To what extent is the product of two random number sequences (0 mean and sd of 1)also random?
139457
The approximate variance of the product of two random numbers (Var_1x2) is a function of their means (Mean_1,Mean_2), their variances (Var_1,Var_2) and Covariance (Cov_12) :\n\nVar_1x2=\nVar_1*(Mean_2)^2 + \nVar_2*(Mean_1)^2 +\n2*Mean_1*Mean_2*Cov_12
773699
You have said that they aren't independent; however, you have no mention of how they depend on each other. Therefore there is no way to calculate the answer.\nIf they *were* independent, then E[X*Y] = E[X]*E[Y]. Each one is an exponential distribution with lambda = 1, so the mean is just 1 (could also integrate e^(-x) from 0 to infinity, you also get 1.)
948825
First answerer correct.  \n\nBut the second set is a function, because there are no repeated x values.  If you have a component of the first set corresponding to two different components of the second, then it isn't a function, e.g. in the second set, if the next pair were (15, 5) then it wouldn't be a function because of the double result for 15.  In the first set, if your misprinted pair were (1,2) then it wouldn't be a function, same reason.
543381
Yes. the RHS of the first is equivalent to xor(P,Q). This becomes an inclusive or when the or(Q,...) is added. Not(P) as suggested above doesn't give the correct result when Q and P both = true.
173519
I guess you are talking about the statistical significance, of the difference of two random variables (or functions of variables).\nThat would be the case, for example if you want to test whethern two means are or not equal.\nThen, the difference of the two means can be tested to be or not zero. For that, you need to know the distribution of the original variables.\nIf the two samples come from the same normal population and are of the same size, then the difference is normal with mean zero and variance = 2*sigma^2/n where n is the sample size.\nThen, if the value (In absolute terms) is greater than X(p) then they are significantly different with probability p.\nX(p) is such P(X>X(p))=1-p.
880673
hint: mean value theorem
822095
a) Both: .1(.02) = .002\n\nb) Exactly one:  .02(1-.1) = .018 + .1(1-.02) = .018 + .098 = .116\n\nc) (1-.02)(1-.1) = (.98)(.9) = .882\n\nThey add up to one.  I'm not a probability person, but I think this is correct.
1079282
If by a(i) and b(i) you mean "a times i" and "b times i", then:\n\nFor i = 0:\n\na*i / (b*i + 1) = 0 / 1 = 0\n\nSo you have:\n\nPo = 1 / (1 + 0 + (0 * (a / (b + 1))) + (0 * (a / (b + 1)) * (2a / (2b + 1))) + ...)\n\nPo = 1 / 1\n\nPo = 1\n\nIf by a(i) and b(i) you mean a and b are infinite sequences (then presumably a / b means "for all i, a(i) / b(1) < 1"), well then that is something else entirely.\n\nThis reminds me of the problem to expand and simplify the product of twenty-six binomial terms:\n\n(x - a) (x - b) (x - c) ... (x - z)
319924
It depends on the distribution.  Some probability distributions have multiple parameters (such as the mean and standard deviation for the Normal distribution).  Phi2 would be one parameter of the particular distribution used in the problem.
1156399
This is a good question even for a statistician.\nThe problem is your set is small and I do not know what the deviation between samples is, however the repeatability and reproducibility information can bind the window \nthrough which the third value will fall through or perhaps even remain in.  \n\nLet me know.
403633
If by infinite randomness you mean true randomness, then I don't think it can. All random generators are programmes that follow an algorithm to produce 'random' results. Whilst for most purposes the result is sufficiently random to be useful, they are not as random as, for example, drawing lottery numbers.
892374
Hi:\n\nWhat the rand() function does is load a number into a resister and through electronic processes generate almost random number list.  However those are not truly a random number like what someone is going to say on the radio, the actual out come of a coin flip. or what specific atom in a large number of atoms in a  pile of radioactive material will split or not. to learn more about this study Chaos Theory.  We call this a pseudo random number meaning because of the electronic process, the numbers value can be repeat, where as true  random number can not. to  know more about this  check computer programming and random numbers generation  in the Wikipedia website they can explain it better then I can here \n\nHope this helps
359986
Student's t-test can be used to determines the probability that 2 random variables have the same mean. This can then be used to test whether two sets of data come from the same source.
258367
.001% Very small probability.
755601
26. The pattern has to start 1, 2, 4, 8, 15. I see the differences are 1, 2, 4, 7, and the differences of the differences are 1, 2, 3. The next difference of the difference has to be 4, so, working my way back up, I figure the next difference has to be 11, making the next number 26.\n\nAs for why this makes sense: Consider the question in two dimensions, i.e. into how many parts will many random lines diivide a plane?  That pattern starts 1, 2, 4, 7, 11 - the same numbers on the differences line above.  If you want to get really abstruse, I would submit that the number parts random spaces divide the fourth dimension is 1, 2, 4, 8, 16, 31, 57.  :)\n\n* * * * * *\n\nHey, you really do ask an interesting question, but it appears I'm the only one willing to answer it.  (Yes, I see you asked the same question a few days ago.)  OK, then, let's look at it more systematically:\n\n\nOne-dimensional question: Into how many parts will random points divide a line?\n\nEasy question, obviously:  Zero points leave one part (i.e. the whole line); one point breaks it into two parts, two points into three parts, etc.  The sequence is 1, 2, 3, 4, 5, etc.  Generally, n random points divide a line into n + 1 parts. \n\n\nTwo-dimensional question: Into how many parts will random lines divide a plane?\n\nHarder question, but you can work this one out on paper.  Zero lines leave one part (the whole plane); one line breaks it in two; two lines breaks it in four.  Then it gets tricky - the third line can't go thru all four parts, but it can go through three of them, so you end up with 7 parts.  The fourth line can go through 4 of the 7 parts.  Now, how you'd prove this rigorously I don't know, but the sequence is 1, 2, 4, 7, 11, 16, etc. and is clearly derived from the one-dimensional sequence: start with 1, then add 1 to get 2, then add 2 to get 4, then add 3 to get 7, then add 4 to get 11, etc.  To put this into a formula, n random lines divide a plane into (n^2 + n + 2)/2 parts.        \n\n\nNow for the three-dimensional question: Into how many parts will random planes divide a three-dimensional space?\n\nTo visualize this one, look at a corner of your room.  Obviously, zero planes leave one part (the whole space), one plane  breaks it in two (visualize the floor of your room as the plane, for example), two planes breaks it in four (visualize the floor plus one wall), and three planes breaks it in eight (visualize the floor plus two adjoining walls).  Now, try to imagine the fourth plane - it can't go through all eight parts, but it can go through seven of them, making a total of 15 parts.  The three-dimensional  sequence starts 1, 2, 4, 8, 15, and again we see it is derived from the sequence for the previous dimension: start with 1, then add 1 to get 2, then add 2 to get 4, then add 4 to get 8, then add 7 to get 15, etc.  To get the number for 5 planes, add 11 to get 26.  To put this into a formula, n random planes divide space into (n^3 + 5n + 6)/6 parts.   \n\nBased on what we are seeing, what the heck - why not carry it forward to the fourth dimension.  Using the numbers from the three-dimensional sequence, the sequence becomes 1, 2, 4, 8, 16, 31, 57, etc.  We can conclude, for example, that 6 random three-dimensional spaces divide the fourth dimension in 57 parts.  Now that should really impress your instructor!
649237
you will get the answer for all sequences BEST WEBSITE \n\nhttp://pirate.shu.edu/projects/reals/numseq/sequence.html
402618
I have no idea about the math, but I'm thinking the assumption is that the data are relatively correlated linearly.  There's a lot of math behind the wiki link.