Imagine that black ink is heavy, and so cards with black sides are heavier than cards with white sides. $\frac{\Pr(\mathrm{rain},\mathrm{Monday})}{\Pr(\mathrm{Monday})} = \Pr(\mathrm{rain}|\mathrm{Monday})$. They look exactly alike and eat the same food, and there is yet no genetic assay capable of telling them apart. Your email address will not be published. Reflecting the need for even minor programming in today's model-based statistics, the book pushes readers to perform step-by … 2 Lecture 07 of the Dec 2018 through March 2019 edition of Statistical Rethinking: A Bayesian Course with R and Stan. $\Pr(A | \mathrm{single}) = \frac{\Pr(\mathrm{single}|A)\Pr(A)}{\Pr(\mathrm{single})} = \frac{0.9(1/3)}{5/6} = 0.36$. Here I work through the practice questions in Chapter 2, “Small Worlds and Large Worlds,” of Statistical Rethinking (McElreath, 2016). We can use the same approach to update the probability again. Lectures. This thread is archived. $\frac{\Pr(\mathrm{rain},\mathrm{Monday})}{\Pr(\mathrm{Monday})} = \frac{\Pr(\mathrm{rain}|\mathrm{Monday})\Pr(\mathrm{Monday})}{\Pr(\mathrm{Monday})}$ Option 3 needs to be converted using the formula on page 36: The rst chapter is a short introduction to statistics and probability. $\Pr(\mathrm{Earth} | \mathrm{land}) = \frac{0.15}{\Pr(\mathrm{land})}=\frac{0.15}{0.65}$. Pretty much everything derives from the simple state- ment that entropy is maximized. Someone reaches into the bag and pulls out a card and places it flat on a table. So the total ways for the first card to be BB is $$3+3=6$$. Not for re-distribution, re-sale or use in derivative works. The probability that it is Monday and that it is raining. \sigma \sim \text{Uniform}(0,10) & [\text{prior for }\sigma] $\Pr(A) = 0.5$ Let’s update the table and include new columns for the prior and the likelihood. You have a new female panda of unknown species, and she has just given birth to twins. 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 the posterior probability of species A (using just the test result) is 0.552. Now we can solve this like we have been solving the other questions: The third card has two white sides. Powered by the The probability of the other side being black is now 4/5. \], $$\mu_{i} = \alpha + \beta_{R}R_{i} + \beta_{A}A_{i}$$. First ignore your previous information from the births and compute the posterior probability that your panda is species A. Before looking at the other side, we draw another card from the bag and lay it face up on the table. Species B births twins 20% of the time, otherwise birthing singleton infants. 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. Suppose you have a deck with only three cards. Suppose there are two globes, one for Earth and one for Mars. These functions are used in the Pluto notebooks projects specifically intended for hands-on use while studying the book or taking the course. They differ however in family sizes. Option 4 would be $$\Pr(\mathrm{Monday}, \mathrm{rain})$$. 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'. Again suppose that a card is pulled and a black side appears face up. Option 3 would be $$\Pr(\mathrm{Monday} | \mathrm{rain})$$. 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). $\Pr(A) = 0.5$ In practice, Bayesian models are fit to data using numerical techniques, like grid approximation, quadratic approximation, and Markov chain Monte Carlo. Statistical Rethinking: A Bayesian Course with Examples in R and Stan Book Description Statistical Rethinking: A Bayesian Course with Examples in R and Stan read ebook Online PDF EPUB KINDLE,Statistical Rethinking: A Bayesian Course with Examples in R and Stan pdf,Statistical Rethinking: A Bayesian Course with Examples in R and Stan read online,Statistical Rethinking: A … Again calculate the probability that the other side is black. The second card has one black and one white side. We just have to calculate the updated marginal probability of twins. Note the discreteness of the predictor groupsize and the invariance of the group-level variables within groups. Note that this probability increased from 0.33 to 0.36 when it was observed that the second birth was not twins. Now compute the probability that the panda we have is from species A, assuming we have observed only the first birth and that it was twins. lecture note include statistical signal processing, digital communications, information theory, and modern con-trol theory. Predictor residual plots. \beta_{A} \sim \text{Normal}(0, 1) & [\text{prior for }\beta_{A}] \\ More mechanically, a Bayesian model is a composite of a likelihood, a choice of parameters, and a prior. To use the previous birth information, we can update our priors of the probability of species A and B. $\Pr(\mathrm{Earth}) = \Pr(\mathrm{Mars}) = 0.5$, Now, we need to use Bayes’ theorem (first formula on page 37) to get the answer: $\Pr(\mathrm{twins}) = \Pr(\mathrm{twins} | A) \Pr(A) + \Pr(\mathrm{twins} | B) \Pr(B) = 0.1(0.5) + 0.2(0.5) = 0.15$, 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): https://github.com/jffist/statistical-rethinking-solutions/blob/master/ch02_hw.R, $$\Pr(\mathrm{rain}, \mathrm{Monday}) / \Pr(\mathrm{Monday})$$. Option 3 is the probability of it being Monday, given rain. So we can calculate this probability by dividing the number of ways given BB by the total number of ways: We can now use algebra and the joint probability formula (page 36) to simplify this: Using the test information only, we go back to the idea that the species are equally likely. This […], This is a tutorial on calculating row-wise means using the dplyr package in R, To show off how R can help you explore interesting and even fun questions using data that is freely available […], Here I work through the practice questions in Chapter 7, “Interactions,” of Statistical Rethinking (McElreath, 2016). 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. Which of the expressions below correspond to the statement: the probability that it is Monday, given that it is raining? The probability it correctly identifies a species B panda is 0.65. The American Statistician has published 43 papers on "A World Beyond p < 0.05." $\Pr(\mathrm{rain}|\mathrm{Monday})\Pr(\mathrm{Monday}) = \Pr(\mathrm{rain}, \mathrm{Monday})$ In order for the other side of the first card to be black, the first card would have had to be BB. Rather, it is named after Stanislaw Ulam (1909–1984). Otherwise they are the same as before. Now we can substitute this value into the formula from before to get our answer: $$\Pr(\mathrm{rain}|\mathrm{Monday})\Pr(\mathrm{Monday})$$, $$\Pr(\mathrm{rain}|\mathrm{Monday})\Pr(\mathrm{Monday})/\Pr(\mathrm{rain})$$, $$\Pr(\mathrm{Monday}|\mathrm{rain})\Pr(\mathrm{rain})/\Pr(\mathrm{Monday})$$. Stu- Statistical Rethinking Chapter 5 Problems John Fox 2016-11-4. share. Use the counting method (Section 2 of the chapter) to approach this problem. The probability of rain, given that it is Monday. So it can be interpreted (repeating all the previous work) as the probability of rain, given 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. $\Pr(A | \mathrm{twins}) = \frac{\Pr(\mathrm{twins} | A) \Pr (A)}{\Pr(\mathrm{twins})} = \frac{0.1(0.5)}{0.15} = \frac{1}{3}$ Required fields are marked *. Again suppose a card is drawn from the bag and a black side appears face up. The probability of the other side being black is indeed 2/3. 40 comments. This reflects the idea that singleton births are more likely in species A than in species B. I do my […], Here I work through the practice questions in Chapter 3, “Sampling the Imaginary,” of Statistical Rethinking (McElreath, 2016). \end{array} Richard McElreath (2016) Statistical Rethinking: A Bayesian Course with Examples in R and Stan. We can use the same formulas as before; we just need to update the numbers: $\Pr(\mathrm{BB})=\frac{\mathrm{BB}}{\mathrm{BB+BW+BW}}=\frac{2+2}{2+1+0+2}=\frac{4}{5}$ "Statistical Rethinking is a fun and inspiring look at the hows, whats, and whys of statistical modeling. \alpha \sim \text{Normal}(10, 10) & [\text{prior for }\alpha] \\ \mu_{i} = \alpha + \beta_{R}R_{i} + \beta_{A}A_{i} & \text{[linear model]}\\ 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. Let’s suppose 1,000 people flip a coin 16 times. Differences to the oringal include: a preference for putting data into containers (data frames, mostly), rather than working with lose vectors. The $$\Pr(\mathrm{Monday})$$ in the numerator and denominator of the right-hand side cancel out: $So the final answer is 0.2307692, which indeed rounds to 0.23. To begin, let’s list all the information provided by the question: \[\Pr(\mathrm{land} | \mathrm{Earth}) = 1 – 0.7 = 0.3$ Here I work through the practice questions in Chapter 2, “Small Worlds and Large Worlds,” of Statistical Rethinking (McElreath, 2016). Statistical Rethinking 2019 Lectures Beginning Anew! In each case, assume a uniform prior for $$p$$. Use the counting method, if you can. $\Pr(+|B) = 0.65$ Option 2 is the probability of rain, given that it is Monday. Partial least squares structural equation modeling (PLS-SEM) is an important statistical technique in the toolbox of methods that researchers in marketing and other social sciences disciplines frequently use in their empirical analyses. Why things are normal. Both are equally common in the wild and live in the same place. What does it mean to say “the probability of water is 0.7”? Thus P(+|B) = 1 – P(-|B) = 0.35. This is much easier to interpret as the probability that it is raining and that it is Monday. Lecture 02 of the Dec 2018 through March 2019 edition of Statistical Rethinking: A Bayesian Course with R and Stan. The probability that it is Monday, given that it is raining. $\Pr(B) = \frac{2}{3}$ Here is the chapter summary from page 45: This chapter introduced the conceptual mechanics of Bayesian data analysis. $\Pr(w,p)=\Pr(w|p)\Pr(p)$ New comments cannot be posted and votes cannot be cast. $\Pr(\mathrm{single}) = \Pr(\mathrm{single}|A)\Pr(A) + \Pr(\mathrm{single}|B)\Pr(B) = 0.9(\frac{1}{3}) + 0.8(\frac{2}{3}) = \frac{5}{6}$ $\Pr(A | +) = \frac{\Pr(+ | A) \Pr(A)}{\Pr(+)} = \frac{0.8(0.5)}{0.725} = 0.552$. That the data are grouped makes the assumption of independence among observations suspect. The probability that the female is from species A, given that her first birth was twins, is 1/3 or 0.33. $\Pr(\mathrm{rain}, \mathrm{Monday})/\Pr(\mathrm{rain})=\Pr(\mathrm{Monday}|\mathrm{rain})$ These plausibilities are updated in light of observations, a process known as Bayesian updating. This audience has had some calculus and linear algebra, and one or two joyless undergraduate courses in statistics. $\Pr(\mathrm{rain}|\mathrm{Monday})\Pr(\mathrm{Monday})/\Pr(\mathrm{rain})=\Pr(\mathrm{rain}, \mathrm{Monday})/\Pr(\mathrm{rain})$ $\Pr(B) = 0.5$, Next, let’s calculate the marginal probability of twins on the first birth (using the formula on page 37): Let’s simulate an experiment. Last updated on May 12, 2020 22 min read Notes, R, Statistical Rethinking. This early draft is free to view and download for personal use only. Let’s update our table to include the new card. 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 suppose now that a veterinarian comes along who has a new genetic test that she claims can identify the species of our mother panda. \begin{array}{lr} P(test says A | A) / ( P(test says A | A) + P(test says A | B) ), Your email address will not be published. Finally, there would be no ways for the first card to have been the second side of BW or either side of WW. So there are three total ways to produce the current observation ($$2+1+0=3$$). Learn how your comment data is processed. If anyone notices any errors (of which there will inevitably be some), I would be … What we see is that any process that adds together random values from the same distribution converges to a normal distribution. Academic theme for Afte we already know age at marriage, what additional value is there in also knowing marriage rate? The correct answers are thus Option 1 and Option 4. After we already know marriage rate, what additional value is there in also knowing age at marriage? If anyone notices any errors (of which there will inevitably be some), I … $\Pr(B | \mathrm{twins}) = \frac{\Pr(\mathrm{twins} | B) \Pr (B)}{\Pr(\mathrm{twins})} = \frac{0.2(0.5)}{0.15} = \frac{2}{3}$, 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. 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. […], Here I work through the practice questions in Chapter 5, “Multivariate Linear Models,” of Statistical Rethinking (McElreath, 2016). This is a rare and valuable book that combines readable explanations, computer code, and active learning." The face that is shown on the new card is white. $\Pr(\mathrm{land} | \mathrm{Mars}) = 1$ His models are re-fit in brms, plots are redone with ggplot2, and the general data wrangling code predominantly follows the tidyverse style. For bonus, to do this in R, we can do the following: Now suppose there are four cards: BB, BW, WW, and another BB. The Mars globe is 100% land. $\Pr(\mathrm{twins} | B) = 0.2$ 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. Species A gives birth to twins 10% of the time, otherwise birthing a single infant. 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. The test says A, given that it is actually A is 0.8. ... Side note … This means counting up the ways that each card could produce the observed data (a black card facing up on the table). So the posterior probability that this panda is species A is 0.36. Sort by. Again compute and plot the grid approximate posterior distribution for each of the sets of observations in the problem just above. 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. $\Pr(\mathrm{BB})=\frac{\mathrm{BB}}{\mathrm{BB+BW+WW}}=\frac{6}{6+2+0}=\frac{6}{8}=0.75$. Rethinking P-Values: Is "Statistical Significance" Useless? P (test says A | A) = 0.8. Here is a super-easy visual guide to setting up and running RStudio Server for Ubuntu 20 on Windows 10. The best intro Bayesian Stats course is beginning its new iteration. As a result, it’s less likely that a card with black sides is pulled from the bag. Statistical Rethinking: Chapter 3. Code from Statistical Rethinking modified by R Pruim is shown below. c Rui M. Castro and Robert D. Nowak, 2017. I do […], Here I work through the practice questions in Chapter 2, “Small Worlds and Large Worlds,” of Statistical Rethinking (McElreath, 2016). 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). Using the approach detailed on page 40, we use the dbinom() function and provide it with arguments corresponding to the number of $W$s and the number of tosses (in this case 3 and 3): We recreate this but update the arguments to 3 $W$s and 4 tosses. D_{i} \sim \text{Normal}(\mu_{i}, \sigma) & \text{[likelihood]} \\ Let’s convert each expression into a statement: Option 1 would be the probability that it is Monday, given that it is raining. So the probability that the female will give birth to twins, given that she has already given birth to twins is 1/6 or 0.17. 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: Although it will be easier to see if we rename $$w$$ to $$\mathrm{rain}$$ and $$p$$ to $$\mathrm{Monday}$$: Statistical physics is a beautiful subject. 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. 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. Use the counting method, as before. PREREQUISITES The reader is assumed to be familiar with basic classical estimation theory as it is presented in [1]. 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. A black side is shown facing up, but you don’t know the color of the side facing down. Here, we describe the meaning of entropy, and show how the tenet of maximum entropy is related to time-reversal via the ergodic theorem. Show that the probability that the other side is also black is 2/3. Compute the posterior probability that this panda is species A. $\Pr(\mathrm{single}|A) = 1 – \Pr(\mathrm{twins}|A) = 1 – 0.1 = 0.9$ Chapman & Hall/CRC Press. Statistical Rethinking is an introduction to applied Bayesian data analysis, aimed at PhD students and researchers in the natural and social sciences. $\Pr(+|B) = 0.65$ $\Pr(+) = \Pr(+ | A) \Pr(A) + \Pr(+ | B)\Pr(B) = 0.8(0.36) + 0.65(0.64) = 0.704$ $\Pr(A | +) = \frac{\Pr(+ | A) \Pr(A)}{\Pr(+)} = \frac{0.8(0.36)}{0.704} = 0.409$. Further suppose that one of these globes–you don’t know which–was tossed in the air and produces a “land” observation. I do my best […], Here I work through the practice questions in Chapter 6, “Overfitting, Regularization, and Information Criteria,” of Statistical Rethinking (McElreath, 2016). Prior beliefs about Bayesian statistics, updated by reading Statistical Rethinking by Richard McElreath. Posterior probabilities state the relative numbers of ways each conjectured cause of the data could have produced the data. These relative numbers indicate plausibilities of the different conjectures. 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 Earth globe is 70% covered in water. Option 2 would be the probability of rain, given that it is Monday. Software. $\Pr(+|A) = 0.8$ The test says B, given that it is actually B is 0.65. $\Pr(\mathrm{single}|B) = 1 – \Pr(\mathrm{twins}|B) = 1 – 0.2 = 0.8$ California Polytechnic State University, San Luis Obispo. Hugo. Option 4 is the same as the previous option but with division added: 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. 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). Now suppose all three cards are placed in a bag and shuffled. 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. Which of the expressions below correspond to the statement: the probability of rain on Monday? So the probability of the first card having black on the other side is indeed 0.75. This results in the posterior 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. The rst part of the book deals with descriptive statistics and provides prob-ability concepts that are required for the interpretation of statistical inference. You will actually get to practice Bayesian statistics while learning about it and the book is incredibly easy to follow. Discuss the globe tossing example from the chapter, in light of this statement. I agree – see https://github.com/jffist/statistical-rethinking-solutions/blob/master/ch02_hw.R. Show that the posterior probability that the globe was the Earth, conditional on seeing “land” ($$\Pr(\mathrm{Earth}|\mathrm{land})$$), is 0.23. One card has two black sides. 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. What is the probability that her next birth will also be twins? Since BB could produce this result from either side facing up, it has two ways to produce it ($$2$$). As before, let’s begin by listing the information provided in the question: $\Pr(\mathrm{twins} | A) = 0.1$ Suppose there are two species of panda bear. The target of inference in Bayesian inference is a posterior probability distribution. NOTE: Descriptive statistics summarize data to make sense or meaning of a list of numeric values. 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. 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’. Option 2 would be $$\Pr(\mathrm{rain} | \mathrm{Monday})$$. So again assume that there are three cards: BB, BW, and WW. 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): $\Pr(B) = 1 – \Pr(A) = 1 – 0.36 = 0.64$, Now we just need to do the same process again using the updated values. If the first card was the second side of BB, then there would be the same 3 ways for the second card to show white. BW could only produce this with its black side facing up ($$1$$), and WW cannot produce it in any way ($$0$$). The bag and lay it face up on the other side, we draw another card from chapter. 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