Hi amur Panda,, Is this for any arbitrary sequence when m happens to equal to 2^M, such that Aor B is true at every instance up to m; for example do you want to know whether there is somme function equation for any given n specific arbitrary n values;or rather given the probability of any such specific n value say 5 (so 16 z coming out true) what then is the probability given that for other values of n, and is then dependent, on both the n values; or rather is any depedence within any given sequence (specific n value) between probabilities for each zi, as approach whatever value happens to be z m. Maybe see if its exchangeabel for any given finite n first; given a probability assigned to each specific possible relative frequency not equal to one (ie whether possible ways of a getting some relative frequency for m=16, n=5, are equi-probable, it may not show independence but if it fails to be the case it may show that is not independent). Or otherwise are you considering the infinite limit (presumably the sequence would not really be a sequence then (there attempts at uncountable strong laws)). It also depends i suppose on what you mean by the probability, the relative frequency, or whether there is some necessary convergence so constrained by these equations, that this only the consistent solution; for example I presume you suspect that it isnt either 1, or zero (and if that is the case, then i need to know what you mean by probability of a sequence in order to help) etc. For example, if you could give a simple example for which you know the answer, it might be helpful for me to see if i can resolve this case, so that sort of figure out (by working backwards) what it is, precisely, what you mean by the probabiility of  a sequence

Yes the sequence is random and it is the combination of 0's and 1's. There are 'm' number of bits in this sequence where m = 2^n and n = 1,2,3 ..... Here, each bit zi in this sequence depend on the xi, yi, pi and qi events which are random integers whose value lies between 0 to (m-1). xi, yi, pi and qi are dependent on xi-1, yi-1, pi-1 and qi-1 .

It means. if the events xi, yi, pi and qi are occurred for m-iteration then z1,z2,.... zm will occur . Therefore, Z = {z1, z2, z3, .... zm}. In this case what is the P(Z=z1), P(PZ=z2) ... ?

and also what is the probability of m-compound event i.e. P(Z=z1z2z3..zm) ?

Here z1, z2, .. zm has only two value 0 and 1.

Obviously, the probability depends on the  parameter n, on  the eight values of  parameters  a  and  b  AND on the starting point (not mentioned in the problem) determined by four initial values  x_0, y_0, p_0  and  q_0. It is  reasonably to restrict them within the interval  [0,1,2, . . .,2^m-1). Thus, for given  n ,  we start with  2^{12 n} sequences. Obviously some of them are equivalent with respect to permutation. For instance if the obtained sequence possesses a period not greater than  2^n. Such sequences exist, which can be seen from the example with  all  a-s equal 1 and all b-s equal  0. Obviously, each sequence of  x-s, y-s, p-s  and  q-s  has a period of length being a divisor of  2^n i.e. of the form  2^k, where  k \in {0,. . . .,n-1}. Therefore, let me suggest to divite  the problem into  parts, each characterized by the four values of  k (for  x-s,  y-s, p-s  qnd  q-s, separately) .  Next steps require a careful analysis, how to collect into one subproblem resulting with the same cyclic order of  Z-s; I understand that the advice is rather taugh, but at a hand I have nothing more. 

Only one simplification is probably possible, if the parameters are chosen at random. Such a global randomness might help in solution, but it could be not what the author needs.

Regards

thanks Domsta for your response. a1, a2, a3, a4, b1, b2, b3 and b4 are randomly chosen constant parameters. Whereas, x0, yo, p0 and q0 are randomly chosen initial seeds. From the equation point of view it looks dependent but when I analyzed numerically fon n=2, the probability value of 1's remain constant for each position of Z . Similarly, the probability value of 0's are also remain constant for different position of Z. It means P(zi=1) = p1 and P(zi = 0) = p0.

I have considered  all the dependency cases during calculation of P(zi = 1) and P(zi = 0).

Thanks, Amit, for your presentation. But, believe me, the variability of the probabilities will be evident for higher values of n, take -if possible-  n=5. Then dependence on  the constant parameters should be substantial. Interesting if this is so!

Best regards, Joachim

for n = 2 there are more than 128 possible cases. If I go for n = 5 then it will have more than 2^63 possible cases. But still I am trying to write a code in Matlab for n=8. Let see what result is coming. I am sure probability will remain constant. If it remains constant for all position then the sequence will behave like an independent.

Hi Amit,,

I have calculated the following probabilities for  n=2:

P{ x_i = given byte) = 1/4, for any given byte from the range {0,1,2,3}

P{X_1 = X_3 = 0} = 3/16 (not equal   1/16).

The reason is that the bi-iterates possess attractor or a fixed point, at  least one. Therefore, if  X_1 enters this basin, then  X_3  is there, too,  which makes equal values strongly correlated.

Unfortunately, I have not enough much time for elaborating the question of your post, but, hopefully these remarks will help you in organizing the program more efficient. Perhaps a theorem of general type is possible?

Best wishes sends Joachim

Thank you for giving time.

Yo have calculated P(X=xi) where xi = {0,1,2,3}. In my case xi is random and the initial value is also a random. Even it depends on a1, b1 which are randomly chosen constant.

Yes if you fix the initial seed and a1,b1  then you can find the probability of xi (P(X=xi)). However it is not fixed. even though probability of xi in X increases then also it is not useful to find the P(Z) bcz you have to find the P(B xor C) to find P(Z). B and C are independent events. If you find P(Z) by considering all the dependency then only you can understand. It is not easy where all variables are unknown.

Let me correct your impression: In my calculations the randomness is caused by randomness of  three terms: a1,b1,x0  (so, I considered all 64 elementary events). As you know, the result for  n-2 is

P{ X_i = given byte) = 1/4, for any given byte from the range {0,1,2,3}

P{X_1 = X_3 = 0} = 3/16 (not equal   1/16)

It  proves that  X_1  and X_3  are not independent. For reasons I have explained in former post,  I  predict the same phenomenon for all other  n. In particular  the events 

{ X_{2^n -3}= x }   and  { X_{2^n -1}=x }  (with the same  x )

are expected to be positively correlated for  all  x \in {0,1,2,..., 2^n - 1}. This would be  a strong indication, that the random variables  Z_i  are also correlated. Of course it is not a proof, so there is a lot to do yet .  Obviously, it  would be nice if a provable theorem appeared as a result of your investigation.

Regards

Hello Sir, I am getting your point and for the case of individual random process X, Y, P and Q, it may true but for the sequence Z which depends on various random processes, will have different approach.

I have calculated the probability, P(Zi) for n =2, please check this. 

For n = 2, m = 2n = 4 and therefore, the size of every individual sequence such as X, Y, P, Q, B C and Z is 4. The value of Xi, Yi , Pi and Qi will remain in the range 0 to 3.  Similarly, the value of Bi , Ci and Zi will remain in the range 0 to 1.

X = {x1, x2, x3, x4}, Y = {y1, y2, y3, y4}, P = {p1, p2, p3, p4} and Q = {q1, q2, q3, q4}

where, xi, yi, pi, qi  ∈ {0, 1, 2, 3}

B = {b1, b2, b3, b4 }, C = {c1, c2, c3, c4}

Z = {z1, z2, z3, z4}       where bi, ci, zi ∈ {0, 1}

Now to get Z1=1, X1 must be less than equal to Y1 (X1 Q1) or X1 must be greater than Y1 (X1 > Y1) along with (P1 Y1), P(P1 > Q1), P(X1 Q1) = 6/16.

P(X1 Q1) + P(X1 Q) as white ball and (X

Hi, at the first glance the first error is (probably) made at the lines

" Similarly, we can find the P(Z2= 1|Z1=z1) and P(Z2= 0|Z1=z1)

P(Z2 = 1|Z1 =z1) = P(Z2 = 1|Z1 = 0) + P(Z2 = 1|Z1 = 1)"

which lead you to stationarity, i.e. the independence of the instantenous distribution on say time  t.  Instead, ,what you need is to compare 

P{Z2=1| Z1=0}  to  P{Z2=1|Z1=1}

If they are not equal, then the sequence of  Z-s does not consist of independent random variables. At least according to my undersatnding of this notion.  Right now I am too busy to verify this to be absolutely sure, Thus please forgive me if  I've get wrong. 

There is also a possibility, that you understand independence of random variables in different way, don't you?

Best regards

sir, in this case so many probability values of Z will come for different 0-1 combinations. But how to represent the probability value of Z in simple mathematical form? Also in my case the 'zi'  in the sequence Z are unknown, If it is unknown then how can I find the probability of m-compound events ?

My objective is to find the P(Z) = P(z1z2z3z4...zm).

for n=2, m= 2^n = 4. I have to find the P(Z) = P(z1z2z3z4). where zi ∈ {0,1}

In this case there are 16 possible outcomes and therefore, 16 probability values will come. But it is also true for randomly chosen initial seed x0, y0, p0, q0 and constant parameters a1, b1, a2, b2, a3, b3, a4, b4 there is only one 4-bit size sequence as outcomes which is a combination of 0-1 out of 16 possible combinations. If we change the initial seeds and other constant parameters then we may get another sequence or same sequence as outcomes. Hence, if we go for all the trials then total of 4^12 = 16 777 216 trials will occur and out of which maximum 16 possible sequences Z will come.

Therefore what is the P(Z) = P(z1z2z3z4) ? and how can we represent with a mathematical form  ?

In other way, what is the probability to predict the sequence Z of size 'm' when initial seeds and constant parameters are unknown ?

Yes, there are 16 probabbilities; i f written correctly, they could look like this:

P{(Z1,Z2,Z3,Z4)= (0,0,0,0)} =   . . .  (I would denote it by  p(0,0,0,0)

P{(Z1,Z2,Z3,Z4)= (0,0,0,1)} = . . . (I would denote it by p(0,0,0,1)

. . .  etc.

And this  is the peoblem , that they are not calculable by independence, which would simply give the folloing wrong numbers like

(15/32)^4,  (15/32)^3 17/32, etc.

The general theorem is probably unattainable, and therefore it  mekes sense of simulation.

Good luck anyway,

Joachim

Thank you sir.

I tried to find the probability P(z1z2z3) = P(001)

I find it in two approaches. In first approach I just used the probability of 1 and 0 in the sequence "001".

P(001) = P(0) * P(0) * P(1) = 17/32 * 17/32 * 15*32 = 0.13229

In second approach, I considered all the cases for getting z1=0, z2=0, z3=1 and then got 6 different probability values for the sequence "001"

P(Z=001) = P(Z1=0) * P(Z2=0|Z1=0) * P(Z3=1|Z1Z2 = 00)

P(Z=001) = 136/256 * 125/225 * 96/196 = 0.14456

P(Z=001) = 136/256 * 125/225 * 90/196 = 0.135523

P(Z=001) = 136/256 * 125/225 * 80/196 = 0.120465

P(Z=001) = 136/256 * 117/225 * 96/196 = 0.13531

P(Z=001) = 136/256 * 117/225 * 90/196 = 0.12685

P(Z=001) = 136/256 * 117/225 * 80/196 = 0.112755

It means, for one sequence "001", it gives 6 different probabilities which are varies from 0.112755 to 0.14456. Therefore, I am not sure which probability value should consider.

These values are approximately equal to the the probability value "0.13229" calculated by independent way.

The reason of coming different probability values is that

P(Z1=0)= 136/256

P(Z2=0|Z1=0) = 125/225 for B1=1, C1=1

P(Z2=0|Z1=0) = 125/225 for B1=0, C1=0

similarly, I calculated for P(Z3=1|Z1Z2 = 00).

P(Z3=1|Z1Z2 = 00) = 96/196 for B1=0, C1=0 and B2=0, C2=0

P(Z3=1|Z1Z2 = 00) = 90/196 for B1=0, C1=0 and B2=1, C2=1

P(Z3=1|Z1Z2 = 00) = 90/196 for B1=1, C1=1 and B2=0, C2=0

P(Z3=1|Z1Z2 = 00) = 80/196 for B1=1, C1=1 and B2=1, C2=1

So for different cases of Bi and CI, the probability values of "001" varies and therefore, it is difficult to choose the exact probability value of one sequence. But if I take the average of these probabilities then the probability value is 0.1292438 which is nearly equal to "0.13229" as calculated by independence case. Similarly, for other sequences we can get similar kind of different probability values which are approximate to the independence case. 

In case of m-compound event where m = 32 bit, I think it is impossible to find the 2(2^32) possible probabilities. For such case how to get the P(Z) ?

So is it possible if I assume that the sequence Z is independent and further calculate P(Z) to represent in the simple mathematical form ?

Well it does not appear to be 'exchangeable' in the sense that the same frequency 001, has six slightly different probabilty values; so they appear not to be even conditionally independent (given the ci bi values); although there be might be some partial exchangeability results out there that cover this.

I presume however that one would not simply want to presume independence, for as m gets bigger I would presume that these disparities and even the average will diverge more from the value calculated, as if it were independent; Are you able to derive that, as m increases, any given z sequence, will diverge/converge from the value calculated as if it were 'independent, and or that the average will diverge/converge from said value?. 

Are the distinct parameter values,  B1=0, C1=0 and B2=1, C2=1 independent of each other or equal in probability value. Which is what I think you presume; ; maybe some distinct weighted average will give you "0.13229? Or maybe there may be a further dependency or difference in the probability values ci, bi etc,  as m goes larger, that will give you the  'as if independence' result of the average.

Although this may not work, as it appears that all three of the last conditional probabilities are smaller

@ Amin Kumar

The following approach is definitely wrong:

P(001) = P(0) * P(0) * P(1) = 17/32 * 17/32 * 15*32 = 0.13229

In the second aproach starting with the correct formula

P(Z=001) = P(Z1=0) * P(Z2=0|Z1=0) * P(Z3=1|Z1Z2 = 00)

its application as follows

P(Z=001) = 136/256 * 125/225 * 96/196 = 0.14456

P(Z=001) = 136/256 * 125/225 * 90/196 = 0.135523 etc.

is also wrong, since the conditional probability P(Z3=1|Z1Z2 = 00) possesses a  well determined value, which cannot be  96/196  and simultaneously  90/196. I do not evaluate correctness of any of the values, I am just stating their inconsistency.

An exception is that I am sure, that you have obtained  correctly P(Zi=0) = 17/32  = 136/256 :)

After I get rid with some works I have to do soon, you will receive further comments.

Best regargs

Hello William ...

I have calculated  for all the sequences and these are probability ranges for each each sequence

P(001) varies in between 0.112755 to 0.14456 (it is 0.13229 for independence case)

P(010) varies in between 0.12287 to 0.143112 (it is 0.13229 for independence case)

P(100) varies in between 0.13010 to 0.140306 (it is 0.13229 for independence case)

P(110) varies in between 0.11607 to 0.118303 (it is 0.116729 for independence case)

P(101) varies in between 0.10804 to 0.119898 (it is 0.116729 for independence case)

P(011) varies in between 0.113237 to 0.12489 (it is 0.116729 for independence case)

P(000) varies in between 0.14904 to 0.17467 (it is 0.149933 for independence case)

P(111) varies in between 0.09598 to 0.102678 (it is 0.10299 for independence case)

if you closely observe then you can find that these probability values are converges to the probability value as computed independently.