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#1 2015-05-09 18:13:41

subait
Guest

independence

let assume that we have two register A and S with 0s and 1s only. Both of the two register give output at the same time (ai,si) .  The output is ai if the output of the si is ‘1’ else the output of ai is discarded. Is the A and S register are independence and prove it is they are independence.

#2 2015-05-09 18:55:57

Agnishom
Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,974
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Re: independence

Can you explain this algorithm with an example?


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#3 2015-05-09 20:11:18

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: independence

hi subait

Welcome to the forum.

I do not understand this at all. 

subait wrote:

let assume that we have two register A and S with 0s and 1s only. Both of the two register give output at the same time (ai,si) .  The output is ai if the output of the si is ‘1’ else the output of ai is discarded. Is the A and S register are independence and prove it is they are independence.

You say the "two registers give output at the same time", but then say "The output is ai ...."

It makes more sense if it reads "Both of the two register have input at the same time (ai,si)

Then (with my choice of entries) we have this:

1EmgN6g.gif

Next, what does "discarded" mean.  You cannot have empty binary digits.  They have to be 1s or 0s.  Did you mean 0s for these outputs?

In that case OUTPUT = A AND S.

And what is independence in this context?  I have never met this word used in computer science.  I have searched the internet and can find no reference to it.  It is implicit in logic theory that the inputs are regarded as independent.  It is assumed rather than proved. 

So I cannot understand what this question is about at all.  I am so sorry sad  .  Would you be able to post the problem exactly as worded when you met it?

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#4 2015-05-09 23:09:33

subait
Member
Registered: 2015-05-09
Posts: 1

Re: independence

Hi
thank you very much and sorry to write a confused question
  I mean independent event probability
  For example I have two register A and S each of them have a random input of 1 or 0 it is exactly as Moderator way of thinking
  let A =[ 0 , 1 , 0, 1 ] and S = [1 , 0 , 0, 1] when S = 1 then I can get output from A so the output will be as follow [0,1]
     
I don't care about the output what I need to prove is that A and S are independent event of probability

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#5 2015-05-10 00:23:30

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: independence

Thanks for clearing that up.

If A = [w,x,y,z] where w,x,y,z ∈ {0,1} then there are 16 possible elements in the space for A ( [0,0,0,0] through to [1,1,1,1]

Similarly for S.

The choice is RANDOM, so the probability of any particular event is 1/16 in each register and 1/16 x 1/16 = 1/256 for each output.

Some outputs may occur from more than one starting point and maybe that is where this question is going.

The definition of independent events is P(A and B) = P(A).P(B)        ( sorry, I cannot find a quick way to make an 'intersection' symbol )

So what does P(A and B) means exactly? 

It should mean the probability that A and B both occur in which case:

eg.  A = [0,1,0,1] and S = [0,1,0,1] gives output = [1,1] and so does A = [1,1,1,1] and S = [0,1,0,1] but that does not effect the independence of the inputs; it just makes  output = [1,1] more likely than eg. output = [1,1,1,1] which can only happen in one way. 

But if P(A and B) means the probability of a particular output, then the above shows that P(A and B) ≠ P(A).P(B)

So, in your position, I'd go back to my teacher and ask what he means by P(A and B).  smile

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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