Let [[.]] denote greatest integer value function.Define RV Yn=(X1+X2+...+Xn) - [[X1+X2+...+Xn]] for any n in {1,2,3,...}.Now it is not a trivial task to demonstrate that Y1, Y2, Y3, ... is a sequence of uniform RVs over [0,1] if the X1, X2, X3, ... are iid uniform RVs over [0,1].Questions: (1) Is the sequence Y1, Y2, Y3, ... a sequence of INDEPENDENT uniform RVs over [0,1] if X1, X2, X3, ... are iid uniform RVs over [0,1]?(2) For what kinds of iid RVs X1, X2, X3, ... is the sequence Y1, Y2, Y3, ... a sequence of uniform RVs over [0,1]?(3) Pose some questions of your own regarding the relationship between the Ys and Xs.

I've looked at #2 for the case of increasing N. One set of RVs that quickly converge to Yn = U[0,1] are those with a continuous PDF and a scale parameter that is not too small. (NOTE: if the scale parameter is small, it just implies a slower convergence to uniformity) A second category of rapidly convergent RVs are those with discontinuities restricted to integer locations in X. But as N increases, I've found that any distribution with a everywhere finite PDF (i.e., no delta functions in the PDF) will converge to Yi = U[0,1]. I've not worked out the formal proof of this, so take that hypothesis with an epsilon of NaCl. I'll also freely admit that I never looked at independence because that was irrelevant to my application at the time.This property could be almost like a "central limit theorem" for anything of the form: Mod(Sum({Xi}),a) where n grows large and with modest restrictions on X.

Quote:(2) For what kinds of iid RVs X1, X2, X3, ... is the sequence Y1, Y2, Y3, ... a sequence of uniform RVs over [0,1].A comment/inquiry: Since Y1=X1, then the only way Y1 can be uniform on [0,1] is if X1 is uniform on [0,1]. So, if Y2 is to be uniform on [0,1], is it necessary that X2 must also be uniform on [0,1]?

@quantyst Y1 = X1-[X1]

QuoteOriginally posted by: maneesh@quantyst Y1 = X1-[X1]True. However, it is most often the case that 0<X1<1 with probability 1. So [[X1]]=0 with probability 1, so Y1= X1-[[X1]]=X1-0=X1.In other words, a random variable X1 that is 0 or 1 with a positive probability is not one we should consider as viable. What do you think?

QuoteOriginally posted by: Traden4AlphaI've looked at #2 for the case of increasing N. One set of RVs that quickly converge to Yn = U[0,1] are those with a continuous PDF and a scale parameter that is not too small. (NOTE: if the scale parameter is small, it just implies a slower convergence to uniformity) A second category of rapidly convergent RVs are those with discontinuities restricted to integer locations in X. But as N increases, I've found that any distribution with a everywhere finite PDF (i.e., no delta functions in the PDF) will converge to Yi = U[0,1]. I've not worked out the formal proof of this, so take that hypothesis with an epsilon of NaCl. I'll also freely admit that I never looked at independence because that was irrelevant to my application at the time.This property could be almost like a "central limit theorem" for anything of the form: Mod(Sum({Xi}),a) where n grows large and with modest restrictions on X.Very nice!

QuoteOriginally posted by: quantystQuote:(2) For what kinds of iid RVs X1, X2, X3, ... is the sequence Y1, Y2, Y3, ... a sequence of uniform RVs over [0,1].A comment/inquiry: Since Y1=X1, then the only way Y1 can be uniform on [0,1] is if X1 is uniform on [0,1]. Actually, a wide variety of X1 != U[0,1] produce Y1 == U[0,1] outcomes. For example, X = U[-0.5,0.5], X = U[-0.9,0.1], and X = {IF -1<x<=0 THEN P(X) = x/2+1/2, IF 0<x<=1 THEN P(x) = 1/2-x/2, ELSE P(x)=0 } all produce Y1 = U[0,1]. Anything of the form X = U[c,c+k] where c is any real number and k is any integer will produce Y = U[0,1].

dup

QuoteOriginally posted by: Traden4AlphaQuoteOriginally posted by: quantystQuote:(2) For what kinds of iid RVs X1, X2, X3, ... is the sequence Y1, Y2, Y3, ... a sequence of uniform RVs over [0,1].A comment/inquiry: Since Y1=X1, then the only way Y1 can be uniform on [0,1] is if X1 is uniform on [0,1]. Actually, a wide variety of X1 != U[0,1] produce Y1 == U[0,1] outcomes. For example, X = U[-0.5,0.5], X = U[-0.9,0.1], and X = {IF -1<x<=0 THEN P(X) = x/2+1/, IF 0<x<=1 THEN P(x) = 1/2-x/2, ELSE P(x)=0 } all produce Y1 = U[0,1]. Anything of the form X = U[c,c+k] where c is any real number and k is any integer will produce Y = U[0,1].You are correct, of course!Objectively speaking, my error is due to my unstated assumption that the Xs range over [0,1] only. So, if we assume that the Xs range over [0,1] only, then would you agree with the observation made earlier and would you reconsider the question?It's notable that in all your examples above, the CDFs are linear. Can you think of a non-linear example?

Yes, if X is confined to [0,1], then only X=U[0,1] produces Y1 = U[0,1] because the function relating X to Y is linear on [0,1].X = {IF -1<x<=0 THEN P(x) = x/2+1/2, IF 0<x<=1 THEN P(x) = 1/2-x/2, ELSE P(x)=0 } has a nonlinear CDF and Y1 = U[0,1]. The interesting issue is that it's unclear whether Y2 = U[0,1] for this X.

just to offer a different perspective (might be a tautology):define {x}=x-[x]. easy to prove the following, if {a} and {b} are two i.i.d ~U[0,1], then {a+b} is ~U[0,1].(essentially n=2 case for question 1).then by induction, i.e., taking a=Y(i), b=X(i+1), we prove 1.similarly for 2 (note Y(1)=X(1), thus {X(1)} is required ~U[0,1]), take a=Y(i+1), b=-Y(i), hence {X(i)} are all ~U[0,1].geometrically one could consider the map Z=exp[i*(2*pi)*X]. if {X}~U[0,1], Z is uniform on a circle (|Z|=1). so one could construct X by drawing random uniform points on the unit circle then take the inverse map: X=log(Z)/[i*(2*pi)]+N, where N is a random integer labeling different branches of log.