Testwiki:Reference desk/Archives/Mathematics/2006 November 28

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November 28

Bayes Theorem Equation

Looking at the equation for Bayes' Theorem.

$\mathrm{Pr}(A|B)=\frac{\mathrm{Pr}(B|A)\mathrm{Pr}(A)}{\mathrm{Pr}(B|A)\mathrm{Pr}(A)+\mathrm{Pr}(B|A^{\text{'}})\mathrm{Pr}(A^{\text{'}})}$

Assume that $\mathrm{Pr}(A^{\text{'}})=1-\mathrm{Pr}(A)$

$\mathrm{Pr}(A|B)=\frac{\mathrm{Pr}(B|A)\mathrm{Pr}(A)}{\mathrm{Pr}(B|A)\mathrm{Pr}(A)+\mathrm{Pr}(B|A^{\text{'}})-\mathrm{Pr}(B|A^{\text{'}})\mathrm{Pr}(A)}$

$\downarrow$

$\mathrm{Pr}(A|B)=\frac{\mathrm{Pr}(B|A)\mathrm{Pr}(A)}{(\mathrm{Pr}(B|A)-\mathrm{Pr}(B|A^{\text{'}}))\mathrm{Pr}(A)+\mathrm{Pr}(B|A^{\text{'}})}$

$\downarrow$

$\frac{\mathrm{Pr}(A|B)}{\mathrm{Pr}(B|A)\mathrm{Pr}(A)}=\frac{1}{(\mathrm{Pr}(B|A)-\mathrm{Pr}(B|A^{\text{'}}))\mathrm{Pr}(A)+\mathrm{Pr}(B|A^{\text{'}})}$

$\downarrow$

$\frac{\mathrm{Pr}(B|A)\mathrm{Pr}(A)}{\mathrm{Pr}(A|B)}=(\mathrm{Pr}(B|A)-\mathrm{Pr}(B|A^{\text{'}}))\mathrm{Pr}(A)+\mathrm{Pr}(B|A^{\text{'}})$

$\downarrow$

$\mathrm{Pr}(A)[\frac{\mathrm{Pr}(B|A)}{\mathrm{Pr}(A|B)}-(\mathrm{Pr}(B|A)-\mathrm{Pr}(B|A^{\text{'}}))]=\mathrm{Pr}(B|A^{\text{'}})$

$\downarrow$

$\mathrm{Pr}(A)=\frac{\mathrm{Pr}(B|A^{\text{'}})}{\frac{\mathrm{Pr}(B|A)}{\mathrm{Pr}(A|B)}-(\mathrm{Pr}(B|A)-\mathrm{Pr}(B|A^{\text{'}}))}$

Now assume that a positive integer number X (between 1 and 1 million) is picked at random.

let $A$ be "X is divisible by 2"

and

let $B$ be "X is divisible by 4"

We have

$\mathrm{Pr}(B|A^{\text{'}})=0$

Thus

$\mathrm{Pr}(A)=\frac{0}{\frac{\mathrm{Pr}(B|A)}{\mathrm{Pr}(A|B)}-(\mathrm{Pr}(B|A)-\mathrm{Pr}(B|A^{\text{'}}))}$

$\downarrow$

$\mathrm{Pr}(A)=0$

Therefore if we pick a positive integer between 1 and 1 million at random, the number we pick will not be an even number.

This is of course WRONG! But I can't see where the mistake is.

You can have fun with this:

let $A$ be "George Bush is an American"

and

let $B$ be "George Bush is the president of the United States"

202.168.50.40 22:59, 28 November 2006 (UTC)

Hint. At some point, rather late in your derivation, you take a step of the form ab = c → a = c/b. Then you conclude that c = 0 implies a = 0. But if you substitute c := 0 in the equation ab = c, it becomes ab = 0. You cannot conclude from there to a = 0.  --Lambiam^Talk 23:32, 28 November 2006 (UTC)