Re: Random numbers

On Dec 23, 2:03 am, quasi <qu...@xxxxxxxx> wrote:
On Sat, 22 Dec 2007 11:17:53 -0800 (PST), vr <simple.pop...@xxxxxxxxx>
wrote:





On Dec 23, 12:00 am, quasi <qu...@xxxxxxxx> wrote:
On Sat, 22 Dec 2007 13:51:57 -0500, quasi <qu...@xxxxxxxx> wrote:
On Sat, 22 Dec 2007 10:43:27 -0800 (PST), vr <simple.pop...@xxxxxxxxx>
wrote:

On Dec 22, 11:35 pm, quasi <qu...@xxxxxxxx> wrote:
On Sat, 22 Dec 2007 10:32:37 -0800 (PST), vr <simple.pop...@xxxxxxxxx>
wrote:

On Dec 22, 11:16 pm, quasi <qu...@xxxxxxxx> wrote:
On Fri, 21 Dec 2007 10:57:00 -0800 (PST), simple.pop...@xxxxxxxxx
wrote:

On Dec 21, 11:37 pm, bill <b92...@xxxxxxxxx> wrote:
On Dec 21, 3:16 am, John <iamach...@xxxxxxxxx> wrote:

Given a function that returns a random number between 1-5, write one
that returns a random number between 1-7 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 + (T-7)*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 non-uniformity 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). As you can see, it's a lot less than 1/28, therefore T does
not yield a uniform distribution.

Alright. Though any number in [1,5] has equal chance of getting hit
for a single hit, multiple hits have bias towards the center of the
interval, due to the normal distribution. I agree. Thanks.
.

Relevant Pages

  • Re: Random numbers
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  • Re: Random numbers
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  • Re: Random numbers
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