hi, One more query on the same topic.
Now add a second bit. assume that the bits are (i) and (ii) so we know that the probability of (i) being 1 is 0.5-e and and being 0 is 0.5+e (there isn't a bias btw in that notation - e could be negative)
so all the possible combinations are
P(i=1, ii=1) =(0.5-e)(0.5-e) P(i=1, ii=0) =(0.5-e)(0.5+e) P(i=0, ii=1) =(0.5+e)(0.5-e) P(i=0, ii=0) =(0.5+e)(0.5+e)
Two events E1 and E2 are said to be independent,if P(E1(intersection)E2)=P(E1).P(E2) As in the above case we assume that the bits under xor are independent. if the inputs are frm different sources then we consider them independent. what about the following case. i open any two text files in binary mode and xor 1 st bit of file 1 with file 2,2 nd bit with 2 nd bit with second bit and so on. The rate of english varies between 1.0 bit /letter & 1.5 bit/letter for large values of N. absolute rate of english is R=log(26)base 2=4.7 bits/letter There is lot of redundancy in the language,0.6 to 0.85 percent redundancy. Since in ASCII we use 8 bits to represent english alphabets,the redundancy is 8-1.3=6.7 bits/charecter of redundancy. Since in both files opened,english charecters are represented in the same set of ASCII charecters. there is redundancy in both the files. Does that mean that such bits we xor are not independent? Regards Sarath. --- David Howe <DaveHowe@gmx.co.uk> wrote:
at Thursday, October 17, 2002 4:38 PM, Sarad AV <jtrjtrjtr2001@yahoo.com> was seen to say:
He wanted to know how I was able to do XOR on P(0) and P(1) when xor is defined only on binary digits. you don't.
P(x) is a probability of digit x in the output. ideally, P(0)=P(1)=0.5 (obviously in binary, only 0 and 1 are defined, so they are the only two possible outcomes. Now assume that one output (1 say) is more probable than the other. If this is true, you can define some value of probability (e) that is the amount a given outcome is more or less probable than the ideal. Now add a second bit. assume that the bits are (i) and (ii) so we know that the probability of (i) being 1 is 0.5-e and and being 0 is 0.5+e (there isn't a bias btw in that notation - e could be negative)
so all the possible combinations are
P(i=1, ii=1) =(0.5-e)(0.5-e) P(i=1, ii=0) =(0.5-e)(0.5+e) P(i=0, ii=1) =(0.5+e)(0.5-e) P(i=0, ii=0) =(0.5+e)(0.5+e)
but of course if you XOR (i) and (ii) together, then (i=1, ii=1) = 0 (i=1, ii=0) = 1 (i=0, ii=1) = 1 (i=0, ii=0) = 0
collecting identical outputs allows you to say
P(0)=P(i=1, ii=1)+P(i=0, ii=0) = (0.5-e)(0.5-e)+(0.5+e)(0.5+e) P(1) P(i=1, ii=0) + P(i=0, ii=1) = (0.5-e)(0.5+e)+(0.5+e)(0.5-e)
reducing P(0) as in the example you gave gives you the probability of P(0) being 0.5+(2*(e^2))
so the answer is - you don't ever apply XOR to anything but binary - you do straight algebraic math on the *probabilities* of a given output (0 or 1)
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