
Re: Random numbers
Posted:
Dec 23, 2007 7:32 AM


On Dec 23, 5:15 pm, vr <simple.pop...@gmail.com> wrote: > On Dec 23, 2:03 am, quasi <qu...@null.set> wrote: > > > > > > > On Sat, 22 Dec 2007 11:17:53 0800 (PST), vr <simple.pop...@gmail.com> > > wrote: > > > >On Dec 23, 12:00 am, quasi <qu...@null.set> wrote: > > >> On Sat, 22 Dec 2007 13:51:57 0500, quasi <qu...@null.set> wrote: > > >> >On Sat, 22 Dec 2007 10:43:27 0800 (PST), vr <simple.pop...@gmail.com> > > >> >wrote: > > > >> >>On Dec 22, 11:35 pm, quasi <qu...@null.set> wrote: > > >> >>> On Sat, 22 Dec 2007 10:32:37 0800 (PST), vr <simple.pop...@gmail.com> > > >> >>> wrote: > > > >> >>> >On Dec 22, 11:16 pm, quasi <qu...@null.set> wrote: > > >> >>> >> On Fri, 21 Dec 2007 10:57:00 0800 (PST), simple.pop...@gmail.com > > >> >>> >> wrote: > > > >> >>> >> >On Dec 21, 11:37 pm, bill <b92...@yahoo.com> wrote: > > >> >>> >> >> On Dec 21, 3:16 am, John <iamach...@gmail.com> wrote: > > > >> >>> >> >> > Given a function that returns a random number between 15, write one > > >> >>> >> >> > that returns a random number between 17 for the case when it should > > >> >>> >> >> > be integer and for the case it can be real. > > > >> >>> >> >> Let S be the function that generates a RN between 1 and 5. Then > > > >> >>> >> >> T = S_1 + S_2 + ... + S_7 > > > >> >>> >> >> For the reals , RN_7 = T/7 > > > >> >>> >> >May be this should fix it: > > > >> >>> >> >For the reals , RN_7 = 1 + (T7)*3/14 > > > >> >>> >> Yes, that fixes the range. > > > >> >>> >> But it's still biased (that is, not a unform distribution). > > > >> >>> >> quasi Hide quoted text  > > > >> >>> >>  Show quoted text  > > > >> >>> >Hmm. Let me simplify it: > > > >> >>> >RN_7 = T*3/14  0.5 > > > >> >>> >If you look at T*3/14, it just scales the sum of random numbers > > >> >>> >uniformly using a constant multiplier. Did I miss to notice any non > > >> >>> >uniformity here? > > > >> >>> Yes, T is not uniformly distributed in its range. > > > >> >>Ok. But if S_n is guaranted to be uniformly distributed in the range 1 > > >> >>to 5, then doesn't it mean the sum of 7 such numbers will also get > > >> >>distributed over 7 to 35? I'm just curious. Thanks. > > > >> >Yes, but not uniformly. > > > >> >What's the chance of getting exactly 35? > > > >> >If it was a uniform distribution it would be 1/29, right? > > > >> Of course, that's assuming S is a uniformly distributed integer > > >> variable on {1,2,3,4,5}. > > > >> If instead, S is a uniformly distributed continuous variable on (1,5) > > >> then we can ask  what's that chance of getting a result more than > > >> 34? It should be at least 1/28, right? But it's easily seen to be a > > >> lot less than that (the probability is bounded above by 1/2^(14)). > > > >I understand the problem with the intger sum. But for the sum of > > >reals, If there is a nonuniformity I think it is due to finite number > > >of double precision numbers available in the given interval. > > > No, nothing to do with that. > > > Assume S is a true continuous, random number generator, with uniform > > distribution (1,5). > > > But T is the _sum_ of 7 independently generated values of S. While > > its range is (7,35), it is definitely not uniformly distributed. The > > bias is towards the mean (21), and away from the ends (7 and 35). > > > If T was uniformly distributed, the probability of T > 34 would be > > exactly 1/28. However the only way T can exceed 34 is if all of the 7 > > generated S values exceed 4 (a necessary but not sufficient > > condition). > > > But in fact, since the probability that an S value exceeds 4 is 1/4, > > the probability that T exceeds 34 must be less than (1/4)^7, or > > 1/2^(14). > > But if you consider that probability that an S value exceeds (4 + 6/7) > is 1/28. And any value of S exceeding (4 + 6/7) results in T value > exceeding 34, isn't it?
Correction: The probablity for all 7 values of S for exceeding (4 + 6/7) is 1/(28^7). And this probability should be same for any 1/28 segment in the range [1,5]. Is this correct?
vr

