Independence and redundncy Re: XORing bits to eliminate skew

[Sarad AV]

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 at gmx.co.uk> wrote:
> at Thursday, October 17, 2002 4:38 PM, Sarad AV
> <jtrjtrjtr2001 at 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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