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# statistical rethinking notes

We can do this using the third formula on page 37. P(test says A | A) / ( P(test says A | A) + P(test says A | B) ), Your email address will not be published. Of these three ways, only the ways produced by the BB card would allow the other side to also be black. 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). 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. $\Pr(\mathrm{rain}, \mathrm{Monday})/\Pr(\mathrm{rain})=\Pr(\mathrm{Monday}|\mathrm{rain})$ $\Pr(A | \mathrm{twins}) = \frac{\Pr(\mathrm{twins} | A) \Pr (A)}{\Pr(\mathrm{twins})} = \frac{0.1(0.5)}{0.15} = \frac{1}{3}$ $\Pr(\mathrm{twins} | B) = 0.2$ "Statistical Rethinking is a fun and inspiring look at the hows, whats, and whys of statistical modeling. So we can calculate this probability by dividing the number of ways given BB by the total number of ways: Like the other BB card, it has $$2$$ ways to produce the observed data. PREREQUISITES The reader is assumed to be familiar with basic classical estimation theory as it is presented in [1]. Again compute and plot the grid approximate posterior distribution for each of the sets of observations in the problem just above. $\Pr(A | \mathrm{twins}) = \frac{\Pr(\mathrm{twins} | A) \Pr (A)}{\Pr(\mathrm{twins})} = \frac{0.1(0.5)}{0.15} = \frac{1}{3}$. $\Pr(+|A) = 0.8$ Suppose there are two species of panda bear. 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. Use the counting method (Section 2 of the chapter) to approach this problem. Compute the posterior probability that this panda is species A. The second card has one black and one white side. Differences to the oringal include: a preference for putting data into containers (data frames, mostly), rather than working with lose vectors. Finally, there would be no ways for the first card to have been the second side of BW or either side of WW. Let’s convert each expression into a statement: Option 1 would be the probability that it is Monday, given that it is raining. Statistical Rethinking is an introduction to applied Bayesian data analysis, aimed at PhD students and researchers in the natural and social sciences. $\Pr(\mathrm{land} | \mathrm{Mars}) = 1$ $\Pr(\mathrm{twins}) = \Pr(\mathrm{twins} | A) \Pr(A) + \Pr(\mathrm{twins} | B) \Pr(B) = 0.1\bigg(\frac{1}{3}\bigg) + 0.2\bigg(\frac{2}{3}\bigg) = \frac{1}{6}$. Here I work through the practice questions in Chapter 2, “Small Worlds and Large Worlds,” of Statistical Rethinking (McElreath, 2016). Posted Mar 22, 2019 This early draft is free to view and download for personal use only. Option 5 is the same as the previous option but with the terms exchanged. Statistical physics is a beautiful subject. Statistical rethinking with brms, ggplot2, and the tidyverse This project is an attempt to re-express the code in McElreath’s textbook. $\Pr(B) = 0.5$ Thus P(+|B) = 1 – P(-|B) = 0.35. Now we can substitute this value into the formula from before to get our answer: Rebel Bayes Day 4. […], Here I work through the practice questions in Chapter 5, “Multivariate Linear Models,” of Statistical Rethinking (McElreath, 2016). So the total ways for the first card to be BB is $$3+3=6$$. BW could only produce this with its black side facing up ($$1$$), and WW cannot produce it in any way ($$0$$). Feb. 21, 2019. Otherwise they are the same as before. What does it mean to say “the probability of water is 0.7”? 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. Continuing on from the previous problem, suppose the same panda mother has a second birth and that it is not twins, but a singleton infant. $\Pr(B) = 0.5$, Next, let’s calculate the marginal probability of twins on the first birth (using the formula on page 37): The Earth globe is 70% covered in water. Which of the expressions below correspond to the statement: the probability of rain on Monday? Now suppose all three cards are placed in a bag and shuffled. $\Pr(\mathrm{single}) = \Pr(\mathrm{single}|A)\Pr(A) + \Pr(\mathrm{single}|B)\Pr(B) = 0.9(\frac{1}{3}) + 0.8(\frac{2}{3}) = \frac{5}{6}$ As before, let’s begin by listing the information provided in the question: $\Pr(\mathrm{twins} | A) = 0.1$ This equivalence can be derived using algebra and the joint probability definition on page 36: $\Pr(\mathrm{rain}|\mathrm{Monday})\Pr(\mathrm{Monday})/\Pr(\mathrm{rain})=\Pr(\mathrm{rain}, \mathrm{Monday})/\Pr(\mathrm{rain})$ For bonus, to do this in R, we can do the following: Now suppose there are four cards: BB, BW, WW, and another BB. As a result, it’s less likely that a card with black sides is pulled from the bag. I do […], Here I work through the practice questions in Chapter 4, “Linear Models,” of Statistical Rethinking (McElreath, 2016). To use the previous birth information, we can update our priors of the probability of species A and B. 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. 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. Academic theme for $\Pr(\mathrm{single}|B) = 1 – \Pr(\mathrm{twins}|B) = 1 – 0.2 = 0.8$ The target of inference in Bayesian inference is a posterior probability distribution. Here I work through the practice questions in Chapter 2, “Small Worlds and Large Worlds,” of Statistical Rethinking (McElreath, 2016). The UNDP Human Development Report 2020 explores how human activity, environmental change, and inequality are changing how we work, live and cooperate. The probability that it is Monday, given that it is raining. The $$\Pr(\mathrm{Monday})$$ in the numerator and denominator of the right-hand side cancel out: 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. Now suppose you are managing a captive panda breeding program. $\frac{\Pr(\mathrm{rain},\mathrm{Monday})}{\Pr(\mathrm{Monday})} = \Pr(\mathrm{rain}|\mathrm{Monday})$. 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. Option 4 is the probability of rain and it being Monday, given that it is Monday. \beta_{A} \sim \text{Normal}(0, 1) & [\text{prior for }\beta_{A}] \\ Reflecting the need for even minor programming in today's model-based statistics, the book pushes readers to perform step-by … Statistical inference is the subject of the second part of the book. So again assume that there are three cards: BB, BW, and WW. In each case, assume a uniform prior for $$p$$. Again suppose that a card is pulled and a black side appears face up. 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 probability it correctly identifies a species B panda is 0.65. Statistical Rethinking Chapter 5 Problems John Fox 2016-11-4. $\Pr(B) = 1 – \Pr(A) = 1 – 0.36 = 0.64$, Now we just need to do the same process again using the updated values. Show that the probability the other side is black is now 0.5. Chapman & Hall/CRC Press. If anyone notices any errors (of which there will inevitably be some), I … The test says A, given that it is actually A is 0.8. The probability of the other side being black is indeed 2/3. Statistical Rethinking is the only resource I have ever read that could successfully bring non-Bayesians of a lower mathematical maturity into the fold. 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). If anyone notices any errors (of which there will inevitably be some), I would be … 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 Your email address will not be published. […], Data Visualization Principles and Practice Tutorial on the principles and practice of data visualization, including an introduction to the layered […]. 40 comments. This is a rare and valuable book that combines readable explanations, computer code, and active learning." The American Statistician has published 43 papers on "A World Beyond p < 0.05." I do my […], Here I work through the practice questions in Chapter 3, “Sampling the Imaginary,” of Statistical Rethinking (McElreath, 2016). As described on pages 26-27, the likelihood for a card is the product of multiplying its ways and its prior: Now we can use the same formula as before, but using the likelihood instead of the raw counts. best. 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. Option 3 would be $$\Pr(\mathrm{Monday} | \mathrm{rain})$$. California Polytechnic State University, San Luis Obispo. One card has two black sides. The rules of probability tell us that the logical way to compute the plausibilities, after accounting for the data, is to use Bayes’ theorem. Imagine that black ink is heavy, and so cards with black sides are heavier than cards with white sides. Hugo. \end{array} His models are re-fit in brms, plots are redone with ggplot2, and the general data wrangling code predominantly follows the tidyverse style. Show that the posterior probability that the globe was the Earth, conditional on seeing “land” ($$\Pr(\mathrm{Earth}|\mathrm{land})$$), is 0.23. 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). What is the probability that her next birth will also be twins? Compute and plot the grid approximate posterior distribution for each of the following sets of observations. $\Pr(\mathrm{rain},\mathrm{Monday})=\Pr(\mathrm{rain}|\mathrm{Monday})\Pr(\mathrm{Monday})$, Now we divide each side by $$\Pr(p)$$ to isolate $$\Pr(\mathrm{rain}|\mathrm{Monday})$$: $\Pr(+) = \Pr(+ | A) \Pr(A) + \Pr(+ | B)\Pr(B) = 0.8(0.36) + 0.65(0.64) = 0.704$ Here is the chapter summary from page 45: This chapter introduced the conceptual mechanics of Bayesian data analysis. $\Pr(\mathrm{twins}) = \Pr(\mathrm{twins} | A) \Pr(A) + \Pr(\mathrm{twins} | B) \Pr(B) = 0.1(0.5) + 0.2(0.5) = 0.15$, 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): 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 The probability that the female is from species A, given that her first birth was twins, is 1/3 or 0.33. Here is a super-easy visual guide to setting up and running RStudio Server for Ubuntu 20 on Windows 10. 99% Upvoted. We can use the same formulas as before; we just need to update the numbers: $\Pr(\mathrm{BB})=\frac{\mathrm{BB}}{\mathrm{BB+BW+BW}}=\frac{2+2}{2+1+0+2}=\frac{4}{5}$ Option 4 is the same as the previous option but with division added: $\Pr(A | +) = \frac{\Pr(+ | A) \Pr(A)}{\Pr(+)} = \frac{0.8(0.5)}{0.725} = 0.552$. Not for re-distribution, re-sale or use in derivative works. 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. Observed that the second birth was twins, is imperfect setting up and RStudio... Only the ways produced by the BB card, it is actually a 0.36! That could successfully bring non-Bayesians of a list of numeric values looking at the other to. Like all tests, is 1/3 or 0.33 having black on the other to... 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