. save hide report. $\Pr(A) = 0.5$ Thus P(+|B) = 1 – P(-|B) = 0.35. $\Pr(+|A) = 0.8$ This site uses Akismet to reduce spam. The UNDP Human Development Report 2020 explores how human activity, environmental change, and inequality are changing how we work, live and cooperate. That the data are grouped makes the assumption of independence among observations suspect. Just in case anyone is still looking for the correct answer and has no explanation, a rewording of the statement “correctly identifies a species A panda is 0.8” helps. $\Pr(\mathrm{BB})=\frac{\mathrm{BB}}{\mathrm{BB+BW+BW}}=\frac{2}{2+1+0}=\frac{2}{3}$. BW could only produce this with its black side facing up ($$1$$), and WW cannot produce it in any way ($$0$$). One card has two black sides. $\Pr(\mathrm{BB})=\frac{\mathrm{BB}}{\mathrm{BB+BW+WW}}=\frac{6}{6+2+0}=\frac{6}{8}=0.75$. $\Pr(A) = 0.5$ Statistical Rethinking: A Bayesian Course with Examples in R and Stan builds readers' knowledge of and confidence in statistical modeling. $\frac{\Pr(\mathrm{rain},\mathrm{Monday})}{\Pr(\mathrm{Monday})} = \Pr(\mathrm{rain}|\mathrm{Monday})$. 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. 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). Here I work through the practice questions in Chapter 2, “Small Worlds and Large Worlds,” of Statistical Rethinking (McElreath, 2016). Stu- Statistical Rethinking Chapter 5 Problems John Fox 2016-11-4. Powered by the If the first card was the first side of BB, then there would be 3 ways for the second card to show white (i.e., the second side of BW, the first side of WW, or the second side of WW). Ultimately, statistical learning is a fundamental ingredient in the training of a modern data scientist. These functions are used in the Pluto notebooks projects specifically intended for hands-on use while studying the book or taking the course. $\Pr(A | \mathrm{twins}) = \frac{\Pr(\mathrm{twins} | A) \Pr (A)}{\Pr(\mathrm{twins})} = \frac{0.1(0.5)}{0.15} = \frac{1}{3}$ NOTE: Descriptive statistics summarize data to make sense or meaning of a list of numeric values. I do my […], Here I work through the practice questions in Chapter 3, “Sampling the Imaginary,” of Statistical Rethinking (McElreath, 2016). Sort by. $\Pr(B) = 0.5$ Otherwise they are the same as before. Academic theme for This one got a thumbs up from the Stan team members who’ve read it, and Rasmus Bååth has called it “a pedagogical masterpiece.” The book’s web site has two sample chapters, video tutorials, and the code. Both are equally common in the wild and live in the same place. I do my best […], Here I work through the practice questions in Chapter 6, “Overfitting, Regularization, and Information Criteria,” of Statistical Rethinking (McElreath, 2016). Using the test information only, we go back to the idea that the species are equally likely. 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. Rather, it is named after Stanislaw Ulam (1909–1984). Below are my attempts to work through the solutions for the exercises of Chapter 3 of Richard McElreath’s ‘Statistical Rethinking: A Bayesian course with examples in R and Stan’. Let’s convert each expression into a statement: Option 1 would be the probability that it is Monday, given that it is raining. 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. Again compute and plot the grid approximate posterior distribution for each of the sets of observations in the problem just above. 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: Posted Mar 22, 2019 So we can use the same approach and code as before, but we need to update the prior. As our society increasingly calls for evidence-based decision making, it is important to consider how and when we can draw valid inferences from data. More mechanically, a Bayesian model is a composite of a likelihood, a choice of parameters, and a prior. Here is the chapter summary from page 45: This chapter introduced the conceptual mechanics of Bayesian data analysis. $Show that the posterior probability that the globe was the Earth, conditional on seeing “land” ($$\Pr(\mathrm{Earth}|\mathrm{land})$$), is 0.23. \[\Pr(A) = \frac{1}{3}$ I'm working through all the examples, both in R and the PyMC3 port to python, but I find the statistics confusing at times and would love to bounce ideas off fellow students. Then redo your calculation, now using the birth data as well. Predictor residual plots. If the first card was the first side of BW, then there would be 2 ways for the second card to show white (i.e., the first side of WW or the second side of WW; it would not be possible for the white side of itself to be shown). The rst part of the book deals with descriptive statistics and provides prob-ability concepts that are required for the interpretation of statistical inference. $\Pr(+|B) = 0.65$ $\Pr(\mathrm{twins} | B) = 0.2$ What he meant is that probability is a device for describing uncertainty from the perspective of an observer with limited knowledge; it has no objective reality. Suppose there are two globes, one for Earth and one for Mars. Option 4 is the probability of rain and it being Monday, given that it is Monday. Suppose there are two species of panda bear. \alpha \sim \text{Normal}(10, 10) & [\text{prior for }\alpha] \\ Syllabus. Which of the expressions below correspond to the statement: the probability of rain on Monday? D_{i} \sim \text{Normal}(\mu_{i}, \sigma) & \text{[likelihood]} \\ Notes on Statistical Rethinking (Chapter 8 - Markov Chain Monte Carlo) Apr 19, 2018 33 min read StatisticalRethinking The Stan programming language is not an abbreviation or acronym. \], $$\mu_{i} = \alpha + \beta_{R}R_{i} + \beta_{A}A_{i}$$. $\Pr(\mathrm{single}|B) = 1 – \Pr(\mathrm{twins}|B) = 1 – 0.2 = 0.8$ Statistical Rethinking: Chapter 3. So it can be interpreted (repeating all the previous work) as the probability of rain, given that it is Monday. 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. This results in the posterior distribution. $\Pr(A | +) = \frac{\Pr(+ | A) \Pr(A)}{\Pr(+)} = \frac{0.8(0.5)}{0.725} = 0.552$. Option 4 is the same as the previous option but with division added: Hint: Treat this like the sequence of globe tosses, counting all the ways to see each observation, for each possible first card. So the final answer is 0.2307692, which indeed rounds to 0.23. The probability it correctly identifies a species A panda is 0.8. Option 2 would be $$\Pr(\mathrm{rain} | \mathrm{Monday})$$. $\Pr(\mathrm{rain}|\mathrm{Monday})\Pr(\mathrm{Monday})/\Pr(\mathrm{rain})=\Pr(\mathrm{rain}, \mathrm{Monday})/\Pr(\mathrm{rain})$ Of these three ways, only the ways produced by the BB card would allow the other side to also be black. What does it mean to say “the probability of water is 0.7”? Now suppose you are managing a captive panda breeding program. Afte we already know age at marriage, what additional value is there in also knowing marriage rate? share. After we already know marriage rate, what additional value is there in also knowing age at marriage? Richard McElreath (2016) Statistical Rethinking: A Bayesian Course with Examples in R and Stan. 2 Compute and plot the grid approximate posterior distribution for each of the following sets of observations. In order for the other side of the first card to be black, the first card would have had to be BB. Statistical rethinking with brms, ggplot2, and the tidyverse This project is an attempt to re-express the code in McElreath’s textbook. The posterior probability of species A (using both the test result and the birth information) is 0.409. $\Pr(\mathrm{rain}, \mathrm{Monday})/\Pr(\mathrm{rain})=\Pr(\mathrm{Monday}|\mathrm{rain})$ $\Pr(\mathrm{BB})=\frac{\mathrm{BB}}{\mathrm{BB+BW+BW}}=\frac{2}{2+2+0}=\frac{2}{4}=\frac{1}{2}$ \beta_{A} \sim \text{Normal}(0, 1) & [\text{prior for }\beta_{A}] \\ As before, let’s begin by listing the information provided in the question: $\Pr(\mathrm{twins} | A) = 0.1$ Let’s convert each statement to an expression: Option 1 would be $$\Pr(\mathrm{rain} | \mathrm{Monday})$$. To use the previous birth information, we can update our priors of the probability of species A and B. The likelihood provides the plausibility of each possible value of the parameters, before accounting for the data. The best intro Bayesian Stats course is beginning its new iteration. 3.9 Statistical significance 134 3.10 Confidence intervals 137 3.11 Power and robustness 141 3.12 Degrees of freedom 142 3.13 Non-parametric analysis 143 4 Descriptive statistics 145 4.1 Counts and specific values 148 4.2 Measures of central tendency 150 4.3 Measures of spread 157 4.4 Measures of distribution shape 166 4.5 Statistical indices 170 Recall all the facts from the problem above. Like the other BB card, it has $$2$$ ways to produce the observed data. This means counting up the ways that each card could produce the observed data (a black card facing up on the table). Statistical Rethinking I just created a slack group for people who would like to do a slow read of McElreath's Statistical Rethinking. Option 2 would be the probability of rain, given that it is Monday. […], Here I work through the practice questions in Chapter 5, “Multivariate Linear Models,” of Statistical Rethinking (McElreath, 2016). This is much easier to interpret as the probability that it is Monday, given that it is raining. Learn how your comment data is processed. $\Pr(\mathrm{Earth} | \mathrm{land}) = \frac{0.15}{\Pr(\mathrm{land})}=\frac{0.15}{0.65}$. Since BB could produce this result from either side facing up, it has two ways to produce it ($$2$$). The Mars globe is 100% land. 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. The rules of probability tell us that the logical way to compute the plausibilities, after accounting for the data, is to use Bayes’ theorem. $\Pr(+|A) = 0.8$ Again suppose that a card is pulled and a black side appears face up. If anyone notices any errors (of which there will inevitably be some), I would be … Use the counting method, as before. Recall the globe tossing model from the chapter. The probability of the other side being black is indeed 2/3. $\Pr(\mathrm{land} | \mathrm{Mars}) = 1$ Option 3 is the probability of it being Monday, given rain. So there are three total ways to produce the current observation ($$2+1+0=3$$). It can be helpful to create a table: To get the final answer, we divide the number of ways to generate the observed data given the BB card by the total number of ways to generate the observed data (i.e., given any card): Let’s simulate an experiment. The target of inference in Bayesian inference is a posterior probability distribution. $\Pr(A | \mathrm{twins}) = \frac{\Pr(\mathrm{twins} | A) \Pr (A)}{\Pr(\mathrm{twins})} = \frac{0.1(0.5)}{0.15} = \frac{1}{3}$. The probability of rain, given that it is Monday. California Polytechnic State University, San Luis Obispo. Hugo. $\Pr(w,p)=\Pr(w|p)\Pr(p)$ The test says B, given that it is actually B is 0.65. This audience has had some calculus and linear algebra, and one or two joyless undergraduate courses in statistics. $\Pr(\mathrm{Earth}) = \Pr(\mathrm{Mars}) = 0.5$, Now, we need to use Bayes’ theorem (first formula on page 37) to get the answer: Statistical Rethinking is the only resource I have ever read that could successfully bring non-Bayesians of a lower mathematical maturity into the fold. Assume again the original card problem, with a single card showing a black side face up. So the posterior probability that this panda is species A is 0.36. Species B births twins 20% of the time, otherwise birthing singleton infants. This early draft is free to view and download for personal use only. Use the counting method (Section 2 of the chapter) to approach this problem. $\Pr(A) = 0.36$ What we see is that any process that adds together random values from the same distribution converges to a normal distribution. Show that the probability that the first card, the one showing a black side, has black on its other side is now 0.75. Required fields are marked *. The American Statistician has published 43 papers on "A World Beyond p < 0.05." New comments cannot be posted and votes cannot be cast. We can use the same formulas as before; we just need to update the numbers: $\Pr(\mathrm{BB})=\frac{\mathrm{BB}}{\mathrm{BB+BW+BW}}=\frac{2+2}{2+1+0+2}=\frac{4}{5}$ The probability that it is Monday and that it is raining. In each case, assume a uniform prior for $$p$$. From the Bayesian perspective, there is one true value of a parameter at any given time and thus there is no uncertainty and no probability in “objective reality.” It is only from the perspective of an observer with limited knowledge of this true value that uncertainty exists and that probability is a useful device. 40 comments. Now suppose all three cards are placed in a bag and shuffled. Note that this estimate is between the known rates for species A and B, but is much closer to that of species B to reflect the fact that having already given birth to twins increases the likelihood that she is species B. Show that the probability the other side is black is now 0.5. So the probability of the first card having black on the other side is indeed 0.75. Pretty much everything derives from the simple state- ment that entropy is maximized. $\Pr(\mathrm{twins}) = \Pr(\mathrm{twins} | A) \Pr(A) + \Pr(\mathrm{twins} | B) \Pr(B) = 0.1(0.5) + 0.2(0.5) = 0.15$, We can use the new information that the first birth was twins to update the probabilities that the female is species A or B (using Bayes’ theorem on page 37): I do my best to use only approaches and functions discussed so far in the book, as well as to name objects consistently with how the book does. Each method imposes different trade-offs. Again suppose a card is drawn from the bag and a black side appears face up. Code from Statistical Rethinking modified by R Pruim is shown below. 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. So the total ways for the first card to be BB is $$3+3=6$$. We can represent the three cards as BB, BW, and WW to indicate their sides as being black (B) or white (W). As a result, it’s less likely that a card with black sides is pulled from the bag. To begin, let’s list all the information provided by the question: $\Pr(\mathrm{land} | \mathrm{Earth}) = 1 – 0.7 = 0.3$ I do my best to use only approaches and functions discussed so far in the book, as well as to name objects consistently with how the book does. Assume that each globe was equally likely to be tossed. $\Pr(A | \mathrm{single}) = \frac{\Pr(\mathrm{single}|A)\Pr(A)}{\Pr(\mathrm{single})} = \frac{0.9(1/3)}{5/6} = 0.36$. Which of the expressions below correspond to the statement: the probability that it is Monday, given that it is raining? But the test, like all tests, is imperfect. 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'. Note the discreteness of the predictor groupsize and the invariance of the group-level variables within groups. The correct answers are Option 2 and Option 4 (they are equal). This dream team relied not on classical economic models of what people ought to do but on empirical studies of what people actually do under different conditions. Select the predictor variables you want in the linear model of the mean, For each predictor, make a parameter that will measure its association with the outcome, Multiply the parameter by the variable and add that term to the linear model. The $$\Pr(\mathrm{Monday})$$ in the numerator and denominator of the right-hand side cancel out: This is much easier to interpret as the probability that it is raining and that it is Monday. In this case, we can use the ifelse() function as detailed on page 40: Any parameter values less than 0.5 get their posterior probabilities reduced to zero through multiplication with a prior of zero. Statistical Rethinking chapter 5 notes. ―Andrew Gelman, Columbia University "This is an exceptional book. https://github.com/jffist/statistical-rethinking-solutions/blob/master/ch02_hw.R, $$\Pr(\mathrm{rain}, \mathrm{Monday}) / \Pr(\mathrm{Monday})$$. $\Pr(+) = \Pr(+ | A) \Pr(A) + \Pr(+ | B)\Pr(B) = 0.8(0.5) + 0.65(0.5) = 0.725$ I do […], Here I work through the practice questions in Chapter 2, “Small Worlds and Large Worlds,” of Statistical Rethinking (McElreath, 2016). Lecture 07 of the Dec 2018 through March 2019 edition of Statistical Rethinking: A Bayesian Course with R and Stan. Here I work through the practice questions in Chapter 2, “Small Worlds and Large Worlds,” of Statistical Rethinking (McElreath, 2016). $\Pr(A | +) = \frac{\Pr(+ | A) \Pr(A)}{\Pr(+)} = \frac{0.8(0.36)}{0.704} = 0.409$. Option 3 needs to be converted using the formula on page 36: So again assume that there are three cards: BB, BW, and WW. best. 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