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text: '[summary: The odds form of Bayes' rule is a form of [1lz Bayes' rule] that uses [1rq relative likelihoods] and [-1rb]. It says that in general, the [1rm prior] odds times the [56t likelihood ratio] equals the [1rp posterior] odds.\n\n[22s Suppose] that the odds of a particular patient being sick are about $(1 : 4),$ i.e., they may be drawn from a population where there is 1 sick patient for every 4 healthy ones. Suppose we administer a test which comes back positive 90% of the time if the patient is _actually_ sick, but also comes back positive 30% of the time even when the patient is healthy; this means relative likelihoods of $(90 : 30) = (3 : 1)$: the test is 3x as likely to come back positive if the patient is sick. Bayes' rule then says that the odds of the patient being sick _after_ observing a positive test are $(1 : 4) \\times (3 : 1) = (3 : 4),$ for a final probability of about $3/(3+4) \\approx 43\\%.$]\n\n%%!if-after([1x1]):\n**This introduction is meant to be read after the introductions to [55z frequency visualizations] and [1x1 waterfall visualizations].**\n%%\n\nIn general, Bayes' rule states:\n\n$$ \\textbf{Prior odds} \\times \\textbf{Relative likelihoods} = \\textbf{Posterior odds}$$\n\nIf we consider the [1wy waterfall visualization] of [22s the Diseasitis example], \nthen we can visualize how *relative odds* are appropriate for thinking about the two rivers at the top of the waterfall. \n\n\n\nThe *proportion* of red vs. blue water at the bottom will be the same whether there's 200 vs. 800 gallons per second of red vs. blue water at the top of the waterfall, or 20,000 vs. 80,000 gallons/sec, or 1 vs. 4 gallons/second. So long as the rest of the waterfall behaves in a proportional way, we'll get the same proportion of red vs blue at the bottom. Thus, we're justified in ignoring the _amount_ of water and considering only the relative proportion between amounts.\n\nSimilarly, what matters is the *relative* proportion between how much of each gallon of red water makes it into the shared pool, and how much of each gallon of blue water, makes it. 45% and 15% of the red and blue water making it to the bottom would give the same *relative proportion* of red and blue water in the bottom pool as 90% and 30%.\n\n\n\nThis justifies throwing away the specific data that 90% of the red stream and 30% of the blue stream make it down, and summarizing this into *relative likelihoods* of (3 : 1).\n\nMore generally, suppose we have a medical test that detects a sickness with a 90% true positive rate (10% false negatives) and a 30% false positive rate (70% true negatives). A positive result on this test represents the same *strength of evidence* as a test with 60% true positives and 20% false positives. A negative result on this test represents the same *strength of evidence* as a test with 9% false negatives and 63% true negatives.\n\nIn general, the strength of evidence is summarized by how *relatively* likely different states of the world make our observations. %%!if-before([1zh]): For more on this idea, see [22x]. %% %%if-before([1zh]): More on this later. %%\n\n# The equation\n\nTo state Bayes' rule in full generality, and prove it as a theorem, we'll need to introduce some new notation.\n\n## Conditional probability\n\nFirst, when $X$ is a proposition, $\\mathbb P(X)$ will stand for the [1rf probability] of $X.$\n\nIn other words, $X$ is something that's either true or false in reality, but we're uncertain about it, and $\\mathbb P(X)$ is a way of expressing our [4y9 degree of belief] that $X$ is true. A patient is, in fact, either sick or healthy; but if you don't know which of these is the case, the evidence might lead you to assign a 43% subjective probability that the patient is sick.\n\n$\\mathbb \\neg X$ will mean "$X$ is false", so $\\mathbb P(\\neg X)$ is the "the probability $X$ is false".\n\nThe Diseasitis involved some more complicated statements than this, though; in particular it involved:\n\n- The 90% chance that a patient blackens the tongue depressor, *given* that they have Diseasitis.\n- The 30% chance that a patient blackens the tongue depressor, *given* that they're healthy.\n- The 3/7 chance that a patient has Diseasitis, *given* that they blackened the tongue depressor.\n\nIn these cases we want to go from some fact that is *assumed* or *known* to be true (on the right), to some other proposition (on the left) whose new probability we want to ask about, taking into account that assumption.\n\nProbability statements like those are known as "conditional probabilities". The standard notation for conditional probability expresses the above quantities as:\n\n- $\\mathbb P(blackened \\mid sick) = 0.9$\n- $\\mathbb P(blackened \\mid \\neg sick) = 0.3$\n- $\\mathbb P(sick \\mid blackened) = 3/7$\n\nThis standard notation for $\\mathbb P(X \\mid Y)$ meaning "the probability of $X$, assuming $Y$ to be true" is a helpfully symmetrical vertical line, to avoid giving you any visual clue to remember that the assumption is on the right and the inferred proposition is on the left. </sarcasm>\n\nConditional probability is *defined* as follows. Using the notation $X \\wedge Y$ to denote "X and Y" or "both $X$ and $Y$ are true":\n\n$$\\mathbb P(X \\mid Y) := \\frac{\\mathbb P(X \\wedge Y)}{\\mathbb P(Y)}$$\n\nE.g. in the Diseasitis example, $\\mathbb P(sick \\mid blackened)$ is calculated by dividing the 18% students who are sick *and* have blackened tongue depressors ($\\mathbb P(sick \\wedge blackened)$), by the total 42% students who have blackened tongue depressors ($\\mathbb P(blackened)$).\n\nOr $\\mathbb P(blackened \\mid \\neg sick),$ the probability of blackening the tongue depressor *given* that you're healthy, is equivalent to the 24 students who are healthy *and* have blackened tongue depressors, divided by the 80 students who are healthy. 24 / 80 = 3/10, so this corresponds to the 30% false positives we were told about at the start.\n\nWe can see the law of conditional probability as saying, "Let us restrict our attention to worlds where $Y$ is the case, or thingies of which $Y$ is true. Looking only at cases where $Y$ is true, how many cases are there *inside* that restriction where $X$ is *also* true - cases with $X$ *and* $Y$?"\n\nFor more on this, see [1rj].\n\n## Bayes' rule\n\nBayes' rule says:\n\n$$\\textbf{Prior odds} \\times \\textbf{Relative likelihoods} = \\textbf{Posterior odds}$$\n\nIn the Diseasitis example, this would state:\n\n$$\\dfrac{\\mathbb P({sick})}{\\mathbb P(healthy)} \\times \\dfrac{\\mathbb P({blackened}\\mid {sick})}{\\mathbb P({blackened}\\mid healthy)} = \\dfrac{\\mathbb P({sick}\\mid {blackened})}{\\mathbb P(healthy\\mid {blackened})}.$$\n\n%todo: apparently the parallel is not super obvious, and maybe we can use slightly different colors in the text to make it clearer that e.g. Prior odds -> sick/healthy%\n\nThe [1rm prior] odds refer to the relative proportion of sick vs healthy patients, which is $1 : 4$. Converting these odds into probabilities gives us $\\mathbb P(sick)=\\frac{1}{4+1}=\\frac{1}{5}=20\\%$.\n\nThe [-1rq] refers to how much more likely each sick patient is to get a positive test result than each healthy patient, which (using [1rj conditional probability notation]) is $\\frac{\\mathbb P(positive \\mid sick)}{\\mathbb P(positive \\mid healthy)}=\\frac{0.90}{0.30},$ aka relative likelihoods of $3 : 1.$\n\nThe [1rp posterior] odds are the relative proportions of sick vs healthy patients among those with positive test results, or $\\frac{\\mathbb P(sick \\mid positive)}{\\mathbb P(healthy \\mid positive)} = \\frac{3}{4}$, aka $3 : 4$ odds.\n\nTo extract the *probability* from the relative odds, we keep in mind that probabilities of [1rd mutually exclusive and exhaustive propositions] need to sum to $1,$ that is, there is a 100% probability of *something* happening. Since everyone is either sick or not sick, we can [1rk normalize] the odd ratio $3 : 4$ by dividing through by the sum of terms:\n\n$$(\\frac{3}{3+4} : \\frac{4}{3+4}) = (\\frac{3}{7} : \\frac{4}{7}) \\approx (0.43 : 0.57)$$\n\n...ending up with the probabilities (0.43 : 0.57), proportional to the original ratio of (3 : 4), but summing to 1. It would be very odd if something had probability $3$ (300% probability) of happening.\n\nUsing the waterfall visualization:\n\n\n\nWe can generalize this to _any_ two **hypotheses** $H_j$ and $H_k$ with **evidence** $e$, in which case Bayes' rule can be written as:\n\n$$\\dfrac{\\mathbb P(H_j)}{\\mathbb P(H_k)} \\times \\dfrac{\\mathbb P(e \\mid H_j)}{\\mathbb P(e \\mid H_k)} = \\dfrac{\\mathbb P(H_j \\mid e)}{\\mathbb P(H_k \\mid e)}$$\n\nwhich says "the posterior odds ratio for hypotheses $H_j$ vs $H_k$ (after seeing the evidence $e$) are equal to the prior odds ratio times the ratio of how well $H_j$ predicted the evidence compared to $H_k.$"\n\nIf $H_j$ and $H_k$ are [-1rd], we can convert the posterior odds into a posterior probability for $H_j$ by [1rk normalizing] the odds - dividing through the odds ratio by the sum of its terms, so that the elements of the new ratio sum to $1.$\n\n## Proof of Bayes' rule\n\nRearranging [1rj the definition of conditional probability], $\\mathbb P(X \\wedge Y) = \\mathbb P(Y) \\cdot \\mathbb P(X|Y).$ E.g. to find "the fraction of all patients that are sick *and* get a positive result", we multiply "the fraction of patients that are sick" times "the probability that a sick patient blackens the tongue depressor".\n\nThen this is a proof of Bayes' rule:\n\n$$\n\\frac{\\mathbb P(H_j)}{\\mathbb P(H_k)}\n\\cdot\n\\frac{\\mathbb P(e_0 | H_j)}{\\mathbb P(e_0 | H_k)}\n=\n\\frac{\\mathbb P(e_0 \\wedge H_j)}{\\mathbb P(e_0 \\wedge H_k)}\n= \n\\frac{\\mathbb P(H_j \\wedge e_0)/\\mathbb P(e_0)}{\\mathbb P(H_k \\wedge e_0)/\\mathbb P(e_0)}\n= \n\\frac{\\mathbb P(H_j | e_0)}{\\mathbb P(H_k | e_0)}\n$$\n\nQED.\n\nIn the Diseasitis example, these proof steps correspond to the operations:\n\n$$\n\\frac{0.20}{0.80}\n\\cdot\n\\frac{0.90}{0.30}\n=\n\\frac{0.18}{0.24}\n= \n\\frac{0.18/0.42}{0.24/0.42}\n= \n\\frac{0.43}{0.57}\n$$\n\nUsing red for sick, blue for healthy, grey for a mix of sick and healthy patients, and + signs for positive test results, the calculation steps can be visualized as follows:\n\n\n\n%todo: maybe replace this diagram with pie-chart circles in exactly right proportions (but still with the correct populations of + signs)%\n\nThis process of observing evidence and using its likelihood ratio to transform a prior belief into a posterior belief is called a "[1ly Bayesian update]" or "belief revision."\n\n%%if-before([21v]):\nCongratulations! You now know (we hope) what Bayes' rule is, and how to apply it to simple setups. After this, the path continues with further implications %if-before([1zg]): and additional forms% of Bayes' rule. This might be a good time to take a break, if you want one--but we hope you continue on this Arbital path after that!\n%%\n\n%%!if-before([1zg]):\n- For the generalization of the odds form of Bayes' rule to multiple hypotheses and multiple items of evidence, see [1zg].\n- For a transformation of the odds form that makes the strength of evidence even more directly visible, see [1zh].\n%%',
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Summary: 'The odds form of Bayes' rule is a form of [1lz Bayes' rule] that uses [1rq relative likelihoods] and [-1rb]. It says that in general, the [1rm prior] odds times the [56t likelihood ratio] equals the [1rp posterior] odds.\n\n[22s Suppose] that the odds of a particular patient being sick are about $(1 : 4),$ i.e., they may be drawn from a population where there is 1 sick patient for every 4 healthy ones. Suppose we administer a test which comes back positive 90% of the time if the patient is _actually_ sick, but also comes back positive 30% of the time even when the patient is healthy; this means relative likelihoods of $(90 : 30) = (3 : 1)$: the test is 3x as likely to come back positive if the patient is sick. Bayes' rule then says that the odds of the patient being sick _after_ observing a positive test are $(1 : 4) \\times (3 : 1) = (3 : 4),$ for a final probability of about $3/(3+4) \\approx 43\\%.$'
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