人工智能08不確定性-資料下載頁(yè)
2025-02-19 12:32本頁(yè)面
【正文】 ? 此時(shí)此刻給你個(gè)機(jī)會(huì)換一個(gè)信封 . 是換呢還是換呢還是換呢 ? 使用貝葉斯法則 : II E: envelope, 1=(R,B), 2=(B,B) B: the event of drawing a black ball P(E|B) = P(B|E)*P(E) / P(B) We want to pare P(E=1|B) vs. P(E=2|B) P(B|E=1) = , P(B|E=2) = 1 P(E=1) = P(E=2) = P(B) = P(B|E=1)P(E=1) + P(B|E=2)P(E=2) = (.5)(.5) + (1)(.5) = .75 P(E=1|B) = P(B|E=1)P(E=1)/P(B) = (.5)(.5)/(.75) = 1/3 P(E=2|B) = P(B|E=2)P(E=2)/P(B) = (1)(.5)/(.75) = 2/3 因此在已發(fā)現(xiàn)一個(gè)黑球后 , 該信封是 1的后驗(yàn)概率 (thus worth $100)比信封是 2的后驗(yàn)概率低 所以還是換吧 課堂測(cè)驗(yàn) ? 一名醫(yī)生做了一個(gè)具有 99% 可靠性的測(cè)試：也就是說(shuō) , 99% 的病人其檢測(cè)呈陽(yáng)性 , 99% 的健康人士檢測(cè)呈陰性 . 該醫(yī)生估計(jì) 1%的人類病了。 ? Question: 一個(gè)患者檢測(cè)呈陽(yáng)性 . 該患者得病的幾率是多少 ? ? 025%, 2575%, 7595%, or 95100%? 課堂測(cè)驗(yàn) ? A doctor performs a test that has 99% reliability, ., 99% of people who are sick test positive, and 99% of people who are healthy test negative. The doctor estimates that 1% of the population is sick. ? Question: A patient tests positive. What is the chance that the patient is sick? ? 025%, 2575%, 7595%, or 95100%? ? Intuitive answer: 99%。 Correct answer: 50% Bayes’ rule with 多重證據(jù)和條件獨(dú)立性 P(Cavity | toothache ∧ catch) = αP(toothache ∧ catch | Cavity) P(Cavity) = αP(toothache | Cavity) P(catch | Cavity) P(Cavity) This is an example of a na239。ve Bayes model（樸素貝葉斯模型） : Total number of parameters（參數(shù)） is linear in n 鏈?zhǔn)椒▌t ? 全聯(lián)合分布 using 鏈?zhǔn)椒▌t : P(Toothache, Catch, Cavity) = P(Toothache | Catch, Cavity) P(Catch, Cavity) = P(Toothache | Catch, Cavity) P(Catch | Cavity) P(Cavity) = P(Toothache | Cavity) P(Catch | Cavity) P(Cavity) 圖模型 : ? Each variable is a node ? The parents of a node are the other variables which the deposed joint conditions on ? MUCH more on this to e! 概率分布從哪來(lái) ? ? Idea One: 人類 , 領(lǐng)域?qū)＜? . what’s P(raining | cold)? ? Idea Two: 簡(jiǎn)單事實(shí)和一些代數(shù)學(xué) ? 使用鏈?zhǔn)椒▌t和獨(dú)立性來(lái)試著計(jì)算聯(lián)合分布 概率分布從哪來(lái) ? ? Idea Three: 從數(shù)據(jù)里學(xué)習(xí)出來(lái)的 ! A good chunk of machine learning research is essentially about various ways of learning various forms of them! Estimation估計(jì) ? 怎樣去估計(jì)一個(gè)隨機(jī)變量 X的分布 ? ? Maximum likelihood （最大似然） : 從現(xiàn)實(shí)世界中收集觀察值 For each value x, look at the empirical rate of that value: This estimate is the one which maximizes the likelihood of the data Estimation估計(jì) ? Problems with 最大似然估計(jì) : 如果我拋一次硬幣 , 是正面 heads, 那么對(duì) P(heads)的估計(jì)是多少 ? What if I flip it 50 times with 27 heads? What if I flip 10M times with 8M heads? ? Basic idea: 我們對(duì)一些參數(shù)有先驗(yàn)期望值 (here, the probability of heads) 在缺少證據(jù)時(shí)，我們傾向先驗(yàn)值 若給定很多證據(jù) , 則應(yīng)該以數(shù)據(jù)為準(zhǔn) Summary Probability is a rigorous formalism for uncertain knowledge 概率是對(duì)不確定知識(shí)一種嚴(yán)密的形式化方法 Joint probability distribution specifies probability of every atomic event 全聯(lián)合概率分布 指定了對(duì)隨機(jī)變量的每種完全賦值，即每個(gè) 原子事件 的概率 Queries can be answered by summing over atomic events 可以通過(guò)把對(duì)應(yīng)于查詢命題的原子事件的條目相加的方式來(lái)回答查詢 For nontrivial domains, we must find a way to reduce the joint size Independence and conditional independence provide the tools 作業(yè) ? ， ， ， (不交 ) 演講完畢，謝謝觀看！
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