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16-17 (1.2 probability theory)

2022-01-26 23:29:34 CNpupil

   Now suppose we choose a box at random , The result is a blue box . then , The probability of choosing an apple is the score of the apple in the blue box , namely 3 / 4 3/4 , therefore p ( F = a B = b ) = 3 / 4 p(F=a|B=b)=3/4 ​. in fact , We can write all four conditional probabilities for the type of fruit , Given the selected box

p ( F = a B = r ) = 1 / 4 (1.16) p(F=a|B=r)=1/4\tag{1.16}
p ( F = o B = r ) = 3 / 4 (1.17) p(F=o|B=r)=3/4\tag{1.17}
p ( F = a B = b ) = 3 / 4 (1.18) p(F=a|B=b)=3/4\tag{1.18}
p ( F = o B = b ) = 1 / 4 (1.19) p(F=o|B=b)=1/4\tag{1.19}

Again, pay attention to , These probabilities are standardized , therefore

p ( F = a B = r ) + p ( F = o B = r ) = 1 (1.20) p(F=a|B=r)+p(F=o|B=r)=1\tag{1.20}

alike

p ( F = a B = b ) + p ( F = o B = b ) = 1 (1.21) p(F=a|B=b)+p(F=o|B=b)=1\tag{1.21}

   We can now use the sum product rule of probability to evaluate the overall probability of choosing Apple

p ( F = a ) = p ( F = a B = r ) p ( B = r ) + p ( F = a B = b ) p ( B = b ) = 1 4 × 4 10 + 3 4 × 6 10 = 11 20 (1.22) p(F=a)=p(F=a|B=r)p(B=r)+p(F=a|B=b)p(B=b)=\frac{1}{4}\times\frac{4}{10}+\frac{3}{4}\times\frac{6}{10}=\frac{11}{20}\tag{1.22}

thus , Use summation rules , p ( F = o ) = 1 11 / 20 = 9 / 20 p(F=o)=1-11/20=9/20 .

   contrary , Suppose we were told to choose a piece of fruit , It's an orange , We want to know which box it comes from . This requires us to evaluate the probability distribution on the box conditional on fruit identity , and (1.16)(1.19) The probability in gives Probability distribution on fruit conditional on box identity . Using Bayes Theorem , We can solve the inverse problem of conditional probability

p ( B = r F = o ) = p ( F = o B = r ) p ( B = r ) p ( F = o ) = 3 4 × 4 10 × 20 9 = 2 3 (1.23) p(B=r|F=o)=\frac{p(F=o|B=r)p(B=r)}{p(F=o)}=\frac{3}{4}\times\frac{4}{10}\times\frac{20}{9}=\frac{2}{3}\tag{1.23}

According to the summation rule , p ( B = b f = o ) = 1 2 / 3 = 1 / 3 p(B=b|f=o)=1-2/3=1/3 .

   We can make the following important explanations for Bayesian Theorem . If we are asked which box we chose before being told the identity of the selected fruit , Then the most complete information we can get is probability p ( B ) p(B) ​. We call it a priori probability , Because it is the probability available before we observe the characteristics of fruit . Once we are told that the fruit is an orange , We can use Bayesian theorem to calculate probability p ( B F ) p(B|F) , We call it a posteriori probability , Because it's when we observe F F Then the probability . Please note that , In this case , The prior probability of choosing the red box is 4 / 10 4/10 , So we are more likely to choose the blue box than the red box . However , Once we observe the selected fruit, the orange , We find that the posterior probability of the red box is now 2 / 3 2/3 , So the box we are more likely to choose now is actually red . This result is in line with our intuition , Because the proportion of oranges in the red box is much higher than that in the blue box , Therefore, the observed fruits when oranges provide important evidence to support the red box . in fact , The evidence is strong enough , Beyond previous evidence , Makes it more likely to choose a red box instead of a blue box .

   Last , We noticed that , If the joint distribution of two variables is decomposed into the product of edges , bring p ( X , Y ) = p ( x ) p ( Y ) p(X,Y)=p(x)p(Y) , that X X and Y Y Called independent . From the product rule , We see p ( Y X ) = p ( Y ) p(Y|X)=p(Y) , So given X X Of Y Y The conditional distribution of is indeed consistent with X X The value of has nothing to do with . for example , In our fruit box example , If each box contains the same proportion of apples and oranges , that p ( F B ) = p ( F ) p(F|B)=p(F) , So choose ( such as ) The probability of an apple has nothing to do with which box to choose .

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